Question
Find the integral of the function $\sin^3 x \cos^3 x$

Answer

Let $I=\int \sin ^{3} x \cos ^{3} x \cdot d x$ 
$\Rightarrow \int \cos ^{3} x \cdot \sin ^{2} x \cdot \sin x \cdot d x$ 
$\Rightarrow \int \cos ^{3} x\left(1-\cos ^{2} x\right) \sin x \cdot d x$
Let $\cos x = t$
$\Rightarrow - \sin x.dx = dt$
$\Rightarrow \mathrm{I}=-\int \mathrm{t}^{3}\left(1-\mathrm{t}^{2}\right) \mathrm{dt}$ 
$=-\int\left(t^{3}-t^{5}\right) d t$ 
$=-\left\{\frac{t^{4}}{4}-\frac{t^{6}}{6}\right\}+C$ 
$=-\left\{\frac{\cos ^{4} x}{4}-\frac{\cos ^{6} x}{6}\right\}+C$ 
$=\frac{\cos ^{6} x}{6}-\frac{\cos ^{4} x}{4}+C$

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