Question

Integrate the function $\sin (ax+b)\cos (ax+b)$

Answer 1

$\sin (ax+b)\cos (ax+b)=\frac{2\sin (ax+b)\cos (ax+b)}{2}=\frac{\sin 2(ax+b)}{2}$
Put $2(ax+b)=t$
$\Rightarrow 2adx=dt$
$=>\int \frac{\sin 2(ax+b)}{2}dx=\frac{1}{2}\int \frac{\sin tdt}{2a}$
$=\frac{1}{4a}[-\cos t]+C$
$=\frac{-1}{4a}\cos 2(ax+b)+C$