${\sin ^3}x{\cos ^3}x$

: Let $I = \int {{{\sin }^3}x{{\cos }^3}xdx} = \int {\sin x \cdot {{\sin }^2}x \cdot {{\cos }^3}x} dx$

$= \int {\sin \,x\left( {1 - {{\cos }^2}x} \right){{\cos }^3}x} dx = \int {\left( {{{\cos }^3}x - {{\cos }^5}x} \right)\sin xdx}$

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

$\therefore$ $I = - \int {\left( {{t^3} - {t^5}} \right)dt} = \cfrac{{ - {t^4}}}{4} + \cfrac{{{t^6}}}{6} + C = \cfrac{1}{6}{\cos ^6}x - \cfrac{1}{4}{\cos ^4}x + C$