Question
Integrate the function: $\sin (a x+b) \cos (a x+b)$

Answer

Let I = $\int \sin (a x+b) \cos (a x+b) d x$ 
We know that,
sin 2A = 2sinA.cosA
Therefore, sin (ax + b) cos (ax + b) = $\frac{2 \sin (a x+b) \cos (a x+b)}{2}=\frac{\sin 2(a x+b)}{2}$ 
Let 2(ax + b) = t
$\Rightarrow$ 2adx = dt
$\Rightarrow$ I = $\int \frac{\sin 2(a x+b)}{2} d x=\frac{1}{2} \int \frac{\sin t}{2 a} d t$ 
= $\frac{1}{4 a}[-\cos t]+C$
=$\frac{-1}{4 a} \cos 2(a x+b)+C$

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