Difference between revisions of "009B Sample Midterm 3, Problem 5"

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(Replaced content with "<span class="exam">Evaluate the indefinite and definite integrals. <span class="exam">(a)   <math>\int x\ln x ~dx</math> <span class="exam">(b)   <math>\int_0...")
 
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<span class="exam">Evaluate the indefinite and definite integrals.
 
<span class="exam">Evaluate the indefinite and definite integrals.
  
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<span class="exam">(a) &nbsp; <math>\int \tan^3x ~dx</math>  
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<span class="exam">(a) &nbsp; <math>\int x\ln x ~dx</math>  
  
 
<span class="exam">(b) &nbsp; <math>\int_0^\pi \sin^2x~dx</math>
 
<span class="exam">(b) &nbsp; <math>\int_0^\pi \sin^2x~dx</math>
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<hr>
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[[009B Sample Midterm 3, Problem 5 Solution|'''<u>Solution</u>''']]
  
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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[[009B Sample Midterm 3, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
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!Foundations: &nbsp;
 
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|'''1.''' Recall the trig identity
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
 
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|'''2.''' Recall the trig identity
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}</math>
 
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|'''3.''' How would you integrate &nbsp;<math style="vertical-align: -1px">\tan x~dx?</math>
 
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&nbsp; &nbsp; &nbsp; &nbsp; You could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 
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|&nbsp; &nbsp; &nbsp; &nbsp; First, write &nbsp;<math style="vertical-align: -14px">\int \tan x~dx=\int \frac{\sin x}{\cos x}~dx.</math>
 
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&nbsp; &nbsp; &nbsp; &nbsp; Now, let &nbsp;<math style="vertical-align: -5px">u=\cos(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">du=-\sin(x)dx.</math>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; Thus,
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int \tan x~dx} & = & \displaystyle{\int \frac{-1}{u}~du}\\
 
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& = & \displaystyle{-\ln(u)+C}\\
 
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&&\\
 
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& = & \displaystyle{-\ln|\cos x|+C.}\\
 
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\end{array}</math>
 
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'''Solution:'''
 
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'''(a)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|We start by writing
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\int \tan^3x~dx=\int \tan^2x\tan x ~dx.</math>
 
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|Since &nbsp;<math style="vertical-align: -5px">\tan^2x=\sec^2x-1,</math>&nbsp; we have
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int (\sec^2x-1)\tan x ~dx}\\
 
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&&\\
 
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& = & \displaystyle{\int \sec^2x\tan x~dx-\int \tan x~dx.}\\
 
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\end{array}</math>
 
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!Step 2: &nbsp;
 
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|Now, we need to use &nbsp;<math>u</math>-substitution for the first integral.
 
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Let &nbsp;<math style="vertical-align: -5px">u=\tan(x).</math>
 
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|Then, &nbsp;<math style="vertical-align: -1px">du=\sec^2x~dx.</math>
 
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|So, we have
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int u~du-\int \tan x~dx}\\
 
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& = & \displaystyle{\frac{u^2}{2}-\int \tan x~dx}\\
 
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&&\\
 
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& = & \displaystyle{\frac{\tan^2x}{2}-\int \tan x~dx.}\\
 
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\end{array}</math>
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
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|For the remaining integral, we also need to use &nbsp;<math>u</math>-substitution.
 
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|First, we write
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\int \tan^3x~dx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}~dx.</math>
 
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|Now, we let &nbsp;<math style="vertical-align: 0px">u=\cos x.</math>
 
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|Then, &nbsp;<math style="vertical-align: -1px">du=-\sin x~dx.</math>
 
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|Therefore, we get
 
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
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\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\frac{\tan^2x}{2}+\int \frac{1}{u}~dx}\\
 
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& = & \displaystyle{\frac{\tan^2x}{2}+\ln |u|+C}\\
 
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& = & \displaystyle{\frac{\tan^2x}{2}+\ln |\cos x|+C.}
 
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\end{array}</math>
 
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'''(b)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|One of the double angle formulas is
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">\cos(2x)=1-2\sin^2(x).</math>
 
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|Solving for &nbsp;<math style="vertical-align: -5px">\sin^2(x),</math>&nbsp; we get
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}.</math>
 
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|Plugging this identity into our integral, we get
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\int_0^\pi \frac{1-\cos(2x)}{2}~dx}\\
 
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& = & \displaystyle{\int_0^\pi \frac{1}{2}~dx-\int_0^\pi \frac{\cos(2x)}{2}~dx.}\\
 
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\end{array}</math>
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|If we integrate the first integral, we get
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\left.\frac{x}{2}\right|_{0}^\pi-\int_0^\pi \frac{\cos(2x)}{2}~dx}\\
 
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& = & \displaystyle{\frac{\pi}{2}-\int_0^\pi \frac{\cos(2x)}{2}~dx.}\\
 
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\end{array}</math>
 
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!Step 3: &nbsp;
 
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|For the remaining integral, we need to use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 
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|Let &nbsp;<math style="vertical-align: -1px">u=2x.</math>
 
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|Then, &nbsp;<math style="vertical-align: -1px">du=2~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -18px">\frac{du}{2}=dx.</math>
 
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|Also, since this is a definite integral and we are using <math style="vertical-align: 0px">u</math>-substitution,
 
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|we need to change the bounds of integration.
 
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|We have &nbsp;<math style="vertical-align: -5px">u_1=2(0)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=2(\pi)=2\pi.</math>
 
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|So, the integral becomes
 
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
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\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\frac{\pi}{2}-\int_0^{2\pi} \frac{\cos(u)}{4}~du}\\
 
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& = & \displaystyle{\frac{\pi}{2}-\left.\frac{\sin(u)}{4}\right|_0^{2\pi}}\\
 
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& = & \displaystyle{\frac{\pi}{2}-\bigg(\frac{\sin(2\pi)}{4}-\frac{\sin(0)}{4}\bigg)}\\
 
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& = & \displaystyle{\frac{\pi}{2}.}\\
 
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\end{array}</math>
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{\tan^2x}{2}+\ln |\cos x|+C</math>
 
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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{\pi}{2}</math>
 
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[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 17:19, 23 November 2017

Evaluate the indefinite and definite integrals.

(a)   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x\ln x ~dx}

(b)   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^\pi \sin^2x~dx}

Solution


Detailed Solution


Return to Sample Exam

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