Question

Evaluate:

$\int {\sin }^{2}x{\cos }^{2}xdx$

Answer 1

$I=\int {\sin }^{2}x{\cos }^{2}xdx$

$I=\frac{1}{4}\int (2\sin x\cos x{)}^{2}dx$

$I=\frac{1}{4}\int (\sin 2x{)}^{2}dx$

$I=\frac{1}{4}\int {\sin }^{2}2xdx$

we know that

$\cos 4x=\cos 2\cdot 2x=1-2{\sin }^{2}x$

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

Therefore,

$I=\int \frac{1}{4}\left(\frac{1-\cos 4x}{2}\right)dx$

$I=\frac{1}{8}\left[\int (1-\cos 4s)dx\right]$

$I=\frac{1}{8}[x-\frac{\sin 4x}{4}]+C$