Question
Find the integral: $\int \sin ^{3} x \cos ^{2} x d x$

Answer

We have
$\int \sin ^{3} x \cos ^{2} x d x=\int \sin ^{2} x \cos ^{2} x(\sin x) d x$ 
= $\int\left(1-\cos ^{2} x\right) \cos ^{2} x(\sin x) d x$ 
Put t = cos x so that dt = -sin x dx
Therefore, $\int \sin ^{2} x \cos ^{2} x(\sin x) d x$ = $-\int\left(1-t^{2}\right) t^{2} d t$ 
= $-\int\left(t^{2}-t^{4}\right) d t=-\left(\frac{t^{3}}{3}-\frac{t^{5}}{5}\right)+\mathrm{C}$ 
= $-\frac{1}{3} \cos ^{3} x+\frac{1}{5} \cos ^{5} x+C$ 

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