009B Sample Final 3, Problem 5 - Revision history

Kayla Murray at 00:45, 3 December 2017

2017-12-03T00:45:42Z

2017-03-12T21:48:21Z

2017-03-01T00:11:23Z

2017-03-01T00:07:08Z

2017-02-28T23:51:09Z

2017-02-28T23:39:11Z

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2017-02-19T02:17:54Z

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp;<math>\int \sin^3(x)\cos^2(x)~dx</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 3, Problem 5 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Final 3, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 69: / Line 69: @@
 |Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|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; and &nbsp;<math style="vertical-align: -5px">-du=\sin(x)dx.</math>
+|Let &nbsp;<math style="vertical-align: -5px">u=\cos(x).</math>&nbsp;
 |-
 |Therefore, we have

@@ Line 55: / Line 55: @@
 !Step 1: &nbsp;
 |-
-|First, we use the identity <math>\sin^2 x=1-\cos^2 x</math> to get
+|First, we use the identity &nbsp;<math style="vertical-align: -1px">\sin^2 x=1-\cos^2 x</math>&nbsp; to get
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
@@ Line 67: / Line 67: @@
 !Step 2: &nbsp;
 |-
-|Now, we use <math>u</math>-substitution.
+|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let <math>u=\cos(x)</math>. Then, <math>du=-\sin(x)dx</math> and <math>-du=\sin(x)dx.</math>
+|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; and &nbsp;<math style="vertical-align: -5px">-du=\sin(x)dx.</math>
 |-
 |Therefore, we have

@@ Line 55: / Line 55: @@
 !Step 1: &nbsp;
 |-
+|First, we use the identity <math>\sin^2 x=1-\cos^2 x</math> to get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 63: / Line 67: @@
 !Step 2: &nbsp;
 |-
+|Now, we use <math>u</math>-substitution.
 |-
+|Let <math>u=\cos(x)</math>. Then, <math>du=-\sin(x)dx</math> and <math>-du=\sin(x)dx.</math>
 |}
@@ Line 74: / Line 90: @@
 |&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>x\sin x +\cos x+C</math>
 |-
-|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;
+|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;<math>\frac{\cos^5 x}{5}-\frac{\cos^3 x}{3}+C</math>
 |}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 29: / Line 29: @@
 |To calculate this integral, we use integration by parts.
 |-
-|Let &nbsp;<math>u=x</math>&nbsp; and &nbsp;<math>dv=\cos xdx.</math>
+|Let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=\cos xdx.</math>
 |-
-|Then, &nbsp;<math>du=dx</math>&nbsp; and &nbsp;<math>v=\sin x.</math>
+|Then, &nbsp;<math style="vertical-align: 0px">du=dx</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">v=\sin x.</math>
 |-
 |Therefore, we have
@@ Line 51: / Line 51: @@
 '''(b)'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 92: / Line 74: @@
 |&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>x\sin x +\cos x+C</math>
 |-
-|'''(b)'''
+|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;
 |}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 35: / Line 35: @@
 |Therefore, we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int x\cos(x)~dx=x\sin x -\int \sin x~dx.</math>
 |}
@@ Line 41: / Line 41: @@
 !Step 2: &nbsp;
 |-
+|Then, we integrate to get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math> \int x\cos(x)~dx=x\sin x +\cos x+C.</math>
 |-
+|
@@ Line 90: / Line 90: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>x\sin x +\cos x+C</math>
 |-
 |'''(b)'''

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
+|'''1.''' Integration by parts tells us that
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
 |-
+|'''2.''' Since &nbsp;<math style="vertical-align: -4px">\sin^2x+\cos^2x=1,</math>&nbsp; we have
 |-
+|

@@ Line 1: / Line 1: @@
 <span class="exam"> Find the following integrals.
-<span class="exam">(a) <math>\int x\cos(x)~dx</math>
+<span class="exam">(a) &nbsp;<math>\int x\cos(x)~dx</math>
-<span class="exam">(b) <math>\int \sin^3(x)\cos^2(x)~dx</math>
+<span class="exam">(b) &nbsp;<math>\int \sin^3(x)\cos^2(x)~dx</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"