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

Evaluate the integral:

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

Foundations:
Recall the trig identity: ${\displaystyle \sin ^{2}x+\cos ^{2}x=1.}$
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, ${\displaystyle \int \sin ^{2}x\cos x~dx\,=\,\int u^{2}~du\,=\,{\frac {u^{3}}{3}}+C\,=\,{\frac {\sin ^{3}x}{3}}+C.}$

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
${\displaystyle \int \sin ^{3}x\cos ^{2}x~dx=\int (\sin x)(1-\cos ^{2}x)\cos ^{2}x~dx=\int (\cos ^{2}x-\cos ^{4}x)\sin(x)~dx}$.
Step 2:
Now, we use ${\displaystyle u}$-substitution. Let ${\displaystyle u=\cos(x)}$. Then, ${\displaystyle du=-\sin(x)dx}$. Therefore,
${\displaystyle \int \sin ^{3}x\cos ^{2}x~dx=\int -(u^{2}-u^{4})~du={\frac {-u^{3}}{3}}+{\frac {u^{5}}{5}}+C={\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C}$.
${\displaystyle {\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C}$