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Tue Aug 23 19:00:47 2011 UTC (20 months, 4 weeks ago) by bmargolius
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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
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") {
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.