add trigonometric_substitution problem set to OPL

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{<xml><trigString>sec</trigString></xml>});
+    $applet->initialState(qq{<xml><trigString>sec</trigString></xml>});
+
+##################################
+# Setup Flash applet -- this does not need to be changed
+###################################
+
+TEXT(MODES(TeX=>"", HTML=><<'END_TEXT'));
+<script>
+if (navigator.appVersion.indexOf("MSIE") > 0) {
+	document.write("<div width='3in' align='center' style='background:yellow'>You seem to be using Internet Explorer.<br/>It is recommended that another browser be used to view this page.</div>");
+}
+</script>
+END_TEXT
+
+BEGIN_TEXT
+
+Evaluate the indefinite integral.
+$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.

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{<xml><trigString>sin</trigString></xml>});
+    $applet->initialState(qq{<xml><trigString>sin</trigString></xml>});
+
+##################################
+# Setup Flash applet -- this does not need to be changed
+###################################
+
+TEXT(MODES(TeX=>"", HTML=><<'END_TEXT'));
+<script>
+if (navigator.appVersion.indexOf("MSIE") > 0) {
+	document.write("<div width='3in' align='center' style='background:yellow'>You seem to be using Internet Explorer.<br/>It is recommended that another browser be used to view this page.</div>");
+}
+</script>
+END_TEXT
+
+
+###################################
+# Main text
+
+BEGIN_TEXT
+
+Evaluate the indefinite integral.
+$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.