Why can't I make the substitution $ u = \sin (ax + b) $ to evaluate $ \int \sin (ax + b) \cos (ax + b) dx$? Evaluate $ \int \sin (ax + b) \cos (ax + b) dx$?
To do this, I started of by substituting $ u = \sin (ax + b) $. That made $ du = cos (ax + b) \cdot a $ and wrote the integral as $ \frac 1a \int u \ du $ to get the final answer as: $$ \frac 1{2a} \sin^2 (ax + b) $$
This answer however, is wrong. My textbook uses a different method and arrives at a different answer. I understand how to arrive at the (right) answer but I want to know why I can't get the same answer by substitution here. 
My textbook starts off by rewriting the expression as $ \frac {\sin 2 (ax + b)}{2} $ and then substitutes $ 2 (ax + b) = u $ to get this answer: $$ - \frac { \cos 2 (ax + b) } {4a } $$

You can solve with the substitution
u=sin(ax+b) then, 
du=acos(ax+b)dx so,
∫sin(ax+b)cos(ax+b)dx=∫(u/a)du=(u^2)/(2a)=sin^2(ax+b)/(2a)
which is same with your answer-1/4a
