Question

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

Answer 1

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

$u=\sin (ax+b)\Rightarrow dx=\frac{1}{acos(ax+b)}du$

$\int \sin (ax+b)\cos (ax+b)$=$\frac{1}{a}\int udu$

$=\frac{u^2}{2a}$

$=\frac{{\sin }^2(ax+b)}{2a}$

$\int \sin (ax+b)\cos (ax+b)dx=\frac{{\sin }^2(ax+b)}{2a}+C$