009B Sample Final 3, Problem 5 - Revision history

MathAdmin: Created page with " Find the following integrals. (a) $\int x\cos(x)~dx$ (b) $\int \sin^3(x)\cos^2(x)~dx$
2017-04-10T16:54:30Z

Created page with "<span class="exam"> Find the following integrals. <span class="exam">(a) <math>\int x\cos(x)~dx</math> <span class="exam">(b) <math>\int \sin^3(x)\cos^2(x)~dx</..."

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<span class="exam"> Find the following integrals.

<span class="exam">(a)  <math>\int x\cos(x)~dx</math>

<span class="exam">(b)  <math>\int \sin^3(x)\cos^2(x)~dx</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
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|'''1.''' Integration by parts tells us that
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|        <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
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|'''2.''' Since  <math style="vertical-align: -4px">\sin^2x+\cos^2x=1,</math>  we have
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|        <math>\sin^2x=1-\cos^2x.</math>
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'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|To calculate this integral, we use integration by parts.
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|Let  <math style="vertical-align: 0px">u=x</math>  and  <math style="vertical-align: 0px">dv=\cos xdx.</math>
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|Then,  <math style="vertical-align: 0px">du=dx</math>  and  <math style="vertical-align: -1px">v=\sin x.</math>
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|Therefore, we have
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|       <math>\int x\cos(x)~dx=x\sin x -\int \sin x~dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|Then, we integrate to get
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|        <math> \int x\cos(x)~dx=x\sin x +\cos x+C.</math>
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|
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'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|First, we use the identity  <math style="vertical-align: -1px">\sin^2 x=1-\cos^2 x</math>  to get
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|        <math>\begin{array}{rcl}
\displaystyle{\int \sin^3(x)\cos^2(x)~dx} & = & \displaystyle{\int \sin^2 x (\cos^2 x) \sin x~dx}\\
&&\\
& = & \displaystyle{\int (1-\cos^2 x)(\cos^2 x) \sin x~dx.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|Now, we use  <math style="vertical-align: 0px">u</math>-substitution.
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|Let  <math style="vertical-align: -5px">u=\cos(x).</math> 
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|Then,  <math style="vertical-align: -5px">du=-\sin(x)dx</math>  and  <math style="vertical-align: -5px">-du=\sin(x)dx.</math>
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|Therefore, we have
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|        <math>\begin{array}{rcl}
\displaystyle{\int \sin^3(x)\cos^2(x)~dx} & = & \displaystyle{\int (-1)(1-u^2)u^2~du}\\
&&\\
& = & \displaystyle{\int u^4-u^2~du}\\
&&\\
& = & \displaystyle{\frac{u^5}{5}-\frac{u^3}{3}+C}\\
&&\\
& = & \displaystyle{\frac{\cos^5 x}{5}-\frac{\cos^3 x}{3}+C.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|   '''(a)'''    <math>x\sin x +\cos x+C</math>
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|   '''(b)'''    <math>\frac{\cos^5 x}{5}-\frac{\cos^3 x}{3}+C</math>
|}
[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]