Question

integrate the function [sin(ax + b) * cos(ax + b)].dx

Answer 1

$I=\int{sin(ax+b).cos(ax+b)}\,dx\\=\int{\frac{2sin(ax+b).cos(ax+b)}{2}}\,dx\\=\int{\frac{sin2(ax+b)}{2}}\,dx\\=\frac{1}{2}\int{sin2(ax+b)}\,dx$

Let 2(ax+ b) = t
differentiate both sides,
2adx = dt
dx = 1/2a dt
now, I = $\frac{1}{4a}\int{sint}\,dt$

= $\frac{1}{4a}[-cost]+C$

put t = 2(ax+ b)

I = $-\frac{1}{4a}cos2(ax+b)+C$

Answer 2

Hope it will help you.