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1 ##DESCRIPTION 2 ##KEYWORDS('integrals', 'trigonometric','substitution') 3 4 ## DBsubject('Calculus') 5 ## DBchapter('Techniques of Integration') 6 ## DBsection('Trigonometric Substitution') 7 ## Date('8/20/11') 8 ## Author('Barbara Margolius') 9 ## Institution('Cleveland State University') 10 ## TitleText1('Pauls Online Math Notes Calc 2') 11 ## EditionText1('2010') 12 ## AuthorText1('Dawkins') 13 ## Section1('') 14 ## Problem1('3') 15 ##ENDDESCRIPTION 16 17 ############################################################################ 18 ## development of this problem is supported in part by the National Science# 19 ## Foundation under the grant DUE-0941388. # 20 ############################################################################ 21 22 DOCUMENT(); # This should be the first executable line in the problem. 23 24 loadMacros( 25 "PGstandard.pl", 26 "AppletObjects.pl", 27 "MathObjects.pl", 28 "parserFormulaUpToConstant.pl", 29 ); 30 31 TEXT(beginproblem()); 32 $showPartialCorrectAnswers = 1; 33 34 $a = random(2,9,1); 35 36 $a2 = $a*$a; 37 $a4 = $a2*$a2; 38 39 $funct = FormulaUpToConstant("-($a2+x^2)^(1/2)/x+ln((x+sqrt($a2+x^2))/$a)"); 40 ################################### 41 # Create link to applet 42 ################################### 43 $appletName = "trigSubWW"; 44 $applet = FlashApplet( 45 codebase => findAppletCodebase("$appletName.swf"), 46 appletName => $appletName, 47 appletId => $appletName, 48 setStateAlias => 'setXML', 49 getStateAlias => 'getXML', 50 setConfigAlias => 'setConfig', 51 maxInitializationAttempts => 10, # number of attempts to initialize applet 52 #answerBoxAlias => 'answerBox', 53 height => '550', 54 width => '595', 55 bgcolor => '#e8e8e8', 56 debugMode => 0, 57 onInit => 'ggbOnInit', 58 ); 59 60 ################################### 61 # Configure applet 62 ################################### 63 64 $applet->configuration(qq{<xml><trigString>tan</trigString></xml>}); 65 $applet->initialState(qq{<xml><trigString>tan</trigString></xml>}); 66 67 ################################## 68 # Setup Flash applet -- this does not need to be changed 69 ################################### 70 71 HEADER_TEXT(qq! 72 73 <script language="javascript"> 74 75 function ggbOnInit(param) { 76 if (param == "$appletName") { 77 applet_loaded(param,1); // report that applet is ready. 78 ww_applet_list[param].safe_applet_initialize(2); 79 } 80 81 } 82 83 </script> 84 85 ! 86 87 ); 88 89 90 TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); 91 <script> 92 if (navigator.appVersion.indexOf("MSIE") > 0) { 93 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>"); 94 } 95 </script> 96 END_TEXT 97 98 ################################### 99 # Main text 100 101 BEGIN_TEXT 102 103 Evaluate the indefinite integral. 104 $BR \[ \int \frac{\sqrt{$a2 + x^2}}{x^2}\; dx \] 105 $BR \{ans_rule( 60) \} 106 107 END_TEXT 108 109 $ans = $funct; 110 ANS( $funct->cmp() ); 111 ################################### 112 TEXT($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); 113 114 $showHint=5; 115 Context()->normalStrings; 116 TEXT(hint( 117 $PAR, MODES(TeX=>'object code', HTML=>$applet->insertAll( 118 debug =>0, reinitialize_button => 0, includeAnswerBox=>0, 119 )) 120 )); 121 ################################## 122 Context()->texStrings; 123 SOLUTION(EV3(<<'END_SOLUTION')); 124 $BBOLD Solution: $EBOLD $PAR 125 To evaluate this integral use a trigonometric substitution. For this problem use the \(\tan\) substitution. \[x = {$a}\tan(\theta)\] 126 So: 127 \[dx = {$a}\sec^2(\theta) \; d\theta\] 128 Therefore: 129 \[\int \frac{\sqrt{$a2 + x^2}}{x^2} \;dx= 130 \int \frac{\sqrt{$a2 + $a2\tan^2\theta}}{$a2\tan^2\theta}($a\sec^2\theta) \; d\theta\] 131 \[= 132 \int \frac{\sec^3\theta}{\tan^2\theta} \; d\theta\] 133 \[= 134 \int \frac{1}{\cos\theta\sin^2\theta} \; d\theta\] 135 136 $BR$BR 137 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. 138 139 Let \(u=\sin\theta\), then \(du=\cos\theta d\theta\). 140 141 \[= 142 \int \frac{1}{\cos\theta\sin^2\theta} \; d\theta 143 =\int \frac{\cos\theta}{\cos^2\theta\sin^2\theta} \; d\theta\] 144 \[=\int \frac{\cos\theta}{(1-\sin^2\theta)\sin^2\theta} \; d\theta 145 =\int \frac{du}{(1-u^2)u^2} \] 146 147 $BR$BR 148 We apply a partial fractions decomposition at this stage and obtain: 149 \[\int \frac{du}{(1-u^2)u^2} 150 = \int \left(\frac{1}{2(1+u)}+\frac{1}{2(1-u)}+\frac{1}{u^2}\right) du\] 151 \[=\frac{1}{2}\ln\left|1+u\right|-\frac{1}{2}\ln\left|{1-u}\right|-\frac{1}{u}+C\] 152 \[=\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|-\frac{1}{u}+C\] 153 154 Since \(u=\sin\theta\), 155 \[\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|-\frac{1}{u}+C 156 =\frac{1}{2}\ln\left|\frac{1+\sin\theta}{1-\sin\theta}\right|+\ln|{\sin\theta}|+C\] 157 \[=\frac{1}{2}\ln\left|\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\right|-\frac{1}{\sin\theta}+C\] 158 \[=\ln\left|\frac{1+\sin\theta}{\cos\theta}\right|-\frac{1}{\sin\theta}+C\] 159 \[=\ln\left|\sec\theta+\tan\theta\right|-\frac{1}{\sin\theta}+C\] 160 $BR$BR 161 \[=\ln\left|\frac{\sqrt{$a2+x^2}}{$a}+\frac{x}{$a}\right|-\frac{\sqrt{$a2+x^2}}{x}+C\] 162 $BR$BR 163 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 \(\sqrt{$a2+x^2}\). 164 165 166 167 END_SOLUTION 168 Context()->normalStrings; 169 ################################## 170 ENDDOCUMENT(); # This should be the last executable line in the problem.
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