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$\int \sin^{3} x \cos^{2} x d x ?$

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

$\frac{\cos^{5} x}{5} + \frac{\cos^{3} x}{3} + C$

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

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

answer is A.

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Detailed Solution

$I = \int \sin x (1 - \cos^{2} x) \cos^{2} x d x$

put $\cos x = t \Rightarrow - \sin x d x = d t$

$\Rightarrow I = - \int (t^{2} - t^{4}) d t = \frac{t^{5}}{5} - \frac{t^{3}}{3} + c = \frac{\cos^{5} x}{5} - \frac{\cos^{3} x}{3} + C$ .

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