Difference between revisions of "009B Sample Midterm 1, Problem 4"

Revision as of 11:52, 31 January 2016

Evaluate the integral:

\int \sin ^{3}x\cos ^{2}x~dx

Foundations:
Review u substitution
Trig identities

Solution:

Step 1:
First, we write $\int \sin ^{3}x\cos ^{2}xdx=\int (\sin x)\sin ^{2}x\cos ^{2}xdx$ .
Using the identity $\sin ^{2}x+\cos ^{2}x=1$ , we get $\sin ^{2}x=1-\cos ^{2}x$ . If we use this identity, we have
$\int \sin ^{3}x\cos ^{2}xdx=\int (\sin x)(1-\cos ^{2}x)\cos ^{2}xdx=\int (\cos ^{2}x-\cos ^{4}x)\sin(x)dx$ .

Step 2:
Now, we use u substitution. Let $u=\cos(x)$ . Then, $du=-\sin(x)dx$ . Therefore,
$\int \sin ^{3}x\cos ^{2}xdx=\int -(u^{2}-u^{4})du={\frac {-u^{3}}{3}}+{\frac {u^{5}}{5}}+C={\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C$

Final Answer:
${\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C$

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|1)
+| Review u substitution
 |-
-|2)
+| Trig identities
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-|Answers:
-|-
-|1)
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-|2)
 |}
@@ Line 23: / Line 16: @@
 !Step 1: &nbsp;
 |-
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+|First, we write <math>\int\sin^3x\cos^2xdx=\int (\sin x) \sin^2x\cos^2xdx</math>.
 |-
-|
+|Using the identity <math>\sin^2x+\cos^2x=1</math>, we get <math>\sin^2x=1-\cos^2x</math>. If we use this identity, we have
 |-
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+|<math>\int\sin^3x\cos^2xdx=\int (\sin x) (1-\cos^2x)\cos^2xdx=\int (\cos^2x-\cos^4x)\sin(x)dx</math>.
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 |
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 !Step 2: &nbsp;
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+|Now, we use u substitution. Let <math>u=\cos(x)</math>. Then, <math>du=-\sin(x)dx</math>. Therefore,
-|-
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-|-
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 |-
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+|<math>\int\sin^3x\cos^2xdx=\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>
 |}
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 !Final Answer: &nbsp;
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+|<math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>
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 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]