I1=∫dxcos2xsin2x
$=\int \dfrac{d(\tan x)}{\sin^2 x} = \int (1+\cot^2 x)d(\tan x) =\int d(\tan x) +\int \dfrac{1}{\tan^2 x}d(\tan x)$
$=\tan x -\dfrac{1}{\tan x} +C$
I2=∫cos3xsinxdx
$=\dfrac{1}{2}\int (\sin 4x - \sin 2x)dx =\dfrac{1}{2}\int\sin 4x dx -\dfrac{1}{2}\int\sin 2x dx= \dfrac{1}{4}\cos 2x -\dfrac{1}{8}\cos 4x +C$