Find the integral: $\int \sin^3 x \cos^2 x \, dx$

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

The correct answer is Option (1) → $\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C$

We have

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

$= \int (1 - \cos^2 x) \cos^2 x (\sin x) \, dx$

Put $t = \cos x$ so that $dt = -\sin x \, dx$

Therefore, $\int \sin^2 x \cos^2 x (\sin x) \, dx = -\int (1 - t^2) t^2 \, dt$

$= -\int (t^2 - t^4) \, dt = -\left( \frac{t^3}{3} - \frac{t^5}{5} \right) + C$

$= -\frac{1}{3} \cos^3 x + \frac{1}{5} \cos^5 x + C$