Question

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

Answer 1

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

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

$=\frac{1}{8}{\int }^{}{\sin }^{3}2xdx$

putting $2x=t$ implies $dx=\frac{dt}{2}$

$=\frac{1}{16}{\int }^{}{\sin }^{3}tdt$

$=\frac{1}{16}{\int }^{}\left(\frac{3\sin t-\sin 3t}{4}\right)dt$

$=\frac{1}{64}\left[-3\cos t+\frac{\cos 3t}{3}\right]+c$

$=\frac{1}{64}\left[-3\cos 2x+\frac{\cos 6x}{3}\right]+c$