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