Question
Evaluate: $\int \sin ^3 x \cos ^3 x d x$

Answer

$\text { (c) : Let } I=\int \sin ^3 x \cos ^3 x d x$
$\Rightarrow I=\frac{1}{8} \int(2 \sin x \cos x)^3 d x$
$\Rightarrow I=\frac{1}{8} \int \sin ^3 2 x d x$
$\Rightarrow I=\frac{1}{8} \int \frac{3 \sin 2 x-\sin 6 x}{4} d x$
$\Rightarrow I=\frac{1}{32}\left\{-\frac{3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$

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