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

Revision as of 15:13, 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 $\int \sin ^{3}x\cos ^{2}x~dx=\int (\sin x)\sin ^{2}x\cos ^{2}x~dx$ .
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
$\int \sin ^{3}x\cos ^{2}x~dx=\int (\sin x)(1-\cos ^{2}x)\cos ^{2}x~dx=\int (\cos ^{2}x-\cos ^{4}x)\sin(x)~dx$ .

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

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

@@ Line 16: / Line 16: @@
 !Step 1: &nbsp;
 |-
-|First, we write <math>\int\sin^3x\cos^2xdx=\int (\sin x) \sin^2x\cos^2xdx</math>.
+|First, we write <math>\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx</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
 |-
-|<math>\int\sin^3x\cos^2xdx=\int (\sin x) (1-\cos^2x)\cos^2xdx=\int (\cos^2x-\cos^4x)\sin(x)dx</math>.
+|<math>\int\sin^3x\cos^2x~dx=\int (\sin x) (1-\cos^2x)\cos^2x~dx=\int (\cos^2x-\cos^4x)\sin(x)~dx</math>.
 |-
 |
@@ Line 30: / Line 30: @@
 |Now, we use u substitution. Let <math>u=\cos(x)</math>. Then, <math>du=-\sin(x)dx</math>. Therefore,
 |-
-|<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>
+|<math>\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>
 |}