009B Sample Midterm 1, Problem 4

Evaluate the integral:

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

Foundations:
Review $u$-substitution, and
Trig identities.

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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\sin^3x\cos^2x~dx=\int -(u^2-u^4)~du=\frac{-u^3}{3}+\frac{u^5}{5}+C=\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C} .

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C}

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