Question

$\int \sin (ax+b)\cos (ax+b)dx$.

Answer 1

$\int \sin (ax+b)\cos (ax+b)dx=\int \frac{1}{2}.2\sin (ax+b)\cos (ax+b)dx$

$=\frac{1}{2}\int \sin 2(ax+b)[\because \sin 2x=2\sin x\cos x]$

$=\frac{1}{2}\int \sin (2ax+2b)dx=-\frac{1}{2}\frac{\cos (2ax+2b)}{2a}+c=\frac{-\cos 2(ax+b)}{4a}+c$