# 009B Sample Midterm 1, Problem 4

Evaluate the integral:

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

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

Solution:

Step 1:
First, we write
${\displaystyle \int \sin ^{3}x\cos ^{2}x~dx=\int (\sin x)\sin ^{2}x\cos ^{2}x~dx.}$
Using the identity ${\displaystyle \sin ^{2}x+\cos ^{2}x=1,}$ we get
${\displaystyle \sin ^{2}x=1-\cos ^{2}x.}$
If we use this identity, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int \sin ^{3}x\cos ^{2}x~dx}&=&\displaystyle {\int (\sin x)(1-\cos ^{2}x)\cos ^{2}x~dx}\\&&\\&=&\displaystyle {\int (\cos ^{2}x-\cos ^{4}x)\sin(x)~dx.}\\\end{array}}}$
Step 2:
Now, we use ${\displaystyle u}$-substitution. Let ${\displaystyle u=\cos(x).}$ Then, ${\displaystyle du=-\sin(x)dx.}$ Therefore,
${\displaystyle {\begin{array}{rcl}\displaystyle {\int \sin ^{3}x\cos ^{2}x~dx}&=&\displaystyle {\int -(u^{2}-u^{4})~du}\\&&\\&=&\displaystyle {{\frac {-u^{3}}{3}}+{\frac {u^{5}}{5}}+C}\\&&\\&=&\displaystyle {{\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C.}\\\end{array}}}$
${\displaystyle {\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C}$