int sin ^3 x cos ^2 x , dx = | vedclass

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Question

(A) Let $I = \int \sin^3 x \cos^2 x \, dx$. We can write $\sin^3 x$ as $\sin^2 x \cdot \sin x = (1 - \cos^2 x) \sin x$. So,$I = \int (1 - \cos^2 x) \c

Vedclass · Accepted Answer

$\frac{\sin ^4 x \cos x}{5} - \frac{\sin ^2 x \cos x}{15} - \frac{2 \cos x}{15} + c$

Vedclass · Answer

(A) Let $I = \int \sin^3 x \cos^2 x \, dx$. We can write $\sin^3 x$ as $\sin^2 x \cdot \sin x = (1 - \cos^2 x) \sin x$. So,$I = \int (1 - \cos^2 x) \cos^2 x \sin x \, dx$. Let $u = \cos x$,then $du = -\sin x \, dx$,or $\sin x \, dx = -du$. Substituting these into the integral: $I = \int (1 - u^2) u^2 (-du) = \int (u^4 - u^2) \, du$. Integrating with respect to $u$: $I = \frac{u^5}{5} - \frac{u^3}{3} + c$. Substituting $u = \cos x$ back: $I = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + c$. Since the provided options are in terms of $\sin x$ and $\cos x$,we check the derivative of option $A$: $\frac{d}{dx} [\frac{\sin^4 x \cos x}{5} - \frac{\sin^2 x \cos x}{15} - \frac{2 \cos x}{15}] = \sin^3 x \cos^2 x$. Thus,option $A$ is the correct integral.

$\int \sin ^3 x \cos ^2 x \, dx =$

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