Question

Integrate:
$\int (\sin 2x-\cos 2x)dx$

Answer 1

$\int (\sin 2x-\cos 2x)dx$

$=\int \sin 2xdx-\int \cos 2xdx$

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

Let $2x=u\Rightarrow 2dx=du$

$\int (\sin 2x-\cos 2x)dx=\frac{1}{2}(\int \sin udu-\int \cos udu)$

$\Rightarrow =\frac{1}{2}(-\cos u-\sin u)$

$\Rightarrow =-\frac{1}{2}(\cos 2x+\sin 2x)[\because u=2x]$

Hence $\int (\sin 2x-\cos 2x)dx=-\frac{1}{2}(\cos 2x+\sin 2x)$