Question

hw part 1: 7.2 trigonometric integrals
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question 1
0/1 pt 3 98 details
evaluate the indefinite integral
\\(\int\sin^{2}x\cos^{3}x dx =\\)

Explanation:

Step1: Rewrite $\cos^{3}x$

We know that $\cos^{3}x=\cos x\cos^{2}x=\cos x(1 - \sin^{2}x)$. So the integral $\int\sin^{2}x\cos^{3}x dx=\int\sin^{2}x\cos x(1 - \sin^{2}x)dx$.

Step2: Use substitution

Let $u = \sin x$, then $du=\cos xdx$. The integral becomes $\int u^{2}(1 - u^{2})du=\int(u^{2}-u^{4})du$.

Step3: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $\int(u^{2}-u^{4})du=\int u^{2}du-\int u^{4}du=\frac{u^{3}}{3}-\frac{u^{5}}{5}+C$.

Step4: Substitute back $u=\sin x$

We get $\frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+C$.

Answer:

$\frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+C$