009B Sample Midterm 1, Problem 4

Evaluate the integral:

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

Foundations:
Recall the trig identity: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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
$\int \sin ^{3}x\cos ^{2}x~dx=\int (\sin x)\sin ^{2}x\cos ^{2}x~dx.$
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,
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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.}

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

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