MathAdmin: Created page with " Evaluate the following integrals: (a) $\int \frac{dx}{x^2\sqrt{x^2-16}}$ (b) $\int_{-\p..."$

2017-04-10T16:50:31Z

Created page with " Evaluate the following integrals: (a) <math>\int \frac{dx}{x^2\sqrt{x^2-16}}</math> (b) <math>\int_{-\p..."

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Evaluate the following integrals:

(a)  <math>\int \frac{dx}{x^2\sqrt{x^2-16}}</math>

(b)  <math>\int_{-\pi}^\pi \sin^3x\cos^3x~dx</math>

(c)  <math>\int_0^1 \frac{x-3}{x^2+6x+5}~dx</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' For  <math>\int \frac{dx}{x^2\sqrt{x^2-16}},</math>  what would be the correct trig substitution?
|-
|       The correct substitution is  <math style="vertical-align: -1px">x=4\sec^2\theta.</math>
|-
|'''2.''' Recall the Pythagorean identity
|-
|       <math style="vertical-align: -5px">\cos^2(x)=1-\sin^2(x).</math>
|-
|'''3.''' Through partial fraction decomposition, we can write the fraction
|-
|       <math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math>
|-
|       for some constants <math style="vertical-align: -4px">A,B.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by using trig substitution.
|-
|Let  <math style="vertical-align: -1px">x=4\sec \theta.</math>
|-
|Then,  <math style="vertical-align: -2px">dx=4\sec \theta \tan \theta ~d\theta.</math>
|-
|So, the integral becomes
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\int \frac{4\sec \theta \tan \theta}{16\sec^2\theta \sqrt{16\sec^2 \theta -16}}~d\theta}\\
&&\\
& = & \displaystyle{\int \frac{4\sec \theta \tan \theta}{16\sec^2\theta (4\tan \theta)} ~d\theta}\\
&&\\
& = & \displaystyle{\int \frac{1}{16\sec \theta} ~d\theta.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we integrate to get
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\int \frac{1}{16}\cos\theta~d\theta}\\
&&\\
& = & \displaystyle{\frac{1}{16}\sin \theta +C}\\
&&\\
& = & \displaystyle{\frac{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C.}
\end{array}</math>
|}

'''(b)'''

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

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we use  <math style="vertical-align: 0px">u</math>-substitution.
|-
|Let  <math style="vertical-align: -1px">u=\sin x.</math>  Then,  <math style="vertical-align: -1px">du=\cos x ~dx.</math>
|-
|Since this is a definite integral, we need to change the bounds of integration.
|-
|Then, we have
|-
|       <math style="vertical-align: -5px">u_1=\sin(-\pi)=0</math>  and  <math style="vertical-align: -5px">u_2=\sin(\pi)=0.</math>
|-
|So, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int_{-\pi}^\pi \sin^3x\cos^3x~dx} & = & \displaystyle{\int_0^0 u^3(1-u^2)~du}\\
&&\\
& = & \displaystyle{0.}
\end{array}</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|       <math>\int_0^1 \frac{x-3}{x^2+6x+5}~dx=\int_0^1 \frac{x-3}{(x+1)(x+5)}~dx.</math>
|-
|Now, we use partial fraction decomposition. Wet set
|-
|       <math>\frac{x-3}{(x+1)(x+5)}=\frac{A}{x+1}+\frac{B}{x+5}.</math>
|-
|If we multiply both sides of this equation by  <math style="vertical-align: -5px">(x+1)(x+5),</math>  we get
|-
|       <math>x-3=A(x+5)+B(x+1).</math>
|-
|If we let  <math style="vertical-align: -4px">x=-1,</math>  we get  <math style="vertical-align: -1px">A=-1.</math>
|-
|If we let  <math style="vertical-align: -4px">x=-5,</math>  we get  <math style="vertical-align: -1px">B=2.</math>
|-
|So, we have
|-
|       <math>\frac{x-3}{(x+1)(x+5)}=\frac{-1}{x+1}+\frac{2}{x+5}.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\int_0^1 \frac{x-3}{(x+1)(x+5)}~dx} & = & \displaystyle{\int_0^1 \frac{-1}{x+1}+\frac{2}{x+5}~dx}\\
&&\\
& = & \displaystyle{\int_0^1 \frac{-1}{x+1}~dx+\int_0^1 \frac{2}{x+5}~dx.}
\end{array}</math>
|-
|Now, we use  <math style="vertical-align: 0px">u</math>-substitution for both of these integrals.
|-
|Let  <math style="vertical-align: -2px">u=x+1.</math>  Then,  <math style="vertical-align: -1px">du=dx.</math>
|-
|Let  <math style="vertical-align: -3px">t=x+5.</math>  Then,  <math style="vertical-align: -1px">dt=dx.</math>
|-
|Since these are definite integrals, we need to change the bounds of integration.
|-
|We have
|-
|       <math style="vertical-align: -4px">u_1=0+1=1</math>  and  <math style="vertical-align: -4px">u_2=1+1=2.</math>
|-
|Also,
|-
|       <math style="vertical-align: -4px">t_1=0+5=5</math>  and  <math style="vertical-align: -4px">t_2=1+5=6.</math>
|-
|Therefore, we get
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int_0^1 \frac{x-3}{(x+1)(x+5)}~dx} & = & \displaystyle{\int_1^2 \frac{-1}{u}~du+\int_5^6 \frac{2}{t}~dt}\\
&&\\
& = & \displaystyle{-\ln|u|\bigg|_1^2+2\ln|t|\bigg|_5^6}\\
&&\\
& = & \displaystyle{-\ln(2)+2\ln(6)-2\ln(5).}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>\frac{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C</math>
|-
|   '''(b)'''    <math>0</math>
|-
|   '''(c)'''    <math>-\ln(2)+2\ln(6)-2\ln(5)</math>
|}
[[009B_Sample_Final_2|'''Return to Sample Exam''']]

009B Sample Final 2, Problem 6 - Revision history

MathAdmin: Created page with " Evaluate the following integrals: (a) \int \frac{dx}{x^2\sqrt{x^2-16}} (b) \int_{-\p..."

MathAdmin: Created page with " Evaluate the following integrals: (a) $\int \frac{dx}{x^2\sqrt{x^2-16}}$ (b) $\int_{-\p..."$