009B Sample Midterm 3, Problem 5 - Revision history

Kayla Murray: Replaced content with "Evaluate the indefinite and definite integrals. (a) $\int x\ln x ~dx$ (b) $\int_0..."$

2017-11-24T00:33:10Z

Replaced content with "<span class="exam">Evaluate the indefinite and definite integrals. <span class="exam">(a) <math>\int x\ln x ~dx</math> <span class="exam">(b) <math>\int_0..."

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Kayla Murray at 00:31, 24 November 2017

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Kayla Murray at 16:55, 6 February 2017

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Kayla Murray at 16:37, 6 February 2017

2017-02-06T16:37:29Z

@@ Line 1: / Line 1: @@
 <span class="exam">Evaluate the indefinite and definite integrals.
-<span class="exam">(a) &nbsp; <math>\int \tan^3x ~dx</math>
+<span class="exam">(a) &nbsp; <math>\int x\ln x ~dx</math>
 <span class="exam">(b) &nbsp; <math>\int_0^\pi \sin^2x~dx</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' Recall the trig identity &nbsp;<math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
+|'''1.''' Recall the trig identity
 |-
-|'''2.''' Recall the trig identity &nbsp;<math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
 |-
 |'''3.''' How would you integrate &nbsp;<math style="vertical-align: -1px">\tan x~dx?</math>
@@ Line 109: / Line 113: @@
 !Step 1: &nbsp;
 |-
-|One of the double angle formulas is &nbsp;<math style="vertical-align: -5px">\cos(2x)=1-2\sin^2(x).</math>
+|One of the double angle formulas is
 |-
 |Solving for &nbsp;<math style="vertical-align: -5px">\sin^2(x),</math>&nbsp; we get

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' Recall the trig identity
+|'''1.''' Recall the trig identity &nbsp;<math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
+|'''2.''' Recall the trig identity &nbsp;<math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}</math>
 |-
-|'''2.''' Recall the trig identity
+|'''3.''' How would you integrate &nbsp;<math style="vertical-align: -1px">\tan x~dx?</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
+&nbsp; &nbsp; &nbsp; &nbsp; You could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; First, write <math style="vertical-align: -14px">\int \tan x~dx=\int \frac{\sin x}{\cos x}~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; First, write &nbsp;<math style="vertical-align: -14px">\int \tan x~dx=\int \frac{\sin x}{\cos x}~dx.</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Now, let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Now, let &nbsp;<math style="vertical-align: -5px">u=\cos(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">du=-\sin(x)dx.</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; Thus,
@@ Line 51: / Line 47: @@
 &nbsp; &nbsp; &nbsp; &nbsp; <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
+|Since &nbsp;<math style="vertical-align: -5px">\tan^2x=\sec^2x-1,</math>&nbsp; we have
 |-
+|
@@ Line 64: / Line 60: @@
 !Step 2: &nbsp;
 |-
-|Now, we need to use <math>u</math>-substitution for the first integral.
+|Now, we need to use &nbsp;<math>u</math>-substitution for the first integral.
 |-
+|
-Let <math style="vertical-align: -5px">u=\tan(x).</math>
+Let &nbsp;<math style="vertical-align: -5px">u=\tan(x).</math>
 |-
-|Then, <math style="vertical-align: -1px">du=\sec^2x~dx.</math>
+|Then, &nbsp;<math style="vertical-align: -1px">du=\sec^2x~dx.</math>
 |-
 |So, we have
@@ Line 86: / Line 82: @@
 !Step 3: &nbsp;
 |-
-|For the remaining integral, we also need to use <math>u</math>-substitution.
+|For the remaining integral, we also need to use &nbsp;<math>u</math>-substitution.
 |-
 |First, we write
@@ Line 93: / Line 89: @@
 &nbsp; &nbsp; &nbsp; &nbsp; <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>
+|Now, we let &nbsp;<math style="vertical-align: 0px">u=\cos x.</math>
 |-
-|Then, <math style="vertical-align: -1px">du=-\sin x~dx.</math>
+|Then, &nbsp;<math style="vertical-align: -1px">du=-\sin x~dx.</math>
 |-
 |Therefore, we get
@@ Line 113: / Line 109: @@
 !Step 1: &nbsp;
 |-
-|One of the double angle formulas is <math style="vertical-align: -5px">\cos(2x)=1-2\sin^2(x).</math>
+|One of the double angle formulas is &nbsp;<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
+|Solving for &nbsp;<math style="vertical-align: -5px">\sin^2(x),</math>&nbsp; we get
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}.</math>
@@ Line 145: / Line 141: @@
 !Step 3: &nbsp;
 |-
-|For the remaining integral, we need to use <math style="vertical-align: 0px">u</math>-substitution.
+|For the remaining integral, we need to use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let <math style="vertical-align: -1px">u=2x.</math>
+|Let &nbsp;<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>
+|Then, &nbsp;<math style="vertical-align: -1px">du=2~dx</math>&nbsp; and &nbsp;<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,
@@ Line 155: / Line 151: @@
 |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>
+|We have &nbsp;<math style="vertical-align: -5px">u_1=2(0)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=2(\pi)=2\pi.</math>
 |-
 |So, the integral becomes

@@ Line 1: / Line 1: @@
 <span class="exam">Evaluate the indefinite and definite integrals.
-::<span class="exam">a) &nbsp; <math>\int \tan^3x ~dx</math>
+<span class="exam">(a) &nbsp; <math>\int \tan^3x ~dx</math>
-::<span class="exam">b) &nbsp; <math>\int_0^\pi \sin^2x~dx</math>

@@ Line 10: / Line 10: @@
 |'''1.''' Recall the trig identity
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
 |-
 |'''2.''' Recall the trig identity
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}</math>
 |-
 |'''3.''' How would you integrate <math style="vertical-align: -1px">\tan x~dx?</math>
 |-
+|
-&nbsp; &nbsp; 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>
+&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
 |-
+|
-&nbsp; &nbsp; Now, let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math> Thus,
+&nbsp; &nbsp; &nbsp; &nbsp; Now, let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math>
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int \tan x~dx} & = & \displaystyle{\int \frac{-1}{u}~du}\\
 &&\\
@@ Line 44: / Line 48: @@
 |-
+|
-&nbsp; &nbsp;<math style="vertical-align: -14px">\int \tan^3x~dx=\int \tan^2x\tan x ~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <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
 |-
+|
-&nbsp; &nbsp;<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int (\sec^2x-1)\tan x ~dx}\\
 &&\\
@@ Line 62: / Line 66: @@
 |-
+|
-Let <math style="vertical-align: -5px">u=\tan(x).</math> Then, <math style="vertical-align: -1px">du=\sec^2x~dx.</math> So, we have
+Let <math style="vertical-align: -5px">u=\tan(x).</math>
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int u~du-\int \tan x~dx}\\
 &&\\
@@ Line 82: / Line 90: @@
 |-
+|
-&nbsp; &nbsp;<math style="vertical-align: -14px">\int \tan^3x~dx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <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>
+|Now, we let <math style="vertical-align: 0px">u=\cos x.</math>
 |-
 |Therefore, we get
 |-
+|
-&nbsp; &nbsp;<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\int \tan^3x~dx} & = & \displaystyle{\frac{\tan^2x}{2}+\int \frac{1}{u}~dx}\\
 &&\\
@@ Line 106: / Line 116: @@
 |Solving for <math style="vertical-align: -5px">\sin^2(x),</math> we get
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}.</math>
 |-
 |Plugging this identity into our integral, we get
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\int_0^\pi \frac{1-\cos(2x)}{2}~dx}\\
 &&\\
@@ Line 124: / Line 134: @@
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <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}\\
 &&\\
@@ Line 136: / Line 146: @@
 |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>
+|Let <math style="vertical-align: -1px">u=2x.</math>
 |-
 |Also, since this is a definite integral and we are using <math style="vertical-align: 0px">u</math>-substitution,
@@ Line 147: / Line 159: @@
 |-
+|
-&nbsp; &nbsp;<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\frac{\pi}{2}-\int_0^{2\pi} \frac{\cos(u)}{4}~du}\\
 &&\\
@@ Line 162: / Line 174: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp; <math>\frac{\tan^2x}{2}+\ln |\cos x|+C</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{\tan^2x}{2}+\ln |\cos x|+C</math>
 |-
-|'''(b)''' &nbsp; <math>\frac{\pi}{2}</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{\pi}{2}</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 14: / Line 14: @@
 |'''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>
+|'''3.''' How would you integrate <math style="vertical-align: -1px">\tan x~dx?</math>
 |-
+|

@@ Line 31: / Line 31: @@
 \end{array}</math>
 |}
 '''Solution:'''
@@ Line 41: / Line 42: @@
 |-
+|
-::<math style="vertical-align: -14px">\int \tan^3x~dx=\int \tan^2x\tan x ~dx.</math>
+&nbsp; &nbsp;<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}
+&nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int (\sec^2x-1)\tan x ~dx}\\
 &&\\
@@ Line 59: / Line 60: @@
 |-
+|
-::Let <math style="vertical-align: -5px">u=\tan(x).</math> Then, <math style="vertical-align: -1px">du=\sec^2x~dx.</math> So, we have
+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}
+&nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int u~du-\int \tan x~dx}\\
 &&\\
@@ Line 79: / Line 80: @@
 |-
+|
-::<math style="vertical-align: -14px">\int \tan^3x~dx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}~dx.</math>
+&nbsp; &nbsp;<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}
+&nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\int \tan^3x~dx} & = & \displaystyle{\frac{\tan^2x}{2}+\int \frac{1}{u}~dx}\\
 &&\\
@@ Line 104: / Line 105: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\int_0^\pi \frac{1-\cos(2x)}{2}~dx}\\
 &&\\
@@ Line 117: / Line 118: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; <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}\\
 &&\\
@@ Line 138: / Line 139: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\frac{\pi}{2}-\int_0^{2\pi} \frac{\cos(u)}{4}~du}\\
 &&\\
@@ Line 148: / Line 149: @@
 \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>
+|'''(a)''' &nbsp; <math>\frac{\tan^2x}{2}+\ln |\cos x|+C</math>
 |-
-|&nbsp;&nbsp; '''(b)''' <math>\frac{\pi}{2}</math>
+|'''(b)''' &nbsp; <math>\frac{\pi}{2}</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]