009B Sample Midterm 1, Problem 4 - Revision history

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MathAdmin: Created page with "Evaluate the integral: :: $\int \sin^3x \cos^2x~dx$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations: |-..."

2016-02-01T22:37:47Z

Created page with "<span class="exam">Evaluate the integral: ::<math>\int \sin^3x \cos^2x~dx</math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations: |-..."

New page

<span class="exam">Evaluate the integral:

::<math>\int \sin^3x \cos^2x~dx</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
| Review <math style="vertical-align: 0px">u</math>-substitution, and
|-
|Trig identities.
|}

'''Solution:'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx</math>.
|-
|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>, we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x</math>. If we use this identity, we have
|-
|     <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) (1-\cos^2x)\cos^2x~dx=\int (\cos^2x-\cos^4x)\sin(x)~dx</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: -5px">u=\cos(x)</math>. Then, <math style="vertical-align: -5px">du=-\sin(x)dx</math>. Therefore,
|-
|    <math style="vertical-align: -14px">\int\sin^3x\cos^2x~dx=\int -(u^2-u^4)~du=\frac{-u^3}{3}+\frac{u^5}{5}+C=\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    <math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>
|}
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' Recall the trig identity: <math style="vertical-align: -2px">\sin^2x+\cos^2x=1.</math>
+|'''1.''' Recall the trig identity
 |-
-|'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>
 |-
+|
-::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=\sin x.</math> Then, <math style="vertical-align: -1px">du=\cos x~dx.</math> Thus,
+&nbsp; &nbsp; &nbsp; &nbsp; You can use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int \sin^2x\cos x~dx} & = & \displaystyle{\int u^2~du}\\
 &&\\
 & = & \displaystyle{\frac{u^3}{3}+C}\\
 &&\\
-& = & \displaystyle{\frac{\sin^3x}{3}+C.} \\
+& = & \displaystyle{\frac{\sin^3x}{3}+C.}
 \end{array}</math>
 |}
 '''Solution:'''
@@ Line 30: / Line 39: @@
 |First, we write
 |-
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math>
-::<math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math>
+|-
 |-
-|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1,</math> we get
+|we get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>
 |-
 |If we use this identity, we have
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int (\sin x) (1-\cos^2x)\cos^2x~dx}\\
 &&\\
-& = & \displaystyle{\int (\cos^2x-\cos^4x)\sin(x)~dx.}\\
+& = & \displaystyle{\int (\cos^2x-\cos^4x)\sin(x)~dx.}
 \end{array}</math>
 |}
@@ Line 51: / Line 60: @@
 !Step 2: &nbsp;
 |-
-|Now, we use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math> Therefore,
+|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
+|Therefore,
-::<math>\begin{array}{rcl}
+|-
 \displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int -(u^2-u^4)~du}\\
 &&\\
 & = & \displaystyle{\frac{-u^3}{3}+\frac{u^5}{5}+C}\\
 &&\\
-& = & \displaystyle{\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C.}\\
+& = & \displaystyle{\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C.}
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; <math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>
+| &nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>
 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|Recall the trig identity: <math style="vertical-align: -2px">\sin^2x+\cos^2x=1.</math>
+|'''1.''' Recall the trig identity: <math style="vertical-align: -2px">\sin^2x+\cos^2x=1.</math>
 |-
-|How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
+|'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
 |-
+|
-::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=\sin x.</math> Then, <math style="vertical-align: -1px">du=\cos x~dx.</math>
+::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=\sin x.</math> Then, <math style="vertical-align: -1px">du=\cos x~dx.</math> Thus,
 |-
+|
-::Thus, <math style="vertical-align: -14px">\int \sin^2x\cos x~dx\,=\,\int u^2~du\,=\,\frac{u^3}{3}+C\,=\,\frac{\sin^3x}{3}+C.</math>
+::<math>\begin{array}{rcl}
 |}
@@ Line 22: / Line 28: @@
 !Step 1: &nbsp;
 |-
-|First, we write <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx</math>.
+|First, we write
 |-
-|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>, we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x</math>. If we use this identity, we have
 |-
-| &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) (1-\cos^2x)\cos^2x~dx=\int (\cos^2x-\cos^4x)\sin(x)~dx</math>.
+|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1,</math> we get
 |-
+|
 |}
@@ Line 34: / Line 51: @@
 !Step 2: &nbsp;
 |-
-|Now, we use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\cos(x)</math>. Then, <math style="vertical-align: -5px">du=-\sin(x)dx</math>. Therefore,
+|Now, we use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math> Therefore,
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -14px">\int\sin^3x\cos^2x~dx=\int -(u^2-u^4)~du=\frac{-u^3}{3}+\frac{u^5}{5}+C=\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>.
 |}

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-| Review <math style="vertical-align: 0px">u</math>-substitution, and
+|Recall the trig identity: <math style="vertical-align: -2px">\sin^2x+\cos^2x=1.</math>
 |-
-|Trig identities.
+|How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
 |}

@@ Line 1: / Line 1: @@
-<span class="exam">Evaluate the integral:
+<span class="exam"> Evaluate the indefinite and definite integrals.
-::<math>\int \sin^3x \cos^2x~dx</math>
+<span class="exam">(a) &nbsp; <math>\int x^2 e^x~dx</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009B Sample Midterm 1, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]