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

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(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

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$\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + c$

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 $\cos x = t$. Then $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$. Substituting these into the integral,we get: $I = \int (1 - t^2) t^2 (-dt) = \int (t^4 - t^2) \, dt$. Integrating with respect to $t$: $I = \frac{t^5}{5} - \frac{t^3}{3} + c$. Substituting back $t = \cos x$: $I = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + c$.

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

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