Question

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

Answer 1

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

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

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

$I=\frac{1}{8}\int \frac{3\sin 2x-\sin 6x}{4}dx$ $[\because \sin 3\theta =3\sin \theta -4{\sin }^3\theta ]$

$I=\frac{1}{32}\int (3\sin 2x-\sin 6x)dx$

$I=\frac{1}{32}[-\frac{3}{2}\cos 2x+\frac{1}{6}\cos 6x]+C$