Question

Evaluate: $\int {\sin }^{3}x\cdot {\cos }^{3}xdx$

Answer 1

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

$\frac{1}{8}\int {\left(2\sin x\cos x\right)}^{3}dx$

$\frac{1}{8}\int {\left(\sin 2x\right)}^{3}dx$

$\frac{1}{8}\int \left({\sin }^{3}2x\right)dx$

$\Rightarrow \sin 3.2x=3\sin 2x-4{\sin }^{3}2x$

$\Rightarrow \sin 6x-3\sin 2x=-4{\sin }^{3}2x$

$4{\sin }^{3}2x=3\sin 2x-\sin 6x$

${\sin }^{3}2x=\frac{3}{4}\sin 2x-\frac{1}{4}\sin 6x$

$=\frac{1}{8}\int (\frac{3}{4}\sin 2x-\frac{1}{4}\sin 6x)dx$

$=\frac{3}{24}\int \sin 2xdx-\frac{1}{32}\int \sin 6xdx$

$=\frac{3}{24}\left(\frac{-\cos 2x}{2}\right)-\frac{1}{32}\left(\frac{-\cos 6x}{6}\right)+c$

$=\frac{-3\cos 2x}{48}+\frac{\cos 6x}{192}+c.$

Hence, the answer is $\frac{-3\cos 2x}{48}+\frac{\cos 6x}{192}+c.$