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| | <span class="exam">(c) <math>\int_0^1 \frac{x-3}{x^2+6x+5}~dx</math> | | <span class="exam">(c) <math>\int_0^1 \frac{x-3}{x^2+6x+5}~dx</math> |
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| | + | [[009B Sample Final 2, Problem 6 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |'''1.''' For <math>\int \frac{dx}{x^2\sqrt{x^2-16}},</math> what would be the correct trig substitution?
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| − | |-
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| − | | The correct substitution is <math>x=4\sec^2\theta.</math>
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| − | |-
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| − | |'''2.''' We have the Pythagorean identity
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| − | |-
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| − | | <math style="vertical-align: -5px">\cos^2(x)=1-\sin^2(x).</math>
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| − | |-
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| − | |'''3.''' Through partial fraction decomposition, we can write the fraction
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| − | |-
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| − | | <math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math>
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| − | |-
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| − | | for some constants <math style="vertical-align: -4px">A,B.</math>
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| − | |}
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| | + | [[009B Sample Final 2, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''Solution:'''
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| − | '''(a)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We start by using trig substitution.
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| − | |-
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| − | |Let <math>x=4\sec \theta.</math>
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| − | |-
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| − | |Then, <math>dx=4\sec \theta \tan \theta ~d\theta.</math>
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| − | |-
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| − | |So, the integral becomes
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \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}\\
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| − | &&\\
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| − | & = & \displaystyle{\int \frac{4\sec \theta \tan \theta}{16\sec^2\theta (4\tan \theta)} ~d\theta}\\
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| − | &&\\
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| − | & = & \displaystyle{\int \frac{1}{16\sec \theta} ~d\theta.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we integrate to get
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\frac{1}{16}\cos\theta~d\theta}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{16}\sin \theta +C}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(b)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we write
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int_{-\pi}^\pi \sin^3x\cos^3x~dx} & = & \displaystyle{\int_{-\pi}^{\pi} \sin^3x \cos^2x \cos x~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_{-\pi}^{\pi} \sin^3x (1-\sin^2x)\cos x~dx.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we use <math>u</math>-substitution.
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| − | |-
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| − | |Let <math>u=\sin x.</math> Then, <math>du=\cos x ~dx.</math>
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| − | |-
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| − | |Since this is a definite integral, we need to change the bounds of integration.
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| − | |-
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| − | |Then, we have
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| − | |-
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| − | | <math>u_1=\sin(-\pi)=0</math> and <math>u_2=\sin(\pi)=0.</math>
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| − | |-
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| − | |So, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int_{-\pi}^\pi \sin^3x\cos^3x~dx} & = & \displaystyle{\int_0^0 u^3(1-u^2)~du}\\
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| − | &&\\
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| − | & = & \displaystyle{0.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(c)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we write
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| − | |-
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| − | | <math>\int_0^1 \frac{x-3}{x^2+6x+5}~dx=\int_0^1 \frac{x-3}{(x+1)(x+5)}~dx.</math>
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| − | |-
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| − | |Now, we use partial fraction decomposition. Wet set
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| − | |-
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| − | | <math>\frac{x-3}{(x+1)(x+5)}=\frac{A}{x+1}+\frac{B}{x+5}.</math>
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| − | |-
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| − | |If we multiply both sides of this equation by <math>(x+1)(x+5),</math> we get
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| − | |-
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| − | | <math>x-3=A(x+5)+B(x+1).</math>
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| − | |-
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| − | |If we let <math>x=-1,</math> we get <math>A=-1.</math>
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| − | |-
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| − | |If we let <math>x=-5,</math> we get <math>B=2.</math>
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| − | |-
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| − | |So, we have
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| − | |-
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| − | | <math>\frac{x-3}{(x+1)(x+5)}=\frac{-1}{x+1}+\frac{2}{x+5}.</math>
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| − | |-
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| − | |
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we have
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \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}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_0^1 \frac{-1}{x+1}~dx+\int_0^1 \frac{2}{x+5}~dx.}
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| − | \end{array}</math>
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| − | |-
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| − | |Now, we use <math>u</math>-substitution for both of these integrals.
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| − | |-
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| − | |Let <math>u=x+1.</math> Then, <math>du=dx.</math>
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| − | |-
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| − | |Let <math>t=x+5.</math> Then, <math>dt=dx.</math>
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| − | |-
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| − | |Since these are definite integrals, we need to change the bounds of integration.
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| − | |-
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| − | |We have <math>u_1=0+1=1</math> and <math>u_2=1+1=2.</math>
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| − | |-
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| − | |Also, <math>t_1=0+5=5</math> and <math>u_2=1+5=6.</math>
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| − | |-
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| − | |Therefore, we get
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \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}\\
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| − | &&\\
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| − | & = & \displaystyle{-\ln|u|\bigg|_1^2+2\ln|t|\bigg|_5^6}\\
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| − | &&\\
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| − | & = & \displaystyle{-\ln(2)+2\ln(6)-2\ln(5).}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' <math>\frac{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C</math>
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| − | |-
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| − | | '''(b)''' <math>0</math>
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| − | |-
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| − | | '''(c)''' <math>-\ln(2)+2\ln(6)-2\ln(5)</math>
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| − | |}
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| | [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
