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

Latest revision as of 17:19, 23 November 2017

Evaluate the indefinite and definite integrals.

(a) $\int x\ln x~dx$

(b) $\int _{0}^{\pi }\sin ^{2}x~dx$

Solution

Detailed Solution

Return to Sample Exam

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

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

Latest revision as of 17:19, 23 November 2017

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