I want to calculate $ \displaystyle\int\sin^2(mx) \,dx$. My steps are the following. Please tell me if I am wrong in it.
So if we substitute $u=m x$ then $du=m \,dx$, so
$$\frac {1} {m}\int \sin^2 u du$$ then we kow that, $ \sin ^2u=\dfrac{1-\cos(2u)}{2}$ and if we put this into original integral and evaluate it,we get $$\dfrac{x}{2}-\dfrac{\sin(2mx)}{4m}+ C$$
Am I correct? Or there is some mistake? Please give me any hint if necessary
Best Answer
Yes, this is correct, as long as $m\neq 0$.
You could have checked your answer though by simply differentiating w.r.t. $x$.