009B Sample Midterm 1, Problem 4

Evaluate the integral:

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

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

Solution:

Step 1:
First, we write $\int \sin ^{3}x\cos ^{2}x~dx=\int (\sin x)\sin ^{2}x\cos ^{2}x~dx$.
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:
${\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C$

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