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

Evaluate the integral:

$\int \sin ^{3}x\cos ^{2}x~dx$ Foundations:
1. Recall the trig identity: $\sin ^{2}x+\cos ^{2}x=1.$ 2. How would you integrate $\int \sin ^{2}x\cos x~dx?$ You could use $u$ -substitution. Let $u=\sin x.$ Then, $du=\cos x~dx.$ Thus,
${\begin{array}{rcl}\displaystyle {\int \sin ^{2}x\cos x~dx}&=&\displaystyle {\int u^{2}~du}\\&&\\&=&\displaystyle {{\frac {u^{3}}{3}}+C}\\&&\\&=&\displaystyle {{\frac {\sin ^{3}x}{3}}+C.}\\\end{array}}$ 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
${\begin{array}{rcl}\displaystyle {\int \sin ^{3}x\cos ^{2}x~dx}&=&\displaystyle {\int (\sin x)(1-\cos ^{2}x)\cos ^{2}x~dx}\\&&\\&=&\displaystyle {\int (\cos ^{2}x-\cos ^{4}x)\sin(x)~dx.}\\\end{array}}$ Step 2:
Now, we use $u$ -substitution. Let $u=\cos(x).$ Then, $du=-\sin(x)dx.$ Therefore,
${\begin{array}{rcl}\displaystyle {\int \sin ^{3}x\cos ^{2}x~dx}&=&\displaystyle {\int -(u^{2}-u^{4})~du}\\&&\\&=&\displaystyle {{\frac {-u^{3}}{3}}+{\frac {u^{5}}{5}}+C}\\&&\\&=&\displaystyle {{\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C.}\\\end{array}}$ ${\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C$ 