From 7e80590461062ac24be7ce2f3dd6d6a6e638f4d3 Mon Sep 17 00:00:00 2001 From: Barbara Margolius Date: Fri, 22 Feb 2013 11:55:40 -0500 Subject: [PATCH] add trigonometric_substitution problem set to CSUOhio --- .../trigonometric_substitution/trigSub1.pg | 217 +++++++++++++++ .../trigonometric_substitution/trigSub10.pg | 149 +++++++++++ .../trigonometric_substitution/trigSub12.pg | 140 ++++++++++ .../trigonometric_substitution/trigSub13.pg | 141 ++++++++++ .../trigonometric_substitution/trigSub15.pg | 151 +++++++++++ .../trigonometric_substitution/trigSub16.pg | 149 +++++++++++ .../trigonometric_substitution/trigSub17.pg | 188 +++++++++++++ .../trigonometric_substitution/trigSub18.pg | 149 +++++++++++ .../trigonometric_substitution/trigSub2.pg | 158 +++++++++++ .../trigonometric_substitution/trigSub20.pg | 172 ++++++++++++ .../trigonometric_substitution/trigSub21.pg | 142 ++++++++++ .../trigonometric_substitution/trigSub22.pg | 251 ++++++++++++++++++ .../trigonometric_substitution/trigSub23.pg | 172 ++++++++++++ .../trigonometric_substitution/trigSub24.pg | 146 ++++++++++ .../trigonometric_substitution/trigSub25.pg | 144 ++++++++++ .../trigonometric_substitution/trigSub26.pg | 178 +++++++++++++ .../trigonometric_substitution/trigSub28.pg | 245 +++++++++++++++++ .../trigonometric_substitution/trigSub29.pg | 139 ++++++++++ .../trigonometric_substitution/trigSub3.pg | 153 +++++++++++ .../trigonometric_substitution/trigSub4.pg | 160 +++++++++++ .../trigonometric_substitution/trigSub5.pg | 150 +++++++++++ .../trigonometric_substitution/trigSub6.pg | 175 ++++++++++++ .../trigonometric_substitution/trigSub7.pg | 246 +++++++++++++++++ .../trigonometric_substitution/trigSub8.pg | 164 ++++++++++++ .../trigonometric_substitution/trigSub9.pg | 148 +++++++++++ .../trigonometric_substitution/trigsub.png | Bin 0 -> 18110 bytes .../CSUOhio/htdocs/applets/trigSubWW.swf | Bin 42769 -> 42868 bytes 27 files changed, 4227 insertions(+) create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub1.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub10.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub12.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub13.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub15.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub16.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub17.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub18.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub2.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub20.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub21.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub22.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub23.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub24.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub25.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub26.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub28.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub29.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub3.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub4.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub5.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub6.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub7.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub8.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub9.pg create mode 100644 OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigsub.png diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub1.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub1.pg new file mode 100644 index 0000000000..0d2542dc2c --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub1.pg @@ -0,0 +1,217 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('1') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; + +$funct = FormulaUpToConstant("x/2*sqrt($a2+x^2)+$a2/2*ln(x+sqrt($a2+x^2))"); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sec}); + $applet->initialState(qq{sec}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \sqrt{$a2 + x^2}\; dx \] +$BR \{ans_rule( 50) \} + +END_TEXT + +ANS( $funct->cmp() ); + +################################## +Context()->texStrings; +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->normalStrings; + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the \(\tan\) substitution. \[x = {$a}\tan(\theta)\] +So: +\[dx = {$a}\sec^2(\theta) \; d\theta\] +Therefore: +\[\int \sqrt{$a2 + x^2} \;dx=\int \sqrt{$a2+$a2\tan\theta}($a\sec^2\theta) \; d\theta\] +\[=\int $a2\sec^3\theta \; d\theta\] + +$BR$BR +Before proceeding to the several methods we can use to evaluate the integral of \(\sec^3\theta\), note that \(\tan\theta=\frac{x}{$a}\), and \(\sec\theta=\frac{$a2+x^2}{$a}\). To see this, label a right triangle so that the tangent is \(x/$a\). We will have the opposite side with length \(x\), and the adjacent side with length \($a\), so the hypotenuse has length \(\sqrt{$a2+x^2}\). + +$BR +From this point, we have several choices as to how to evaluate the integral of \(\sec^3\theta\). + +$BR$BR +Method 1: + +$BR +Look up the antiderivative of \(\int\sec^3\theta d\theta\). It is + +\[\int\sec^3\theta d\theta=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec\theta+\tan\theta\right|+C\] + +$BR$BR +so +\[ \int \sqrt{$a2 + x^2}\; dx=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln\left|\sec\theta+\tan\theta\right|+C \] + +$BR +Substituting back in terms of \(\theta\) yields: +\[\frac{$a2}{2}\frac{\sqrt{$a2+x^2}}{$a}\frac{x}{$a}+\frac{$a2}{2}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \] +Simplifying: +\[\frac{x\sqrt{$a2+x^2}}{2}+\frac{$a2}{2}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \] + + +$BR$BR +Method 2: + +$BR +Use the rule for substitution for products of sines and cosines. If sine is to an odd power, use a cosine substitution. If cosine is to an odd power, use a sine substitution. \(\sec^3\theta=\cos^{-3}\theta\), and -3 is an odd number, so we use a sine substitution. + +$BR +Let \(u=\sin\theta\), then \(du=\cos\theta d\theta\) so +\[\int $a2\sec^3\theta \; d\theta=$a2\int\frac{1}{\cos^3\theta}d\theta\] +\[=$a2\int\frac{\cos\theta}{\cos^4\theta}d\theta\] +\[=$a2\int\frac{\cos\theta}{(1-\sin^2\theta)^2}d\theta\] +\[=$a2\int\frac{du}{(1-u^2)^2}du\] +From here, we use a partial fractions decomposition: +\[$a2\int\frac{du}{(1-u^2)^2}du=$a2\int\left[\frac{A}{(1-u)^2}+\frac{B}{1-u}+\frac{C}{(1+u)^2}+\frac{D}{1+u}\right]du\] + +$BR +Solving for the constants \(A\), \(B\), \(C\) and \(D\) we find that \(A=B=C=D=\frac{1}{4}\). + +$BR$BR +\[\int $a2\sec^3\theta \; d\theta=\frac{$a2}{4}\int\frac{1}{(1-u)^2}+\frac{1}{1-u}+\frac{1}{(1+u)^2}+\frac{1}{1+u}du \] +\[=\frac{$a2}{4}\left(\frac{1}{1-u}-\ln|1-u|-\frac{1}{1+u}+\ln|1+u|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2u}{1-u^2}+\ln\left|\frac{1+u}{1-u}\right|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2\sin\theta}{1-\sin^2\theta}+\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2\sin\theta}{\cos^2\theta}+\ln\left|\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\right|\right)+C \] +\[=\frac{$a2}{4}\left(2\sec\theta\tan\theta+\ln\left|\frac{(1+\sin\theta)^2}{\cos^2\theta}\right|\right)+C \] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|+C \] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln|\sec\theta+\tan\theta|+C \] + +$BR +Substituting back in terms of \(\theta\) yields: +\[\frac{$a2}{2}\frac{\sqrt{$a2+x^2}}{$a}\frac{x}{$a}+\frac{$a2}{2}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \] +Simplifying: +\[\frac{x\sqrt{$a2+x^2}}{2}+\frac{$a2}{2}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \] + +$BR$BR +Method 3: + +$BR +Use the secant reduction formula. The secant reduction formula is given by +\[\int \sec^n\theta d\theta=\frac{1}{n-1}\tan\theta\sec^{n-2}\theta+\frac{n-2}{n-1}\int\sec^{n-2}\theta d\theta.\] + +$BR$BR +so +\[$a2\int\sec^3\theta d\theta = \frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\int\sec\theta d\theta.\] + +$BR$BR +We can evaluate \(\int\sec\theta d\theta\) by looking the antiderivative up in a table (\(\int\sec\theta d\theta=\ln|\sec\theta+\tan\theta|+C\)) or we can derive this antiderivative using the substitution rule for products of sines and cosines. We have an odd power of the cosine, we use a sine substitution. + +$BR$BR +\[\int\sec\theta d\theta=\int\frac{\cos\theta}{\cos^2\theta}d\theta\] +\[=\int\frac{\cos\theta}{1-\sin^2\theta}d\theta\] +\[=\int\frac{1}{1-u^2}du\] + +$BR +Do a partial fractions decomposition. +\[\int\frac{1}{1-u^2}du=\frac{1}{2}\int\left(\frac{1}{1-u}+\frac{1}{1+u}\right)du\] +\[=\frac{1}{2}\left(-\ln|1-u|+\ln|1+u|\right)+C\] +\[=\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{\cos^2\theta}\right|+C\] +\[=\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|+C\] +\[=\ln\left|\sec\theta+\tan\theta\right|+C\] + +so \[\int\sec\theta d\theta=\ln\left|\sec\theta+\tan\theta\right|+C\] and +\[$a2\int\sec^3\theta d\theta = \frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\int\sec\theta d\theta.\] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln\left|\sec\theta+\tan\theta\right|+C.\] + + +END_SOLUTION +Context()->normalStrings; +################################## + + +COMMENT('MathObject version. Uses Flash applet.'); +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub10.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub10.pg new file mode 100644 index 0000000000..4c2e1f1cbb --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub10.pg @@ -0,0 +1,149 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('10') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("x/({$a2}*($a2-x^2)^(1/2))"); + ################################### + # Create link to applet + ################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sin}); + $applet->initialState(qq{sin}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + + +################################### +# Main text + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{ dx}{($a2 - x^2)^{3/2}} \] +$BR \{ans_rule( 50) \} + +END_TEXT + +$ans = $funct; +ANS( $funct->cmp() ); +#ANS(fun_cmp($ans, limits=>[$lend,$rend], mode=>"antider", vars=>"x")); +################################## +Context()->texStrings; + +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the \(\tan\) substitution. \[x = {$a}\sin(\theta)\] + +$BR$BR +Before proceeding note that \(\sin\theta=\frac{x}{$a}\), and \(\cos\theta=\frac{\sqrt{$a2-x^2}}{$a}\). To see this, label a right triangle so that the sine is \(x/$a\). We will have the opposite side with length \(x\), and the hypotenuse with length \($a\), so the adjacent side has length \(\sqrt{$a2-x^2}\). + +$BR$BR +With the substitution \[x = {$a}\sin\theta\] +\[dx = {$a}\cos\theta \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{ dx}{($a2 - x^2)^{3/2}}= +\int \frac{{$a}\cos\theta }{($a2 - $a2\sin^2\theta)^{3/2}}\;d\theta\] +\[= +\int \frac{{$a}\cos\theta }{$a3\cos^3\theta}\;d\theta\] +\[= +\frac{1 }{$a2}\int \sec^2\theta\;d\theta\] +\[= +\frac{1 }{$a2}\tan\theta+C\] + +$BR +Substituting back in terms of \(\theta\) yields: +\[= +\frac{1 }{$a2}\tan\theta+C += +\frac{ x}{{$a2}\sqrt{$a2-x^2} } +C \] +so +\[ \int \frac{ dx}{($a2- x^2)^{3/2}}=\frac{ x}{{$a2}\sqrt{$a2-x^2} } +C \] + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub12.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub12.pg new file mode 100644 index 0000000000..389e724ae5 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub12.pg @@ -0,0 +1,140 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('12') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("-sqrt{$a2 - x^2}/($a2*x)"); + ################################### + # Create link to applet + ################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sin}); + $applet->initialState(qq{sin}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{ dx}{x^2\sqrt{$a2 - x^2}} \] +$BR \{ans_rule( 50) \} + +END_TEXT + +$ans = $funct; +ANS( $funct->cmp() ); +#ANS(fun_cmp($ans, limits=>[$lend,$rend], mode=>"antider", vars=>"x")); +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the sine substitution. \[x = {$a}\sin(\theta)\] + +$BR$BR +Before proceeding note that \(\sin\theta=\frac{x}{$a}\), and \(\cos\theta=\frac{\sqrt{$a2-x^2}}{$a}\). To see this, label a right triangle so that the sine is \(x/$a\). We will have the opposite side with length \(x\), and the hypotenuse with length \($a\), so the adjacent side has length \(\sqrt{$a2-x^2}\). + +$BR$BR +With the substitution \[x = {$a}\sin\theta\] +\[dx = {$a}\cos\theta \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{ dx}{x^2\sqrt{$a2 - x^2}}= +\int \frac{ {$a}\cos\theta \; d\theta}{{$a2}\sin^2\theta\sqrt{$a2 - {$a2}\sin^2\theta}}\] +\[=\int \frac{ {$a}\cos\theta \; d\theta}{{$a3}\sin^2\theta\cos\theta}\] +\[= +\int\frac{ d\theta}{{$a2}\sin^2\theta}\] +\[= +-\frac{1 }{$a2}\cot\theta+C\] + +$BR +Substituting back in terms of \(\theta\) yields: +\[ +-\frac{1 }{$a2}\cot\theta+C += +-\frac{\sqrt{$a2-x^2} }{{$a2}x} +C \] +so +\[ \int\frac{ dx}{x^2\sqrt{$a2 - x^2}}=-\frac{\sqrt{$a2-x^2} }{{$a2}x} +C \] + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub13.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub13.pg new file mode 100644 index 0000000000..4378a1e724 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub13.pg @@ -0,0 +1,141 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('13') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("1/($a)*ln(abs($a/x -sqrt($a2-x^2)/x))"); +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sin}); + $applet->initialState(qq{sin}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{ dx}{x\sqrt{$a2 - x^2}} \] +$BR \{ans_rule( 50) \} + +END_TEXT + +$ans = $funct; +ANS( $funct->cmp() ); +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the sine substitution. \[x = {$a}\sin(\theta)\] + +$BR$BR +Before proceeding note that \(\sin\theta=\frac{x}{$a}\), and \(\cos\theta=\frac{\sqrt{$a2-x^2}}{$a}\). To see this, label a right triangle so that the sine is \(x/$a\). We will have the opposite side with length \(x\), and the hypotenuse with length \($a\), so the adjacent side has length \(\sqrt{$a2-x^2}\). + +$BR$BR +With the substitution \[x = {$a}\sin\theta\] +\[dx = {$a}\cos\theta \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{ dx}{x\sqrt{$a2 - x^2}}= +\int \frac{ {$a}\cos\theta \; d\theta}{{$a}\sin\theta\sqrt{$a2 - {$a2}\sin^2\theta}}\] +\[=\int \frac{ {$a}\cos\theta \; d\theta}{{$a2}\sin\theta\cos\theta}\] +\[= +\int\frac{ d\theta}{{$a}\sin\theta}\] +\[= +\frac{1}{$a}\ln|\csc\theta-\cot\theta|+C\] + +$BR +Substituting back in terms of \(\theta\) yields: +\[ +\frac{1}{$a}\ln|\csc\theta-\cot\theta|+C += +\frac{1}{$a}\ln\left|\frac{$a}{x}-\frac{\sqrt{$a2-x^2}}{x}\right|+C\] +so +\[ \int \frac{ dx}{x\sqrt{$a2 - x^2}} +=\frac{1}{$a}\ln\left|\frac{$a}{x}-\frac{\sqrt{$a2-x^2}}{x}\right|+C \] + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub15.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub15.pg new file mode 100644 index 0000000000..3462901efa --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub15.pg @@ -0,0 +1,151 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('15') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("($a2)/2*asin(x/($a))-x*sqrt($a2-x^2)/2"); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sin}); + $applet->initialState(qq{sin}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{x^2 dx}{\sqrt{$a2 - x^2}} \] +$BR \{ans_rule( 50) \} + +END_TEXT + +$ans = $funct; +ANS( $funct->cmp()); +################################## +Context()->texStrings; + +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the sine substitution. \[x = {$a}\sin(\theta)\] + +$BR$BR +Before proceeding note that \(\sin\theta=\frac{x}{$a}\), and \(\cos\theta=\frac{\sqrt{$a2-x^2}}{$a}\). To see this, label a right triangle so that the sine is \(x/$a\). We will have the opposite side with length \(x\), and the hypotenuse with length \($a\), so the adjacent side has length \(\sqrt{$a2-x^2}\). + +$BR$BR +With the substitution \[x = {$a}\sin\theta\] +\[dx = {$a}\cos\theta \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{x^2 dx}{\sqrt{$a2 - x^2}}= +\int \frac{ {$a3}\sin^2\theta\cos\theta \; d\theta}{\sqrt{$a2 - {$a2}\sin^2\theta}}\] +\[=\int \frac{ {$a3}\sin^2\theta\cos\theta \; d\theta}{$a\cos\theta}\] +\[= +\int {$a2}\sin^2\theta \; d\theta\] +\[= +\frac{$a2}{2}\int (1-\cos(2\theta)) \; d\theta\] +\[= +\frac{$a2\theta}{2}-\frac{$a2\sin(2\theta)}{4} +C\] +\[= +\frac{$a2\theta}{2}-\frac{$a2\sin\theta\cos\theta}{2} +C\] + +$BR +Substituting back in terms of \(x\) yields: +\[= +\frac{$a2\theta}{2}-\frac{$a2\sin\theta\cos\theta}{2} +C\ +=\frac{$a2}{2}\sin^{-1}\left(\frac{x}{$a}\right)- +\frac{$a2}{2}\frac{x}{$a}\frac{\sqrt{$a2-x^2}}{$a}+C\] +\[=\frac{$a2}{2}\sin^{-1}\left(\frac{x}{$a}\right)- +\frac{x\sqrt{$a2-x^2}}{2}+C\] +so +\[ \int \frac{x^2 dx}{\sqrt{$a2 - x^2}} +=\frac{$a2}{2}\sin^{-1}\left(\frac{x}{$a}\right)- +\frac{x\sqrt{$a2-x^2}}{2}+C\] +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub16.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub16.pg new file mode 100644 index 0000000000..b06869a25b --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub16.pg @@ -0,0 +1,149 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('16') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("-(1/4)*x*($a2-x^2)^(3/2)+(1/8)*$a2*x*sqrt($a2-x^2)+(1/8)*$a4*arctan(x/sqrt($a2-x^2))"); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sin}); + $applet->initialState(qq{sin}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int x^2 \sqrt{$a2 - x^2}dx \] +$BR \{ans_rule( 60) \} + +END_TEXT + +ANS( $funct->cmp() ); +################################## +Context()->texStrings; + +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the sine substitution. \[x = {$a}\sin(\theta)\] + +$BR$BR +Before proceeding note that \(\sin\theta=\frac{x}{$a}\), and \(\cos\theta=\frac{\sqrt{$a2-x^2}}{$a}\). To see this, label a right triangle so that the sine is \(x/$a\). We will have the opposite side with length \(x\), and the hypotenuse with length \($a\), so the adjacent side has length \(\sqrt{$a2-x^2}\). + +$BR$BR +With the substitution \[x = {$a}\sin\theta\] +\[dx = {$a}\cos\theta \; d\theta\] +$BR$BR +Therefore: +\[\int x^2 \sqrt{$a2 - x^2}dx= +\int {$a3}\sin^2\theta\cos\theta\sqrt{$a2 - {$a2}\sin^2\theta} \; d\theta\] +\[=\int {$a4}\sin^2\theta\cos^2\theta \; d\theta\] +\[= +\frac{$a4}{4}\int \sin^2(2\theta) \; d\theta\] +\[= +\frac{$a4}{8}\int(1- \cos(4\theta)) \; d\theta\] +\[= +\frac{$a4\theta}{8}- \frac{$a4}{32}\sin(4\theta) +C\] +\[= +\frac{$a4\theta}{8}- \frac{$a4}{16}\sin(2\theta)\cos(2\theta) +C\] +\[= +\frac{$a4\theta}{8}- \frac{$a4}{8}\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta) +C\] + +$BR +Substituting back in terms of \(x\) yields: +\[\frac{$a4\theta}{8}- \frac{$a4}{8}\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta) +C +=\frac{$a4\theta}{8}- \frac{$a4}{8}\frac{x}{$a}\frac{\sqrt{$a2-x^2}}{$a}\left(\frac{{$a2-x^2}}{$a2}-\frac{x^2}{$a2}\right) +C\] +\[=\frac{$a4\theta}{8}- \frac{x\sqrt{$a2-x^2}}{8}\left($a2-2x^2\right) +C\] + +so +\[ \int x^2 \sqrt{$a2 - x^2}dx +=\frac{$a4\theta}{8}- \frac{x\sqrt{$a2-x^2}}{8}\left($a2-2x^2\right) +C\] +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub17.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub17.pg new file mode 100644 index 0000000000..577156a1cf --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub17.pg @@ -0,0 +1,188 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('17') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; +$a4_3 = 3*$a4; +$a2_5 = 5*$a2; + +$funct = FormulaUpToConstant("(1/4)*x*($a2-x^2)^(3/2)+(3/8)*{$a2}*x*sqrt({$a2}-x^2)+(3/8)*{$a4}*atan(x/sqrt({$a2}-x^2))"); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sin}); + $applet->initialState(qq{sin}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int($a2 - x^2)^{3/2}dx \] +$BR \{ans_rule( 70) \} + +END_TEXT + +ANS( $funct->cmp() ); +################################## +Context()->texStrings; +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the sine substitution. \[x = {$a}\sin(\theta)\] + +$BR$BR +Before proceeding note that \(\sin\theta=\frac{x}{$a}\), and \(\cos\theta=\frac{\sqrt{$a2-x^2}}{$a}\). To see this, label a right triangle so that the sine is \(x/$a\). We will have the opposite side with length \(x\), and the hypotenuse with length \($a\), so the adjacent side has length \(\sqrt{$a2-x^2}\). + +$BR$BR +With the substitution \[x = {$a}\sin\theta\] +\[dx = {$a}\cos\theta \; d\theta\] +$BR$BR +Therefore: +\[\int($a2 - x^2)^{3/2}dx= +\int {$a}\cos\theta($a2 - {$a2}\sin^2\theta)^{3/2} \; d\theta\] +\[=\int {$a4}\cos^4\theta \; d\theta\] +\[=\int +\frac{$a4}{4}(1+\cos(2\theta))^2 \; d\theta\] +\[= +int\frac{$a4}{4}(1+2\cos(2\theta)+\cos^2(2\theta)) \; d\theta\] +\[= +\int\frac{$a4}{4}\left(1+2\cos(2\theta)+\frac{1}{2}+\frac{\cos(4\theta)}{2}\right) \; d\theta\] +\[= +\int\frac{$a4}{4}\left(1+2\cos(2\theta)+\frac{1}{2}+\frac{\cos(4\theta)}{2}\right) \; d\theta\] +\[= +\int\frac{$a4}{4}\left(\frac{3}{2}+2\cos(2\theta)+\frac{\cos(4\theta)}{2}\right) \; d\theta\] +\[= +\frac{$a4(3)\theta}{8}+\frac{$a4}{4}\sin(2\theta)+\frac{$a4}{32}\sin(4\theta) ++C\] + +$BR$BR +We can use Euler's formula: +\[{\rm e}^{i\theta}=\cos\theta+i\sin\theta\] +to get \(\sin(4\theta)\) in terms of products of sines and cosines of \(\theta\) so that we can get our antiderivative back in terms of \(x\). + +\[{\rm e}^{4i\theta}=\cos(4\theta)+i\sin(4\theta)\] +\[{\rm e}^{4i\theta}=\left(\cos\theta_i\sin\theta\right)^4\] +\[=\cos^4\theta+4i\cos^3\theta\sin\theta-6\cos^2\theta\sin^2\theta-4i\cos\theta\sin^3\theta+\sin^4\theta\] + +Since both of these quantities equal \({\rm e}^{4i\theta}\), they are equal to each other. This means that the real parts equal the real parts and the imaginary parts equal the imaginary parts, so + +\[\sin(4\theta)=4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta\] + +\[ +\frac{$a4(3)\theta}{8}+\frac{$a4}{4}\sin(2\theta)+\frac{$a4}{32}\sin(4\theta) ++C\] +\[= +\frac{$a4(3)\theta}{8}+\frac{$a4}{2}\sin\theta\cos\theta+\frac{$a4}{8}(\cos^3\theta\sin\theta-\sin^3\theta\cos\theta) ++C\] + +$BR$BR +Substituting back in terms of \(x\) yields: +\[ +\frac{$a4(3)\theta}{8}+\frac{$a4}{2}\sin\theta\cos\theta+\frac{$a4}{8}(\cos^3\theta\sin\theta-\sin^3\theta\cos\theta) ++C\] +\[ +=\frac{$a4(3)}{8}\sin^{-1}\left(\frac{x}{$a}\right) ++\frac{$a4}{2}\frac{x}{$a}\frac{\sqrt{$a2-x^2}}{$a}\] +\[+\frac{$a4}{8}\left(\left(\frac{\sqrt{$a2-x^2}}{$a}\right)^3\left(\frac{x}{$a}\right)-\left(\frac{x}{$a}\right)^3\left(\frac{\sqrt{$a2-x^2}}{$a}\right)\right) ++C\] +\[ +=\frac{$a4(3)}{8}\sin^{-1}\left(\frac{x}{$a}\right) ++\frac{{$a2}x\sqrt{$a2-x^2}}{2}\] +\[+\frac{x\sqrt{$a2-x^2}}{8}\left($a2-2x^2\right) ++C\] +\[ +=\frac{$a4_3}{8}\sin^{-1}\left(\frac{x}{$a}\right) ++\frac{{$a2_5}x\sqrt{$a2-x^2}}{8}-\frac{x^3\sqrt{$a2-x^2}}{4} ++C\] + +so +\[ \int($a2 - x^2)^{3/2}dx +=\frac{$a4_3}{8}\sin^{-1}\left(\frac{x}{$a}\right) ++\frac{{$a2_5}x\sqrt{$a2-x^2}}{8}-\frac{x^3\sqrt{$a2-x^2}}{4} ++C\] +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub18.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub18.pg new file mode 100644 index 0000000000..64f4ef8d36 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub18.pg @@ -0,0 +1,149 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('18') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; +$a4_3 = 3*$a4; +$a2_5 = 5*$a2; + +$funct = FormulaUpToConstant("x/2*($a2-x^2)^(1/2)+{$a2}/2*asin(x/{$a})"); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sin}); + $applet->initialState(qq{sin}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int\sqrt{$a2 - x^2}dx \] +$BR \{ans_rule( 60) \} + +END_TEXT + +ANS( $funct->cmp() ); +################################## +Context()->texStrings; +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the sine substitution. \[x = {$a}\sin(\theta)\] + +$BR$BR +Before proceeding note that \(\sin\theta=\frac{x}{$a}\), and \(\cos\theta=\frac{\sqrt{$a2-x^2}}{$a}\). To see this, label a right triangle so that the sine is \(x/$a\). We will have the opposite side with length \(x\), and the hypotenuse with length \($a\), so the adjacent side has length \(\sqrt{$a2-x^2}\). + +$BR$BR +With the substitution \[x = {$a}\sin\theta\] +\[dx = {$a}\cos\theta \; d\theta\] +$BR$BR +Therefore: +\[\int\sqrt{$a2 - x^2}dx= +\int {$a}\cos\theta\sqrt{$a2 - {$a2}\sin^2\theta} \; d\theta\] +\[=\int {$a2}\cos^2\theta \; d\theta\] +\[=\frac{$a2}{2}\int (1+ \cos(2\theta)) \; d\theta\] +\[= +\frac{$a2\theta}{2}+\frac{$a2}{4}\sin(2\theta)) +C\] +\[= +\frac{$a2\theta}{2}+\frac{$a2}{2}\sin\theta\cos\theta +C\] + + +$BR$BR +Substituting back in terms of \(x\) yields: +\[\frac{$a2\theta}{2}+\frac{$a2}{2}\sin\theta\cos\theta +C\] +\[ +=\frac{$a2}{2}\sin^{-1}\left(\frac{x}{$a}\right) ++\frac{$a2}{2}\frac{x}{$a}\frac{\sqrt{$a2-x^2}}{$a}+C\] +\[=\frac{$a2}{2}\sin^{-1}\left(\frac{x}{$a}\right) ++\frac{ x\sqrt{$a2-x^2}}{2}+C\] + +so +\[ \int\sqrt{$a2 - x^2}dx +=\frac{$a2}{2}\sin^{-1}\left(\frac{x}{$a}\right) ++\frac{ x\sqrt{$a2-x^2}}{2}+C\] +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub2.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub2.pg new file mode 100644 index 0000000000..dbcc4565b2 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub2.pg @@ -0,0 +1,158 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('2') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("x/4*($a2+x^2)^(3/2)-$a4/8*ln(x+sqrt($a2+x^2))-$a2/8*x*sqrt(x^2+$a2)")->with(limits => [$a+1,$a+2]); + + ################################### + # Create link to applet + ################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{tan}); + $applet->initialState(qq{tan}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT +################################### +# Main text + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int x^2\sqrt{$a2 + x^2}\; dx \] +$BR \{ans_rule( 50) \} + +END_TEXT + +$ans = $funct; +ANS( $funct->cmp() ); +################################## +Context()->texStrings; + +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the \(\tan\) substitution. \[x = {$a}\tan(\theta)\] +So: +\[dx = {$a}\sec^2(\theta) \; d\theta\] +Therefore: +\[\int x^2\sqrt{$a2 + x^2} \;dx=\int $a2\tan^2\theta\sqrt{$a2+$a2\tan\theta}($a\sec^2\theta) \; d\theta\] +\[=\int $a4\tan^2\theta\sec^3\theta \; d\theta\] +\[=$a4\int (\sec^2\theta-1)\sec^3\theta \; d\theta\] +\[=$a4\int (\sec^5\theta-\sec^3\theta) \; d\theta\] + +$BR$BR +Before proceeding to the several methods we can use to evaluate the integral of \(\sec^3\theta\), note that \(\tan\theta=\frac{x}{$a}\), and \(\sec\theta=\frac{\sqrt{$a2+x^2}}{$a}\). To see this, label a right triangle so that the tangent is \(x/$a\). We will have the opposite side with length \(x\), and the adjacent side with length \($a\), so the hypotenuse has length \(\sqrt{$a2+x^2}\). + +$BR$BR +From this point, we apply the secant reduction formula to evaluate the integral. + +$BR +Use the secant reduction formula. The secant reduction formula is given by +\[\int \sec^n\theta d\theta=\frac{1}{n-1}\tan\theta\sec^{n-2}\theta+\frac{n-2}{n-1}\int\sec^{n-2}\theta d\theta.\] + +$BR$BR +so +\[\int\sec^5\theta d\theta=\frac{1}{5-1}\tan\theta\sec^{5-2}\theta+\frac{5-2}{5-1}\int\sec^{5-2}\theta d\theta\] +\[=\frac{1}{4}\tan\theta\sec^{3}\theta+\frac{3}{4}\int\sec^{3}\theta d\theta\] +and +\[\int\sec^3\theta d\theta=\frac{1}{3-1}\tan\theta\sec^{3-2}\theta+\frac{3-2}{3-1}\int\sec^{3-2}\theta d\theta\] +\[=\frac{1}{2}\tan\theta\sec\theta+\frac{1}{2}\int\sec\theta d\theta\] + +$BR +Substituting back in, +\[$a4\int (\sec^5\theta-\sec^3\theta) \; d\theta +=\frac{{$a4}}{4}\tan\theta\sec^{3}\theta-\frac{{$a4}}{4}\int\sec^{3}\theta d\theta \] +\[=\frac{{$a4}}{4}\tan\theta\sec^{3}\theta-\frac{{$a4}}{4}\left(\frac{1}{2}\tan\theta\sec\theta+\frac{1}{2}\int\sec\theta d\theta\right) \] + +\[=\frac{{$a4}}{4}\tan\theta\sec^{3}\theta-\frac{{$a4}}{8}\tan\theta\sec\theta-\frac{{$a4}}{8}\int\sec\theta d\theta \] + +\[=\frac{{$a4}}{4}\left(\frac{x}{$a}\right)\left(\frac{\sqrt{$a2+x^2}}{$a}\right)^3-\frac{{$a4}}{8}\left(\frac{x}{$a}\right)\left(\frac{\sqrt{$a2+x^2}}{$a}\right)-\frac{{$a4}}{8}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \] + +\[=\frac{x($a2+x^2)^{3/2}}{4}-\frac{{$a2 x\sqrt{$a2+x^2}}}{8}-\frac{{$a4}}{8}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \] + + + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub20.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub20.pg new file mode 100644 index 0000000000..cfc783a843 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub20.pg @@ -0,0 +1,172 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('20') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; +$a4_3 = 3*$a4; +$a2_5 = 5*$a2; + +$funct = FormulaUpToConstant("sqrt{$a2-x^2}+{$a}*ln(abs({$a}/{x}-sqrt{$a2-x^2}/{x}))")->with(limits => [0,$a]); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sin}); + $applet->initialState(qq{sin}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int\frac{\sqrt{$a2 - x^2}}{x}dx \] +$BR \{ans_rule( 60) \} + +END_TEXT + +ANS( $funct->cmp() ); +################################## +Context()->texStrings; +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the sine substitution. \[x = {$a}\sin(\theta)\] + +$BR$BR +Before proceeding note that \(\sin\theta=\frac{x}{$a}\), and \(\cos\theta=\frac{\sqrt{$a2-x^2}}{$a}\). To see this, label a right triangle so that the sine is \(x/$a\). We will have the opposite side with length \(x\), and the hypotenuse with length \($a\), so the adjacent side has length \(\sqrt{$a2-x^2}\). + +$BR$BR +With the substitution \[x = {$a}\sin\theta\] +\[dx = {$a}\cos\theta \; d\theta\] +$BR$BR +Therefore: +\[\int\frac{\sqrt{$a2 - x^2}}{x}dx= +\int \frac{{$a}\cos\theta\sqrt{$a2 - {$a2}\sin^2\theta}}{{$a}\sin\theta} \; d\theta\] +\[=\int \frac{{$a}\cos^2\theta}{\sin\theta} \; d\theta\] + +$BR$BR +For integrals that are products of sines and cosines if sine is raised to an odd power, we use a cosine substitution. Let \(u=\cos\theta\), then \(du=-\sin\theta d\theta\). + + +\[\int \frac{{$a}\cos^2\theta}{\sin\theta} \; d\theta\] +\[= +\int \frac{{$a}\cos^2\theta\sin\theta}{\sin^2\theta} \; d\theta\] +\[= +\int \frac{{$a}\cos^2\theta\sin\theta}{1-\cos^2\theta} \; d\theta\] +\[= +-\int\frac{{$a}u^2}{1-u^2} \; du\] + +$BR$BR +We use a partial fractions decomposition. +\[ +-\int\frac{{$a}u^2}{1-u^2} \; du +=$a\int \left(1-\frac{1/2}{1-u}-\frac{1/2}{1+u}\right)du\] +so +\[ +-\int\frac{{$a}u^2}{1-u^2} \; du +=$a \left(u+\frac{1}{2}\ln\left|\frac{1-u}{1+u}\right|\right)+C\] +\[ +=$a \left(\cos\theta+\frac{1}{2}\ln\left|\frac{1-\cos\theta}{1+\cos\theta}\right|\right)+C\] +\[ +=$a \left(\cos\theta+\frac{1}{2}\ln\left|\frac{(1-\cos\theta)^2}{1-\cos^2\theta}\right|\right)+C\] +\[ +=$a \left(\cos\theta+\frac{1}{2}\ln\left|\frac{(1-\cos\theta)^2}{\sin^2\theta}\right|\right)+C\] +\[ +=$a \left(\cos\theta+\ln\left|\frac{1-\cos\theta}{\sin\theta}\right|\right)+C\] +\[ +=$a \left(\cos\theta+\ln\left|\csc\theta-\cot\theta\right|\right)+C\] + +$BR$BR +Substituting back in terms of \(x\) yields: +\[$a \left(\cos\theta+\ln\left|\csc\theta-\cot\theta\right|\right)+C\] +\[ +=\sqrt{$a2-x^2}+{$a}\ln\left|\frac{$a}{x}-\frac{\sqrt{$a2-x^2}}{x}\right|+C +\] + +so +\[ \int\frac{\sqrt{$a2 - x^2}}{x}dx +=\sqrt{$a2-x^2}+{$a}\ln\left|\frac{$a}{x}-\frac{\sqrt{$a2-x^2}}{x}\right|+C\] +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub21.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub21.pg new file mode 100644 index 0000000000..cdf5a65e94 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub21.pg @@ -0,0 +1,142 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('21') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; +$a4_3 = 3*$a4; +$a2_5 = 5*$a2; + +$funct = FormulaUpToConstant("-sqrt{$a2-x^2}/{x}-asin({x}/{$a})")->with(limits => [$a-1,$a]); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sin}); + $applet->initialState(qq{sin}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int\frac{\sqrt{$a2 - x^2}}{x^2}dx \] +$BR \{ans_rule( 60) \} + +END_TEXT + +ANS( $funct->cmp() ); +################################## +Context()->texStrings; +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the sine substitution. \[x = {$a}\sin(\theta)\] + +$BR$BR +Before proceeding note that \(\sin\theta=\frac{x}{$a}\), and \(\cos\theta=\frac{\sqrt{$a2-x^2}}{$a}\). To see this, label a right triangle so that the sine is \(x/$a\). We will have the opposite side with length \(x\), and the hypotenuse with length \($a\), so the adjacent side has length \(\sqrt{$a2-x^2}\). + +$BR$BR +With the substitution \[x = {$a}\sin\theta\] +\[dx = {$a}\cos\theta \; d\theta\] +$BR$BR +Therefore: +\[\int\frac{\sqrt{$a2 - x^2}}{x^2}dx= +\int \frac{{$a}\cos\theta\sqrt{$a2 - {$a2}\sin^2\theta}}{{$a2}\sin^2\theta} \; d\theta\] +\[=\int \frac{\cos^2\theta}{\sin^2\theta} \; d\theta\] +\[=\int \cot^2\theta \; d\theta\] +\[=\int \csc^2\theta-1 \; d\theta\] +\[=-\cot\theta-\theta+C\] + +$BR$BR +Substituting back in terms of \(x\) yields: +\[-\cot\theta-\theta+C +=-\frac{\sqrt{$a2-x^2}}{x}-\sin^{-1}\left(\frac{x}{$a}\right)+C +\] + +so +\[ \int\frac{\sqrt{$a2 - x^2}}{x^2}dx +=-\frac{\sqrt{$a2-x^2}}{x}-\sin^{-1}\left(\frac{x}{$a}\right)+C\] +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub22.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub22.pg new file mode 100644 index 0000000000..bac5a256ad --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub22.pg @@ -0,0 +1,251 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('22') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; +$a4_3 = 3*$a4; +$a2_5 = 5*$a2; + +$func = FormulaUpToConstant("x/2*sqrt{x^2-$a2}-{$a2}/{2}*ln(abs(sqrt{x^2-$a2}/{$a}+{x}/{$a}))")->with(limits => [$a+1,$a+2]); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sec}); + $applet->initialState(qq{sec}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int\sqrt{ x^2-$a2}dx \] +$BR \{ans_rule( 60) \} + +END_TEXT +################################### +# Answers +################################### + +ANS($func->cmp()); +################################## +Context()->texStrings; +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the secant substitution. \[x = {$a}\sec\theta\] + +$BR$BR +We are motivated by the trigonometric identity +\[\sec^2\theta-1=\tan^2\theta.\] +With the substitution \(x = {$a}\sec\theta\), \(\sqrt{x^2-{$a2}}=\sqrt{{$a2}\sec^2\theta-{$a2}}={$a}\tan\theta\) for \(x>{$a}\), where \(0\le \theta<\pi/2\) and \(\sqrt{x^2-{$a2}}=\sqrt{{$a2}\sec^2\theta-{$a2}}=-{$a}\tan\theta\) for \(x<-{$a}\), where \(\pi/2<\theta\le\pi\). +Note that \(\sec\theta=\frac{x}{$a}\), and \(\sin\theta=\frac{\sqrt{x^2-$a2}}{x}\). To see this, label a right triangle so that the secant is \(x/$a\). We will have the adjacent side of length \($a\), and the hypotenuse with length \(x\), so the opposite side has length \(\sqrt{x^2-$a2}\). + +$BR$BR +With the substitution \[x = {$a}\sec\theta\] +\[dx = {$a}\sec\theta\tan\theta \; d\theta\] +$BR$BR +Therefore: +\[\int\sqrt{ x^2-$a2}dx= +\int {$a}\sec\theta\tan\theta\sqrt{{$a2}\sec^2\theta - {$a2}} \; d\theta\] +\[=\int {$a2}\tan^2\theta\sec\theta\; d\theta\] +\[=\int {$a2}(\sec^2\theta-1)\sec\theta\; d\theta\] +\[=\int {$a2}(\sec^3\theta-\sec\theta)\; d\theta\] + +$BR +From this point, we have several choices as to how to evaluate the integral of \(\sec^3\theta\). + +$BR$BR +Method 1: + +$BR +Look up the antiderivative of \(\int\sec^3\theta d\theta\). It is + +\[\int\sec^3\theta d\theta=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec\theta+\tan\theta\right|+C\] + +$BR +The antiderivative of \(\sec\theta\) is given by +\[\int\sec\theta d\theta=\ln\left|\sec\theta+\tan\theta\right|+C\] + + +$BR$BR +so +\[=\int {$a2}(\sec^3\theta-\sec\theta)\; d\theta\] + +\[ \int \sqrt{ x^2-$a2}\; dx= +\frac{$a2}{2}\sec\theta\tan\theta-\frac{$a2}{2}\ln\left|\sec\theta+\tan\theta\right|+C \] + +$BR +Substituting back in terms of \(x\) yields: +\[\frac{$a2}{2}\sec\theta\tan\theta-\frac{$a2}{2}\ln\left|\sec\theta+\tan\theta\right|+C +=\frac{$a2}{2}\frac{x}{$a}\frac{\sqrt{x^2-{$a2}}}{$a}-\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-{$a2}}}{$a}\right|+C \] +Simplifying: +\[\frac{1}{2}x\sqrt{x^2-{$a2}}-\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-{$a2}}}{$a}\right|+C \] + + +$BR$BR +Method 2: + +$BR +Use the rule for substitution for products of sines and cosines. If sine is to an odd power, use a cosine substitution. If cosine is to an odd power, use a sine substitution. \(\sec^3\theta=\cos^{-3}\theta\), and -3 is an odd number, so we use a sine substitution. + +$BR +Let \(u=\sin\theta\), then \(du=\cos\theta d\theta\) so +\[\int $a2\sec^3\theta \; d\theta=$a2\int\frac{1}{\cos^3\theta}d\theta\] +\[=$a2\int\frac{\cos\theta}{\cos^4\theta}d\theta\] +\[=$a2\int\frac{\cos\theta}{(1-\sin^2\theta)^2}d\theta\] +\[=$a2\int\frac{du}{(1-u^2)^2}du\] +From here, we use a partial fractions decomposition: +\[$a2\int\frac{du}{(1-u^2)^2}du=$a2\int\left[\frac{A}{(1-u)^2}+\frac{B}{1-u}+\frac{C}{(1+u)^2}+\frac{D}{1+u}\right]du\] + +$BR +Solving for the constants \(A\), \(B\), \(C\) and \(D\) we find that \(A=B=C=D=\frac{1}{4}\). + +$BR$BR +\[\int $a2\sec^3\theta \; d\theta=\frac{$a2}{4}\int\frac{1}{(1-u)^2}+\frac{1}{1-u}+\frac{1}{(1+u)^2}+\frac{1}{1+u}du \] +\[=\frac{$a2}{4}\left(\frac{1}{1-u}-\ln|1-u|-\frac{1}{1+u}+\ln|1+u|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2u}{1-u^2}+\ln\left|\frac{1+u}{1-u}\right|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2\sin\theta}{1-\sin^2\theta}+\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2\sin\theta}{\cos^2\theta}+\ln\left|\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\right|\right)+C \] +\[=\frac{$a2}{4}\left(2\sec\theta\tan\theta+\ln\left|\frac{(1+\sin\theta)^2}{\cos^2\theta}\right|\right)+C \] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|+C \] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln|\sec\theta+\tan\theta|+C \] + +$BR +Substituting back in terms of \(x\) yields: +\[ \int \sqrt{ x^2-$a2}\; dx= +\frac{$a2}{2}\sec\theta\tan\theta-\frac{$a2}{2}\ln\left|\sec\theta+\tan\theta\right|+C \] + +\[\frac{$a2}{2}\frac{\sqrt{x^2-{$a2}}}{$a}\frac{x}{$a}+\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-{$a2}}}{$a}\right|+C \] +Simplifying: +\[\frac{1}{2}x\sqrt{x^2-{$a2}}-\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-{$a2}}}{$a}\right|+C \] + +$BR$BR +Method 3: + +$BR +Use the secant reduction formula. The secant reduction formula is given by +\[\int \sec^n\theta d\theta=\frac{1}{n-1}\tan\theta\sec^{n-2}\theta+\frac{n-2}{n-1}\int\sec^{n-2}\theta d\theta.\] + +$BR$BR +so +\[$a2\int\sec^3\theta d\theta = \frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\int\sec\theta d\theta.\] + +$BR$BR +We can evaluate \(\int\sec\theta d\theta\) by looking the antiderivative up in a table (\(\int\sec\theta d\theta=\ln|\sec\theta+\tan\theta|+C\)) or we can derive this antiderivative using the substitution rule for products of sines and cosines. We have an odd power of the cosine, we use a sine substitution. + +$BR$BR +\[\int\sec\theta d\theta=\int\frac{\cos\theta}{\cos^2\theta}d\theta\] +\[=\int\frac{\cos\theta}{1-\sin^2\theta}d\theta\] +\[=\int\frac{1}{1-u^2}du\] + +$BR +Do a partial fractions decomposition. +\[\int\frac{1}{1-u^2}du=\frac{1}{2}\int\left(\frac{1}{1-u}+\frac{1}{1+u}\right)du\] +\[=\frac{1}{2}\left(-\ln|1-u|+\ln|1+u|\right)+C\] +\[=\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{\cos^2\theta}\right|+C\] +\[=\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|+C\] +\[=\ln\left|\sec\theta+\tan\theta\right|+C\] + +so \[\int\sec\theta d\theta=\ln\left|\sec\theta+\tan\theta\right|+C\] and +\[$a2\int\sec^3\theta d\theta = \frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\int\sec\theta d\theta.\] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln\left|\sec\theta+\tan\theta\right|+C.\] + +$BR$BR +Putting this all together, we have that +\[= +\int {$a2}(\sec^3\theta-\sec\theta) \;d\theta\] +\[= +\frac{1}{2}x\sqrt{x^2-{$a2}}-\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-{$a2}}}{$a}\right|+C +\] + +so +\[ \int\sqrt{ x^2-$a2}dx +=\frac{1}{2}x\sqrt{x^2-{$a2}}-\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-{$a2}}}{$a}\right|+C\] +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub23.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub23.pg new file mode 100644 index 0000000000..e8112b7566 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub23.pg @@ -0,0 +1,172 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('23') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; +$a4_3 = 3*$a4; +$a2_5 = 5*$a2; + +$funct = FormulaUpToConstant("x^3/4*sqrt{x^2-$a2}-{$a2}*x/{8}*sqrt{x^2-$a2}-{$a4}/{8}*ln(abs(sqrt{x^2-$a2}/{$a}+{x}/{$a}))")->with(limits => [$a+1,$a+2]); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sec}); + $applet->initialState(qq{sec}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int x^2\sqrt{ x^2-$a2}dx \] +$BR \{ans_rule( 60) \} + +END_TEXT + +################################### +# Answers +################################### + +ANS($funct->cmp()); +################################## +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the secant substitution. \[x = {$a}\sec\theta\] + +$BR$BR +We are motivated by the trigonometric identity +\[\sec^2\theta-1=\tan^2\theta.\] +With the substitution \(x = {$a}\sec\theta\), \(\sqrt{x^2-{$a2}}=\sqrt{{$a2}\sec^2\theta-{$a2}}={$a}\tan\theta\) for \(x>{$a}\), where \(0\le \theta<\pi/2\) and \(\sqrt{x^2-{$a2}}=\sqrt{{$a2}\sec^2\theta-{$a2}}=-{$a}\tan\theta\) for \(x<-{$a}\), where \(\pi/2<\theta\le\pi\). +Note that \(\sec\theta=\frac{x}{$a}\), and \(\sin\theta=\frac{\sqrt{x^2-$a2}}{x}\). To see this, label a right triangle so that the secant is \(x/$a\). We will have the adjacent side of length \($a\), and the hypotenuse with length \(x\), so the opposite side has length \(\sqrt{x^2-$a2}\). + +$BR$BR +With the substitution \[x = {$a}\sec\theta\] +\[dx = {$a}\sec\theta\tan\theta \; d\theta\] +$BR$BR +Therefore: +\[\int x^2\sqrt{ x^2-$a2}dx= +\int {$a3}\sec^3\theta\tan\theta\sqrt{{$a2}\sec^2\theta - {$a2}} \; d\theta\] +\[=\int {$a4}\tan^2\theta\sec^3\theta\; d\theta\] +\[=\int {$a4}(\sec^2\theta-1)\sec^3\theta\; d\theta\] +\[=\int {$a4}(\sec^5\theta-\sec^3\theta)\; d\theta\] + +$BR$BR +From this point, we apply the secant reduction formula to evaluate the integral. + +$BR +The secant reduction formula is given by +\[\int \sec^n\theta d\theta=\frac{1}{n-1}\tan\theta\sec^{n-2}\theta+\frac{n-2}{n-1}\int\sec^{n-2}\theta d\theta.\] + +$BR$BR +so +\[\int\sec^5\theta d\theta=\frac{1}{5-1}\tan\theta\sec^{5-2}\theta+\frac{5-2}{5-1}\int\sec^{5-2}\theta d\theta\] +\[=\frac{1}{4}\tan\theta\sec^{3}\theta+\frac{3}{4}\int\sec^{3}\theta d\theta\] +and +\[\int\sec^3\theta d\theta=\frac{1}{3-1}\tan\theta\sec^{3-2}\theta+\frac{3-2}{3-1}\int\sec^{3-2}\theta d\theta\] +\[=\frac{1}{2}\tan\theta\sec\theta+\frac{1}{2}\int\sec\theta d\theta\] + +$BR +Substituting back in, +\[$a4\int (\sec^5\theta-\sec^3\theta) \; d\theta +=\frac{{$a4}}{4}\tan\theta\sec^{3}\theta-\frac{{$a4}}{4}\int\sec^{3}\theta d\theta \] +\[=\frac{{$a4}}{4}\tan\theta\sec^{3}\theta-\frac{{$a4}}{4}\left(\frac{1}{2}\tan\theta\sec\theta+\frac{1}{2}\int\sec\theta d\theta\right) \] + +\[=\frac{{$a4}}{4}\tan\theta\sec^{3}\theta-\frac{{$a4}}{8}\tan\theta\sec\theta-\frac{{$a4}}{8}\int\sec\theta d\theta \] + +$BR$BR +Substituting back in terms of \(x\): + +\[\frac{{$a4}}{4}\tan\theta\sec^{3}\theta-\frac{{$a4}}{8}\tan\theta\sec\theta-\frac{{$a4}}{8}\ln\left|\sec\theta+\tan\theta\right|+C \] +\[=\frac{{$a4}}{4}\frac{\sqrt{x^2-$a2}}{$a}\left(\frac{x}{$a}\right)^{3}-\frac{{$a4}}{8}\frac{\sqrt{x^2-$a2}}{$a}\left(\frac{x}{$a}\right)-\frac{{$a4}}{8}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C \] +\[=\frac{x^{3}}{4}\sqrt{x^2-$a2}-\frac{{$a2}x}{8}\sqrt{x^2-$a2}-\frac{{$a4}}{8}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C \] + +$BR$BR +so +\[\int x^2\sqrt{ x^2-$a2}dx = +\frac{x^{3}}{4}\sqrt{x^2-$a2}-\frac{{$a2}x}{8}\sqrt{x^2-$a2}-\frac{{$a4}}{8}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C\] + + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub24.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub24.pg new file mode 100644 index 0000000000..ca62a4eb36 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub24.pg @@ -0,0 +1,146 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('24') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; +$a4_3 = 3*$a4; +$a2_5 = 5*$a2; + +$funct = FormulaUpToConstant("sqrt{x^2-{$a2}}-{$a}*atan(sqrt{x^2-{$a2}}/{$a})")->with(limits => [$a+1,$a+2]); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sec}); + $applet->initialState(qq{sec}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{\sqrt{ x^2-$a2}}{x}dx \] +$BR \{ans_rule( 60) \} + +END_TEXT + +ANS( $funct->cmp() ); +################################## +Context()->texStrings; +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the secant substitution. \[x = {$a}\sec\theta\] + +$BR$BR +We are motivated by the trigonometric identity +\[\sec^2\theta-1=\tan^2\theta.\] +With the substitution \(x = {$a}\sec\theta\), \(\sqrt{x^2-{$a2}}=\sqrt{{$a2}\sec^2\theta-{$a2}}={$a}\tan\theta\) for \(x>{$a}\), where \(0\le \theta<\pi/2\) and \(\sqrt{x^2-{$a2}}=\sqrt{{$a2}\sec^2\theta-{$a2}}=-{$a}\tan\theta\) for \(x<-{$a}\), where \(\pi/2<\theta\le\pi\). +Note that \(\sec\theta=\frac{x}{$a}\), and \(\sin\theta=\frac{\sqrt{x^2-$a2}}{x}\). To see this, label a right triangle so that the secant is \(x/$a\). We will have the adjacent side of length \($a\), and the hypotenuse with length \(x\), so the opposite side has length \(\sqrt{x^2-$a2}\). + +$BR$BR +With the substitution \[x = {$a}\sec\theta\] +\[dx = {$a}\sec\theta\tan\theta \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{\sqrt{ x^2-$a2}}{x}dx= +\int {$a}\sec\theta\tan\theta\frac{\sqrt{{$a2}\sec^2\theta - {$a2}}}{{$a}\sec\theta} \; d\theta\] +\[=\int {$a}\tan^2\theta \; d\theta\] +\[=\int {$a}(\sec^2\theta-1)\; d\theta\] +\[= {$a}\tan\theta-{$a}\theta+C\] + +$BR$BR +Substituting back in terms of \(x\): +\[{$a}\tan\theta-{$a}\theta+C\] +\[=\sqrt{x^2-{$a2}}-{$a}\sec^{-1}\left(\frac{x}{$a}\right)+C\] + + +$BR$BR +so +\[\int \frac{\sqrt{ x^2-$a2}}{x}dx =\sqrt{x^2-{$a2}}-{$a}\sec^{-1}\left(\frac{x}{$a}\right)+C\] + + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub25.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub25.pg new file mode 100644 index 0000000000..9e2009016b --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub25.pg @@ -0,0 +1,144 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('25') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; +$a4_3 = 3*$a4; +$a2_5 = 5*$a2; + +$funct = FormulaUpToConstant("1/{$a}*atan(sqrt{x^2-{$a2}}/{$a})")->with(limits => [$a+1,$a+2]); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sec}); + $applet->initialState(qq{sec}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{dx}{x\sqrt{ x^2-$a2}} \] +$BR \{ans_rule( 60) \} + +END_TEXT + +ANS( $funct->cmp() ); +################################## +Context()->texStrings; +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the secant substitution. \[x = {$a}\sec\theta\] + +$BR$BR +We are motivated by the trigonometric identity +\[\sec^2\theta-1=\tan^2\theta.\] +With the substitution \(x = {$a}\sec\theta\), \(\sqrt{x^2-{$a2}}=\sqrt{{$a2}\sec^2\theta-{$a2}}={$a}\tan\theta\) +for \(x>{$a}\), where \(0\le \theta<\pi/2\) and \(\sqrt{x^2-{$a2}}=\sqrt{{$a2}\sec^2\theta-{$a2}}=-{$a}\tan\theta\) +for \(x<-{$a}\), where \(\pi/2<\theta\le\pi\). + +$BR$BR +With the substitution \[x = {$a}\sec\theta\] +\[dx = {$a}\sec\theta\tan\theta \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{dx}{x\sqrt{ x^2-$a2}}= +\int \frac{{$a}\sec\theta\tan\theta d\theta}{{$a2}\tan\theta\sec\theta}\] +\[=\frac{1}{$a}\int \; d\theta\] +\[=\frac{\theta}{$a}+C\] + +$BR$BR +Substituting back in terms of \(x\): +\[\frac{\theta}{$a}+C\] +\[=\frac{1}{$a}\sec^{-1}\left(\frac{x}{$a}\right)+C\] + +$BR$BR +so +\[\int \frac{\sqrt{ x^2-$a2}}{x}dx =\frac{1}{$a}\sec^{-1}\left(\frac{x}{$a}\right)+C\] + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub26.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub26.pg new file mode 100644 index 0000000000..8d6fd0591f --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub26.pg @@ -0,0 +1,178 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('26') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; +$a4_3 = 3*$a4; +$a2_5 = 5*$a2; + +$funct = FormulaUpToConstant("-sqrt{x^2-$a2}/{x}+ln(abs({x}+sqrt{x^2-$a2}))")->with(limits => [$a+1,$a+2]); + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sec}); + $applet->initialState(qq{sec}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{\sqrt{ x^2-$a2}}{x^2}dx \] +$BR \{ans_rule( 60) \} + +END_TEXT + +ANS( $funct->cmp() ); +################################## +Context()->texStrings; +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the secant substitution. \[x = {$a}\sec\theta\] + +$BR$BR +We are motivated by the trigonometric identity +\[\sec^2\theta-1=\tan^2\theta.\] +With the substitution \(x = {$a}\sec\theta\), \(\sqrt{x^2-{$a2}}=\sqrt{{$a2}\sec^2\theta-{$a2}}={$a}\tan\theta\) for \(x>{$a}\), where \(0\le \theta<\pi/2\) and \(\sqrt{x^2-{$a2}}=\sqrt{{$a2}\sec^2\theta-{$a2}}=-{$a}\tan\theta\) for \(x<-{$a}\), where \(\pi/2<\theta\le\pi\). +Note that \(\sec\theta=\frac{x}{$a}\), and \(\sin\theta=\frac{\sqrt{x^2-$a2}}{x}\). To see this, label a right triangle so that the secant is \(x/$a\). We will have the adjacent side of length \($a\), and the hypotenuse with length \(x\), so the opposite side has length \(\sqrt{x^2-$a2}\). + +$BR$BR +With the substitution \[x = {$a}\sec\theta\] +\[dx = {$a}\sec\theta\tan\theta \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{\sqrt{ x^2-$a2}}{x^2}dx= +\int {$a}\sec\theta\tan\theta\frac{\sqrt{{$a2}\sec^2\theta - {$a2}}}{{$a2}\sec^2\theta} \; d\theta\] +\[=\int \frac{\sin^2\theta}{\cos\theta} \; d\theta\] + +$BR$BR +Since this integral is the product of sines and cosines and the cosine is raised to an odd power, we use a sine substitution. + +$BR$BR +Let \(u=\sin\theta\), then \(du=\cos\theta d\theta.\) + +$BR$BR +\[\int \frac{\sin^2\theta}{\cos\theta} \; d\theta\] +\[=\int \frac{\sin^2\theta\cos\theta}{\cos^2\theta} \; d\theta\] +\[=\int \frac{\sin^2\theta\cos\theta}{1-\sin^2\theta} \; d\theta\] +\[=\int \frac{u^2}{1-u^2} \; du\] + +$BR$BR +We use a partial fractions decomposition: +\[\int \frac{u^2}{1-u^2} \; du\] +\[=\int\left(-1-\frac{1}{2(u-1)}+\frac{1}{2(u+1)}\right)du\] +\[=-u+\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C\] + +$BR$BR +Substituting back in terms of \(\theta\): +\[-u+\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C\] +\[=-\sin\theta+\frac{1}{2}\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|+C\] +\[=-\sin\theta+\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\right|+C\] +\[=-\sin\theta+\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{\cos^2\theta}\right|+C\] +\[=-\sin\theta+\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|+C\] +\[=-\sin\theta+\ln\left|\sec\theta+\tan\theta\right|+C\] + +$BR$BR +Substituting back in terms of \(x\): +\[-\sin\theta+\ln\left|\sec\theta+\tan\theta\right|+C\] +\[=-\frac{\sqrt{x^2-$a2}}{x}+\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C\] + +$BR$BR +Note that +\[-\frac{\sqrt{x^2-$a2}}{x}+\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C\] +is equivalent to +\[-\frac{\sqrt{x^2-$a2}}{x}+\ln\left|{x}+{\sqrt{x^2-$a2}}\right|+C\] +with a different constant of integration since +\[\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|=\ln\left|{x}+{\sqrt{x^2-$a2}}\right|-\ln\left|{$a}\right|\] +and \(\ln\left|{$a}\right|\) is just a constant. + +$BR$BR +so +\[\int \frac{\sqrt{ x^2-$a2}}{x^2}dx =-\frac{\sqrt{x^2-$a2}}{x}+\ln\left|x+\sqrt{x^2-$a2}\right|+C\] + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub28.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub28.pg new file mode 100644 index 0000000000..35a549a449 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub28.pg @@ -0,0 +1,245 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('8/20/11') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('2010') +## AuthorText1('') +## Section1('') +## Problem1('28') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; +$a4_3 = 3*$a4; +$a2_5 = 5*$a2; + +$funct = FormulaUpToConstant("x/2*sqrt(x^2-{$a2})+{$a2}/2*ln(abs({x}+sqrt{x^2-$a2}))")->with(limits => [$a+1,$a+2]);; + +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sec}); + $applet->initialState(qq{sec}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{x^2 dx}{\sqrt{ x^2-$a2}} \] +$BR \{ans_rule( 60) \} + +END_TEXT + +ANS( $funct->cmp() ); +################################## +Context()->texStrings; +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the secant substitution. \[x = {$a}\sec\theta\] + +$BR$BR +We are motivated by the trigonometric identity +\[\sec^2\theta-1=\tan^2\theta.\] +With the substitution \(x = {$a}\sec\theta\), \(\sqrt{x^2-{$a2}}\)\(=\sqrt{{$a2}\sec^2\theta-{$a2}}\)\(={$a}\tan\theta\) for \(x>{$a}\), where \(0\le \theta<\pi/2\) and \(\sqrt{x^2-{$a2}}\)\(=\sqrt{{$a2}\sec^2\theta-{$a2}}\) +\(=-{$a}\tan\theta\) for \(x<-{$a}\), where \(\pi/2<\theta\le\pi\). +Note that \(\sec\theta=\frac{x}{$a}\), and \(\sin\theta=\frac{\sqrt{x^2-$a2}}{x}\). To see this, label a right triangle so that the secant is \(x/$a\). We will have the adjacent side of length \($a\), and the hypotenuse with length \(x\), so the opposite side has length \(\sqrt{x^2-$a2}\). + +$BR$BR +With the substitution \[x = {$a}\sec\theta\] +\[dx = {$a}\sec\theta\tan\theta \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{x^2 dx}{\sqrt{ x^2-$a2}}= +\int \frac{{$a3}\sec^3\theta\tan\theta}{\sqrt{{$a2}\sec^2\theta - {$a2}}} \; d\theta\] +\[=\int {$a2}\sec^3\theta \; d\theta\] + +$BR +From this point, we have several choices as to how to evaluate the integral of \(\sec^3\theta\). + +$BR$BR +Method 1: + +$BR +Look up the antiderivative of \(\int\sec^3\theta d\theta\). It is + +\[\int\sec^3\theta d\theta=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec\theta+\tan\theta\right|+C\] + +$BR$BR +so +\[\int \frac{x^2 dx}{\sqrt{ x^2-$a2}}= +\frac{$a2}{2}\frac{x}{$a}\frac{\sqrt{x^2-$a2}}{$a}+\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C\] +\[= +\frac{x}{2}\sqrt{x^2-$a2}+\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C\] + + +$BR$BR +Method 2: + +$BR +Use the rule for substitution for products of sines and cosines. If sine is to an odd power, use a cosine substitution. If cosine is to an odd power, use a sine substitution. \(\sec^3\theta=\cos^{-3}\theta\), and -3 is an odd number, so we use a sine substitution. + +$BR +Let \(u=\sin\theta\), then \(du=\cos\theta d\theta\) so +\[\int $a2\sec^3\theta \; d\theta=$a2\int\frac{1}{\cos^3\theta}d\theta\] +\[=$a2\int\frac{\cos\theta}{\cos^4\theta}d\theta\] +\[=$a2\int\frac{\cos\theta}{(1-\sin^2\theta)^2}d\theta\] +\[=$a2\int\frac{du}{(1-u^2)^2}du\] +From here, we use a partial fractions decomposition: +\[$a2\int\frac{du}{(1-u^2)^2}du=$a2\int\left[\frac{A}{(1-u)^2}+\frac{B}{1-u}+\frac{C}{(1+u)^2}+\frac{D}{1+u}\right]du\] + +$BR +Solving for the constants \(A\), \(B\), \(C\) and \(D\) we find that \(A=B=C=D=\frac{1}{4}\). + +$BR$BR +\[\int $a2\sec^3\theta \; d\theta=\frac{$a2}{4}\int\frac{1}{(1-u)^2}+\frac{1}{1-u}+\frac{1}{(1+u)^2}+\frac{1}{1+u}du \] +\[=\frac{$a2}{4}\left(\frac{1}{1-u}-\ln|1-u|-\frac{1}{1+u}+\ln|1+u|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2u}{1-u^2}+\ln\left|\frac{1+u}{1-u}\right|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2\sin\theta}{1-\sin^2\theta}+\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2\sin\theta}{\cos^2\theta}+\ln\left|\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\right|\right)+C \] +\[=\frac{$a2}{4}\left(2\sec\theta\tan\theta+\ln\left|\frac{(1+\sin\theta)^2}{\cos^2\theta}\right|\right)+C \] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|+C \] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln|\sec\theta+\tan\theta|+C \] + +$BR +Substituting back in terms of \(x\) yields: +\[\frac{$a2}{2}\frac{x}{$a}\frac{\sqrt{x^2-$a2}}{$a}+\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C\] + +Simplifying: +\[ +\frac{x}{2}\sqrt{x^2-$a2}+\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C\] + +$BR$BR +Method 3: + +$BR +Use the secant reduction formula. The secant reduction formula is given by +\[\int \sec^n\theta d\theta=\frac{1}{n-1}\tan\theta\sec^{n-2}\theta+\frac{n-2}{n-1}\int\sec^{n-2}\theta d\theta.\] + +$BR$BR +so +\[$a2\int\sec^3\theta d\theta = \frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\int\sec\theta d\theta.\] + +$BR$BR +We can evaluate \(\int\sec\theta d\theta\) by looking the antiderivative up in a table (\(\int\sec\theta d\theta=\ln|\sec\theta+\tan\theta|+C\)) or we can derive this antiderivative using the substitution rule for products of sines and cosines. We have an odd power of the cosine, we use a sine substitution. + +$BR$BR +\[\int\sec\theta d\theta=\int\frac{\cos\theta}{\cos^2\theta}d\theta\] +\[=\int\frac{\cos\theta}{1-\sin^2\theta}d\theta\] +\[=\int\frac{1}{1-u^2}du\] + +$BR +Do a partial fractions decomposition. +\[\int\frac{1}{1-u^2}du=\frac{1}{2}\int\left(\frac{1}{1-u}+\frac{1}{1+u}\right)du\] +\[=\frac{1}{2}\left(-\ln|1-u|+\ln|1+u|\right)+C\] +\[=\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{\cos^2\theta}\right|+C\] +\[=\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|+C\] +\[=\ln\left|\sec\theta+\tan\theta\right|+C\] + +so \[\int\sec\theta d\theta=\ln\left|\sec\theta+\tan\theta\right|+C\] and +\[$a2\int\sec^3\theta d\theta = \frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\int\sec\theta d\theta.\] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln\left|\sec\theta+\tan\theta\right|+C.\] + +$BR$BR +Substituting back in terms of \(\theta\) yields: +\[\frac{$a2}{2}\frac{x}{$a}\frac{\sqrt{x^2-$a2}}{$a}+\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C\] + +Simplifying: +\[ +\frac{x}{2}\sqrt{x^2-$a2}+\frac{$a2}{2}\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C\] + +$BR$BR +Note that +\[\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|+C\] +is equivalent to +\[\ln\left|{x}+{\sqrt{x^2-$a2}}\right|+C\] +with a different constant of integration since +\[\ln\left|\frac{x}{$a}+\frac{\sqrt{x^2-$a2}}{$a}\right|=\ln\left|{x}+{\sqrt{x^2-$a2}}\right|-\ln\left|{$a}\right|\] +and \(\ln\left|{$a}\right|\) is just a constant. + +$BR$BR +so +\[\int \frac{x^2 dx}{\sqrt{ x^2-$a2}} =\frac{x}{2}\sqrt{x^2-$a2}+\ln\left|x+\sqrt{x^2-$a2}\right|+C\] + + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub29.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub29.pg new file mode 100644 index 0000000000..a5286d04d8 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub29.pg @@ -0,0 +1,139 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('8/20/11') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('2010') +## AuthorText1('') +## Section1('') +## Problem1('29') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; +$a4_3 = 3*$a4; +$a2_5 = 5*$a2; + +$funct = FormulaUpToConstant("sqrt(x^2-{$a2})/x/{$a2}")->with(limits => [$a+1,$a+2]); +################################### +# Create link to applet +################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{sec}); + $applet->initialState(qq{sec}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{ dx}{x^2\sqrt{ x^2-$a2}} \] +$BR \{ans_rule( 60) \} + +END_TEXT + +ANS( $funct->cmp()); +################################## +Context()->texStrings; +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the secant substitution. \[x = {$a}\sec\theta\] + +$BR$BR +We are motivated by the trigonometric identity +\[\sec^2\theta-1=\tan^2\theta.\] +With the substitution \(x = {$a}\sec\theta\), \(\sqrt{x^2-{$a2}}\)\(=\sqrt{{$a2}\sec^2\theta-{$a2}}\)\(={$a}\tan\theta\) for \(x>{$a}\), where \(0\le \theta<\pi/2\) and \(\sqrt{x^2-{$a2}}\)\(=\sqrt{{$a2}\sec^2\theta-{$a2}}\) +\(=-{$a}\tan\theta\) for \(x<-{$a}\), where \(\pi/2<\theta\le\pi\). +Note that \(\sec\theta=\frac{x}{$a}\), and \(\sin\theta=\frac{\sqrt{x^2-$a2}}{x}\). To see this, label a right triangle so that the secant is \(x/$a\). We will have the adjacent side of length \($a\), and the hypotenuse with length \(x\), so the opposite side has length \(\sqrt{x^2-$a2}\). + +$BR$BR +With the substitution \[x = {$a}\sec\theta\] +\[dx = {$a}\sec\theta\tan\theta \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{ dx}{x^2\sqrt{ x^2-$a2}}= +\int \frac{{$a}\sec\theta\tan\theta}{{$a2}\sec^2\theta\sqrt{{$a2}\sec^2\theta - {$a2}}} \; d\theta\] +\[=\int \frac{1}{{$a2}\sec\theta} \; d\theta\] +\[=\frac{1}{$a2}\int \cos\theta \; d\theta\] +\[=\frac{1}{$a2} \sin\theta+C\] + +$BR +Substituting back in terms of \(x\) yields: +\[\int \frac{ dx}{x^2\sqrt{ x^2-$a2}}= +\frac{\sqrt{x^2-$a2}}{{$a2}x} +C\] + +END_SOLUTION +Context()->normalStrings; +################################## +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub3.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub3.pg new file mode 100644 index 0000000000..e29d962ca7 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub3.pg @@ -0,0 +1,153 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('3') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("-($a2+x^2)^(1/2)/x+ln((x+sqrt($a2+x^2))/$a)"); + ################################### + # Create link to applet + ################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{tan}); + $applet->initialState(qq{tan}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +################################### +# Main text + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{\sqrt{$a2 + x^2}}{x^2}\; dx \] +$BR \{ans_rule( 60) \} + +END_TEXT + +$ans = $funct; +ANS( $funct->cmp() ); +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the \(\tan\) substitution. \[x = {$a}\tan(\theta)\] +So: +\[dx = {$a}\sec^2(\theta) \; d\theta\] +Therefore: +\[\int \frac{\sqrt{$a2 + x^2}}{x^2} \;dx= +\int \frac{\sqrt{$a2 + $a2\tan^2\theta}}{$a2\tan^2\theta}($a\sec^2\theta) \; d\theta\] +\[= +\int \frac{\sec^3\theta}{\tan^2\theta} \; d\theta\] +\[= +\int \frac{1}{\cos\theta\sin^2\theta} \; d\theta\] + +$BR$BR +When we have an integrand which is the product of sines and cosines, we let \(u=\sin\theta\) if the cosine is raised to an odd power and let \(u=\cos\theta\) if the sine is raised to an odd power. + +Let \(u=\sin\theta\), then \(du=\cos\theta d\theta\). + +\[= +\int \frac{1}{\cos\theta\sin^2\theta} \; d\theta +=\int \frac{\cos\theta}{\cos^2\theta\sin^2\theta} \; d\theta\] +\[=\int \frac{\cos\theta}{(1-\sin^2\theta)\sin^2\theta} \; d\theta +=\int \frac{du}{(1-u^2)u^2} \] + +$BR$BR +We apply a partial fractions decomposition at this stage and obtain: +\[\int \frac{du}{(1-u^2)u^2} += \int \left(\frac{1}{2(1+u)}+\frac{1}{2(1-u)}+\frac{1}{u^2}\right) du\] +\[=\frac{1}{2}\ln\left|1+u\right|-\frac{1}{2}\ln\left|{1-u}\right|-\frac{1}{u}+C\] +\[=\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|-\frac{1}{u}+C\] + +Since \(u=\sin\theta\), +\[\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|-\frac{1}{u}+C +=\frac{1}{2}\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|+\ln|{\sin\theta}|+C\] +\[=\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\right|-\frac{1}{\sin\theta}+C\] +\[=\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|-\frac{1}{\sin\theta}+C\] +\[=\ln\left|\sec\theta+\tan\theta\right|-\frac{1}{\sin\theta}+C\] +$BR$BR +\[=\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|-\frac{\sqrt{$a2+x^2}}{x}+C\] +$BR$BR +Before proceeding to the method we can use to evaluate the integral of \(\sec^3\theta\), note that \(\tan\theta=\frac{x}{$a}\), and \(\sec\theta=\frac{\sqrt{$a2+x^2}}{$a}\). To see this, label a right triangle so that the tangent is \(x/$a\). We will have the opposite side with length \(x\), and the adjacent side with length \($a\), so the hypotenuse has length \(\sqrt{$a2+x^2}\). + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub4.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub4.pg new file mode 100644 index 0000000000..bd5eeceb57 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub4.pg @@ -0,0 +1,160 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('4') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("($a2+x^2)^(1/2)-$a*ln(($a+sqrt($a2+x^2))/x)"); + + ################################### + # Create link to applet + ################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{tan}); + $applet->initialState(qq{tan}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +################################### +# Main text + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{\sqrt{$a2 + x^2}}{x}\; dx \] +$BR \{ans_rule( 50) \} + +END_TEXT + +$ans = $funct; +ANS( $funct->cmp() ); +#ANS(fun_cmp($ans, limits=>[$lend,$rend], mode=>"antider", vars=>"x")); +################################## +Context()->texStrings; + +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the \(\tan\) substitution. \[x = {$a}\tan(\theta)\] +So: +\[dx = {$a}\sec^2(\theta) \; d\theta\] +Therefore: +\[\int \frac{\sqrt{$a2 + x^2}}{x} \;dx= +\int \frac{\sqrt{$a2 + $a2\tan^2\theta}}{$a\tan\theta}($a\sec^2\theta) \; d\theta\] +\[= +\int \frac{$a\sec^3\theta}{\tan\theta} \; d\theta\] +\[= +\int \frac{$a}{\cos^2\theta\sin\theta} \; d\theta\] + +$BR$BR +When we have an integrand which is the product of sines and cosines, we let \(u=\sin\theta\) if the cosine is raised to an odd power and let \(u=\cos\theta\) if the sine is raised to an odd power. + +Let \(u=\cos\theta\), then \(du=-\sin\theta d\theta\). + +\[= +\int \frac{$a}{\cos^2\theta\sin\theta} \; d\theta +=\int \frac{$a\sin\theta}{\cos^2\theta\sin^2\theta} \; d\theta\] +\[=$a\int \frac{\sin\theta}{(1-\cos^2\theta)\cos^2\theta} \; d\theta +=-$a\int \frac{du}{(1-u^2)u^2} \] + +$BR$BR +We apply a partial fractions decomposition at this stage and obtain: +\[-$a\int \frac{du}{(1-u^2)u^2} += -$a\int \left(\frac{1}{2(1+u)}+\frac{1}{2(1-u)}+\frac{1}{u^2}\right) du\] +\[=-\frac{$a}{2}\ln\left|1+u\right|+\frac{$a}{2}\ln\left|{1-u}\right|+\frac{$a}{u}+C\] +\[=\frac{$a}{2}\ln\left|\frac{1-u}{1+u}\right|+\frac{$a}{u}+C\] + +Since \(u=\cos\theta\), +\[\frac{$a}{2}\ln\left|\frac{1-u}{1+u}\right|+\frac{$a}{u}+C +=\frac{$a}{2}\ln\left|\frac{1-\cos\theta}{1+\cos\theta}\right|+\frac{$a}{\cos\theta}+C\] +\[=\frac{$a}{2}\ln\left|\frac{(1-\cos\theta)^2}{1-\cos^2\theta}\right|+\frac{$a}{\cos\theta}+C\] +\[=$a\ln\left|\frac{1-\cos\theta}{\sin\theta}\right|+\frac{$a}{\cos\theta}+C\] +\[=$a\ln\left|\csc\theta-\cot\theta\right|+$a\sec\theta+C\] +$BR$BR +\[=$a\ln\left|\frac{\sqrt{$a2+x^2}}{x}-\frac{$a}{x}\right|+\sqrt{$a2+x^2}+C\] +$BR$BR +Before proceeding to the method we can use to evaluate the integral of \(\sec^3\theta\), note that \(\tan\theta=\frac{x}{$a}\), and \(\sec\theta=\frac{\sqrt{$a2+x^2}}{$a}\). To see this, label a right triangle so that the tangent is \(x/$a\). We will have the opposite side with length \(x\), and the adjacent side with length \($a\), so the hypotenuse has length \(\sqrt{$a2+x^2}\). + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub5.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub5.pg new file mode 100644 index 0000000000..4e71100769 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub5.pg @@ -0,0 +1,150 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('5') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; + +$funct = FormulaUpToConstant("ln(abs(($a2+x^2)^(1/2)+x))"); + + ################################### + # Create link to applet + ################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{tan}); + $applet->initialState(qq{tan}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT +################################### +# Main text + + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{dx}{\sqrt{$a2 + x^2}} \] +$BR \{ans_rule( 50) \} + + +END_TEXT + + +$ans = $funct; + +ANS( $ans->cmp() ); +#ANS(fun_cmp($ans, limits=>[$lend,$rend], mode=>"antider", vars=>"x")); +################################## +Context()->texStrings; + +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the \(\tan\) substitution. \[x = {$a}\tan(\theta)\] + +$BR$BR +Before proceeding note that \(\tan\theta=\frac{x}{$a}\), and \(\sec\theta=\frac{\sqrt{$a2+x^2}}{$a}\). To see this, label a right triangle so that the tangent is \(x/$a\). We will have the opposite side with length \(x\), and the adjacent side with length \($a\), so the hypotenuse has length \(\sqrt{$a2+x^2}\). + +$BR$BR +With the substitution \[x = {$a}\tan(\theta)\] +\[dx = {$a}\sec^2(\theta) \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{dx}{\sqrt{$a2 + x^2}}= +\int \frac{{$a}\sec^2\theta d\theta}{\sqrt{$a2 + $a2\tan^2\theta}}\] +\[= +\int \frac{{$a}\sec^2\theta d\theta}{$a\sec\theta}\] +\[= +\int \sec\theta d\theta\] + +$BR$BR +\[= +\ln\left| \sec\theta+\tan\theta\right|+C\] + +\[=\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \] + +$BR$BR +Notice that since \(\ln\left|\frac{a}{b}\right|=\ln|a|-\ln|b|\), so +\(\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \) is equivalent to the solution \(\ln\left|\sqrt{$a2+x^2}+x\right|+C^\prime \). The \(-\ln|$a|\) is incorporated into the constant of integration. + + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub6.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub6.pg new file mode 100644 index 0000000000..6fc17d55c7 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub6.pg @@ -0,0 +1,175 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('6') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("-1/($a)*ln(abs(($a2+x^2)^(1/2)/x+$a/x))"); + + ################################### + # Create link to applet + ################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{tan}); + $applet->initialState(qq{tan}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + + +################################### +# Main text + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{dx}{x\sqrt{$a2 + x^2}} \] +$BR \{ans_rule( 50) \} + +END_TEXT + +$ans = $funct; +ANS( $funct->cmp() ); + +################################## +Context()->texStrings; + +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the \(\tan\) substitution. \[x = {$a}\tan(\theta)\] + +$BR$BR +Before proceeding note that \(\tan\theta=\frac{x}{$a}\), and \(\sec\theta=\frac{\sqrt{$a2+x^2}}{$a}\). To see this, label a right triangle so that the tangent is \(x/$a\). We will have the opposite side with length \(x\), and the adjacent side with length \($a\), so the hypotenuse has length \(\sqrt{$a2+x^2}\). + +$BR$BR +With the substitution \[x = {$a}\tan(\theta)\] +\[dx = {$a}\sec^2(\theta) \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{dx}{x\sqrt{$a2 + x^2}}= +\int \frac{{$a}\sec^2\theta d\theta}{$a\tan\theta\sqrt{$a2 + $a2\tan^2\theta}}\] +\[= +\int \frac{\sec^2\theta d\theta}{$a\tan\theta\sec\theta}\] +\[= +\int \frac{\sec\theta d\theta}{$a\tan\theta}\] +\[= +\int \frac{1 }{$a\sin\theta}d\theta\] + +$BR$BR +This is an odd power of the sine so we use a cosine substitution. Let +\[u=\cos\theta\] +then +\[du=-\sin\theta d\theta\] +\[ +\int \frac{1 }{$a\sin\theta}d\theta +=\int \frac{\sin\theta }{$a\sin^2\theta}d\theta\] +\[=\int \frac{\sin\theta }{$a(1-\cos^2\theta)}d\theta\] +\[=-\frac{1}{$a}\int \frac{du }{1-u^2}\] + +$BR$BR +We do a partial fractions decomposition: +\[-\frac{1}{$a}\int \frac{du }{1-u^2}=-\frac{1}{2($a)}\int \frac{1}{1-u}+\frac{1}{1+u}du\] + +$BR$BR +So +\[-\frac{1}{2($a)}\int \frac{1}{1-u}+\frac{1}{1+u}du +=\frac{1}{2($a)}\ln|1-u|-\frac{1}{2($a)}\ln|1+u|+C\] +\[=\frac{1}{2($a)}\ln\left|\frac{1-u}{1+u}\right|+C\] + +$BR$BR +Substituting back in +\[\frac{1}{2($a)}\ln\left|\frac{1-u}{1+u}\right|+C\] +\[=\frac{1}{2($a)}\ln\left|\frac{1-\cos\theta}{1+\cos\theta}\right|+C\] +\[=\frac{1}{2($a)}\ln\left|\frac{(1-\cos\theta)^2}{1-\cos^2\theta}\right|+C\] +\[=\frac{1}{2($a)}\ln\left|\frac{(1-\cos\theta)^2}{\sin^2\theta}\right|+C\] +\[=\frac{1}{$a}\ln\left|\frac{1-\cos\theta}{\sin\theta}\right|+C\] +\[=\frac{1}{$a}\ln\left|\csc\theta-\cot\theta\right|+C\] + + + +$BR$BR +\[\frac{1}{$a}\ln\left|\csc\theta-\cot\theta\right|+C=\frac{1}{$a}\ln\left|\frac{\sqrt{$a2+x^2}}{x}-\frac{$a}{x}\right|+C\] + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub7.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub7.pg new file mode 100644 index 0000000000..0c4851050b --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub7.pg @@ -0,0 +1,246 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('7') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("1/2*x*($a2+x^2)^(1/2)-1/2*($a2)*ln(x+($a2+x^2)^(1/2))"); + + ################################### + # Create link to applet + ################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{tan}); + $applet->initialState(qq{tan}); + +################################## +# Setup Geogebra applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +################################### +# Main text +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{x^2 dx}{\sqrt{$a2 + x^2}} \] +$BR \{ans_rule( 50) \} + +END_TEXT + +$ans = $funct; +ANS( $funct->cmp() ); +################################## +Context()->texStrings; + + +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +In this problem, we have an expression of the form \(\sqrt{a^2+u^2}\) +where \(a=$a\), and \(u=x\), so we use the tangent substitution, +\[x = {$a}\tan(\theta)\] and we refer to the middle triangle. + +To evaluate this integral use a trigonometric substitution. For this problem use the \(\tan\) substitution. \[x = {$a}\tan(\theta)\] + +$BR$BR +Before proceeding note that \(\tan\theta=\frac{x}{$a}\), and \(\sec\theta=\frac{\sqrt{$a2+x^2}}{$a}\). To see this, label a right triangle so that the tangent is \(x/$a\). We will have the opposite side with length \(x\), and the adjacent side with length \($a\), so the hypotenuse has length \(\sqrt{$a2+x^2}\). + +$BR$BR +With the substitution \[x = {$a}\tan(\theta)\] +\[dx = {$a}\sec^2(\theta) \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{x^2 dx}{\sqrt{$a2 + x^2}}= +\int \frac{{$a3}\tan^2\theta\sec^2\theta }{\sqrt{$a2 + $a2\tan^2\theta}}\;d\theta\] +\[= +\int \frac{{$a3}\tan^2\theta\sec^2\theta }{{$a}\sec\theta}\;d\theta\] +\[= +\int {$a2}\tan^2\theta\sec\theta \;d\theta\] +\[= +\int {$a2}(\sec^2\theta-1)\sec\theta \;d\theta\] +\[= +\int {$a2}(\sec^3\theta-\sec\theta) \;d\theta\] + +$BR +From this point, we have several choices as to how to evaluate the integral of \(\sec^3\theta\). + +$BR$BR +Method 1: + +$BR +Look up the antiderivative of \(\int\sec^3\theta d\theta\). It is + +\[\int\sec^3\theta d\theta=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec\theta+\tan\theta\right|+C\] + +$BR$BR +so +\[ \int \sqrt{$a2 + x^2}\; dx=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln\left|\sec\theta+\tan\theta\right|+C \] + +$BR +Substituting back in terms of \(\theta\) yields: +\[\frac{$a2}{2}\frac{\sqrt{$a2+x^2}}{$a}\frac{x}{$a}+\frac{$a2}{2}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \] +Simplifying: +\[\frac{x\sqrt{$a2+x^2}}{2}+\frac{$a2}{2}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \] + + +$BR$BR +Method 2: + +$BR +Use the rule for substitution for products of sines and cosines. If sine is to an odd power, use a cosine substitution. If cosine is to an odd power, use a sine substitution. \(\sec^3\theta=\cos^{-3}\theta\), and -3 is an odd number, so we use a sine substitution. + +$BR +Let \(u=\sin\theta\), then \(du=\cos\theta d\theta\) so +\[\int $a2\sec^3\theta \; d\theta=$a2\int\frac{1}{\cos^3\theta}d\theta\] +\[=$a2\int\frac{\cos\theta}{\cos^4\theta}d\theta\] +\[=$a2\int\frac{\cos\theta}{(1-\sin^2\theta)^2}d\theta\] +\[=$a2\int\frac{du}{(1-u^2)^2}du\] +From here, we use a partial fractions decomposition: +\[$a2\int\frac{du}{(1-u^2)^2}du=$a2\int\left[\frac{A}{(1-u)^2}+\frac{B}{1-u}+\frac{C}{(1+u)^2}+\frac{D}{1+u}\right]du\] + +$BR +Solving for the constants \(A\), \(B\), \(C\) and \(D\) we find that \(A=B=C=D=\frac{1}{4}\). + +$BR$BR +\[\int $a2\sec^3\theta \; d\theta=\frac{$a2}{4}\int\frac{1}{(1-u)^2}+\frac{1}{1-u}+\frac{1}{(1+u)^2}+\frac{1}{1+u}du \] +\[=\frac{$a2}{4}\left(\frac{1}{1-u}-\ln|1-u|-\frac{1}{1+u}+\ln|1+u|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2u}{1-u^2}+\ln\left|\frac{1+u}{1-u}\right|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2\sin\theta}{1-\sin^2\theta}+\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|\right)+C \] +\[=\frac{$a2}{4}\left(\frac{2\sin\theta}{\cos^2\theta}+\ln\left|\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\right|\right)+C \] +\[=\frac{$a2}{4}\left(2\sec\theta\tan\theta+\ln\left|\frac{(1+\sin\theta)^2}{\cos^2\theta}\right|\right)+C \] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|+C \] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln|\sec\theta+\tan\theta|+C \] + +$BR +Substituting back in terms of \(\theta\) yields: +\[\frac{$a2}{2}\frac{\sqrt{$a2+x^2}}{$a}\frac{x}{$a}+\frac{$a2}{2}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \] +Simplifying: +\[\frac{x\sqrt{$a2+x^2}}{2}+\frac{$a2}{2}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C \] + +$BR$BR +Method 3: + +$BR +Use the secant reduction formula. The secant reduction formula is given by +\[\int \sec^n\theta d\theta=\frac{1}{n-1}\tan\theta\sec^{n-2}\theta+\frac{n-2}{n-1}\int\sec^{n-2}\theta d\theta.\] + +$BR$BR +so +\[$a2\int\sec^3\theta d\theta = \frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\int\sec\theta d\theta.\] + +$BR$BR +We can evaluate \(\int\sec\theta d\theta\) by looking the antiderivative up in a table (\(\int\sec\theta d\theta=\ln|\sec\theta+\tan\theta|+C\)) or we can derive this antiderivative using the substitution rule for products of sines and cosines. We have an odd power of the cosine, we use a sine substitution. + +$BR$BR +\[\int\sec\theta d\theta=\int\frac{\cos\theta}{\cos^2\theta}d\theta\] +\[=\int\frac{\cos\theta}{1-\sin^2\theta}d\theta\] +\[=\int\frac{1}{1-u^2}du\] + +$BR +Do a partial fractions decomposition. +\[\int\frac{1}{1-u^2}du=\frac{1}{2}\int\left(\frac{1}{1-u}+\frac{1}{1+u}\right)du\] +\[=\frac{1}{2}\left(-\ln|1-u|+\ln|1+u|\right)+C\] +\[=\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\right|+C\] +\[=\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{\cos^2\theta}\right|+C\] +\[=\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|+C\] +\[=\ln\left|\sec\theta+\tan\theta\right|+C\] + +so \[\int\sec\theta d\theta=\ln\left|\sec\theta+\tan\theta\right|+C\] and +\[$a2\int\sec^3\theta d\theta = \frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\int\sec\theta d\theta.\] +\[=\frac{$a2}{2}\sec\theta\tan\theta+\frac{$a2}{2}\ln\left|\sec\theta+\tan\theta\right|+C.\] + +$BR$BR +Putting this all together, we have that +\[= +\int {$a2}(\sec^3\theta-\sec\theta) \;d\theta\] +\[= +\frac{x\sqrt{$a2+x^2}}{2}+\frac{$a2}{2}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|-{$a2}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C\] +\[= +\frac{x\sqrt{$a2+x^2}}{2}-\frac{$a2}{2}\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|+C +\] + + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub8.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub8.pg new file mode 100644 index 0000000000..324a74f3c3 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub8.pg @@ -0,0 +1,164 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('8') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("-($a2+x^2)^(1/2)/{x*{$a2}}"); + + + ################################### + # Create link to applet + ################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{tan}); + $applet->initialState(qq{tan}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +################################### +# Main text + + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{ dx}{x^2\sqrt{$a2 + x^2}} \] +$BR \{ans_rule( 50) \} + +END_TEXT + +$ans = $funct; +ANS( $funct->cmp() ); +#ANS(fun_cmp($ans, limits=>[$lend,$rend], mode=>"antider", vars=>"x")); +################################## +Context()->texStrings; + +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->normalStrings; + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the \(\tan\) substitution. \[x = {$a}\tan(\theta)\] + +$BR$BR +Before proceeding note that \(\tan\theta=\frac{x}{$a}\), and \(\sec\theta=\frac{\sqrt{$a2+x^2}}{$a}\). To see this, label a right triangle so that the tangent is \(x/$a\). We will have the opposite side with length \(x\), and the adjacent side with length \($a\), so the hypotenuse has length \(\sqrt{$a2+x^2}\). + +$BR$BR +With the substitution \[x = {$a}\tan(\theta)\] +\[dx = {$a}\sec^2(\theta) \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{ dx}{x^2\sqrt{$a2 + x^2}}= +\int \frac{{$a}\sec^2\theta }{{$a2}\tan^2\theta\sqrt{$a2 + $a2\tan^2\theta}}\;d\theta\] +\[= +\int \frac{{$a}\sec^2\theta }{{$a3}\tan^2\theta\sec\theta}\;d\theta\] +\[= +\int \frac{\sec\theta }{{$a2}\tan^2\theta}\;d\theta\] +\[= +\int \frac{\cos\theta }{{$a2}\sin^2\theta}\;d\theta\] + +$BR +We have an odd power of the cosine, so we use a sine substitution. +$BR +Let \(u=\sin\theta\), then \(du=\cos\theta d\theta\). +\[ +\int \frac{\cos\theta }{{$a2}\sin^2\theta}\;d\theta +=\int \frac{du }{{$a2}u^2}\] +\[=-\frac{1}{{$a2}u}+C +\] +\[=-\frac{1}{{$a2}\sin\theta}+C +\] + +$BR +Substituting back in terms of \(\theta\) yields: +\[-\frac{1}{{$a2}\sin\theta}+C= +-\frac{\sqrt{$a2+x^2}}{{$a2} x}+C \] +so +\[ \int \frac{ dx}{x^2\sqrt{$a2 + x^2}}=-\frac{\sqrt{$a2+x^2}}{{$a2} x}+C \] + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. Uses Flash applet.'); + +ENDDOCUMENT(); # This should be the last executable line in the problem. \ No newline at end of file diff --git a/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub9.pg b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub9.pg new file mode 100644 index 0000000000..cb1d93f4e4 --- /dev/null +++ b/OpenProblemLibrary/CSUOhio/calculus/trigonometric_substitution/trigSub9.pg @@ -0,0 +1,148 @@ +##DESCRIPTION +##KEYWORDS('integrals', 'trigonometric','substitution') + +## DBsubject('Calculus') +## DBchapter('Techniques of Integration') +## DBsection('Trigonometric Substitution') +## Date('2/22/13') +## Author('Barbara Margolius') +## Institution('Cleveland State University') +## TitleText1('') +## EditionText1('') +## AuthorText1('') +## Section1('') +## Problem1('9') +##ENDDESCRIPTION + +############################################################################ +## development of this problem is supported in part by the National Science# +## Foundation under the grant DUE-0941388. # +############################################################################ + +DOCUMENT(); # This should be the first executable line in the problem. + +loadMacros( + "PGstandard.pl", + "AppletObjects.pl", + "MathObjects.pl", + "parserFormulaUpToConstant.pl", +); + +TEXT(beginproblem()); +$showPartialCorrectAnswers = 1; + +$a = random(2,9,1); + +$a2 = $a*$a; +$a3 = $a2*$a; +$a4 = $a2*$a2; + +$funct = FormulaUpToConstant("x/({$a2}*($a2+x^2)^(1/2))"); + ################################### + # Create link to applet + ################################### + $appletName = "trigSubWW"; + $applet = FlashApplet( + codebase => findAppletCodebase("$appletName.swf"), + appletName => $appletName, + appletId => $appletName, + setStateAlias => 'setXML', + getStateAlias => 'getXML', + setConfigAlias => 'setConfig', + maxInitializationAttempts => 10, # number of attempts to initialize applet + #answerBoxAlias => 'answerBox', + height => '550', + width => '595', + bgcolor => '#e8e8e8', + debugMode => 0, + ); + +################################### +# Configure applet +################################### + + $applet->configuration(qq{tan}); + $applet->initialState(qq{tan}); + +################################## +# Setup Flash applet -- this does not need to be changed +################################### + +TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); + +END_TEXT + +################################### +# Main text + +BEGIN_TEXT + +Evaluate the indefinite integral. +$BR \[ \int \frac{ dx}{($a2 + x^2)^{3/2}} \] +$BR \{ans_rule( 50) \} + +END_TEXT + +ANS( $funct->cmp() ); +################################## +Context()->texStrings; + +################################### +TEXT($PAR, $BBOLD, $BITALIC, "Hi $studentLogin, If you don't get this in 5 tries I'll give you a hint with an applet to help you out.", $EITALIC, $EBOLD, $PAR); + +$im = image( "trigsub.png", +width=>486, height=>306, tex_size=>900 ); + +$showHint=5; +Context()->normalStrings; +TEXT(hint( + $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( + debug =>0, reinitialize_button => 0, includeAnswerBox=>0, + ), TeX=>$im),"Follow the step-by-step questions in the hint in the online version of this problem." +)); + +################################## +Context()->texStrings; +SOLUTION(EV3(<<'END_SOLUTION')); +$BBOLD Solution: $EBOLD $PAR +To evaluate this integral use a trigonometric substitution. For this problem use the \(\tan\) substitution. \[x = {$a}\tan(\theta)\] + +$BR$BR +Before proceeding note that \(\tan\theta=\frac{x}{$a}\), and \(\sec\theta=\frac{\sqrt{$a2+x^2}}{$a}\). To see this, label a right triangle so that the tangent is \(x/$a\). We will have the opposite side with length \(x\), and the adjacent side with length \($a\), so the hypotenuse has length \(\sqrt{$a2+x^2}\). + +$BR$BR +With the substitution \[x = {$a}\tan(\theta)\] +\[dx = {$a}\sec^2(\theta) \; d\theta\] +$BR$BR +Therefore: +\[\int \frac{ dx}{($a2 + x^2)^{3/2}}= +\int \frac{{$a}\sec^2\theta }{($a2 + $a2\tan^2\theta)^{3/2}}\;d\theta\] +\[= +\int \frac{{$a}\sec^2\theta }{$a3\sec^3\theta}\;d\theta\] +\[= +\int \frac{1 }{$a2\sec\theta}\;d\theta\] +\[= +\frac{1 }{$a2}\int \cos\theta\;d\theta\] +\[= +\frac{1 }{$a2} \sin\theta+C\] + +$BR +Substituting back in terms of \(\theta\) yields: +\[= +\frac{1 }{$a2} \sin\theta+C += +\frac{ x}{{$a2}\sqrt{$a2+x^2} } +C \] +so +\[ \int \frac{ dx}{($a2 + x^2)^{3/2}}=\frac{ x}{{$a2}\sqrt{$a2+x^2} } +C \] + +END_SOLUTION +Context()->normalStrings; +################################## + +COMMENT('MathObject version. 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