Why can't I make the substitution $ u = \sin (ax + b) $ to evaluate $ \int \sin (ax + b) \cos (ax + b) dx$?

Question

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: $$ - \frac { \cos 2 (ax + b) } {4a } $$

Answer 1

Note that $$1-\cos2(ax+b)=2\sin^2(ax+b)$$

So the textbook gives almost the same as your outcome and the difference is constant.

That is allowed because to be found is an antiderivative (not an integral).

Answer 2

The other method probably is that $\int \sin (ax+b) \cos (ax+b) \ dx=\frac{1}{2}\int \sin (2ax+2b) \ dx= \frac{1}{2} \cdot \frac{1}{2a}(-\cos(2ax+2b))+c=\frac{1}{4a}(2 \sin^2(ax+b)-1)+c)= \frac{1}{2a} \sin^2(ax+b)+c$

using $\cos(2x)=1-2\sin^2(x)$ and putting $-\frac{1}{4a}$ inside the constant

Answer 3

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