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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

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up vote 7 down vote accepted

Yes, this is correct, as long as $m\neq 0$.

You could have checked your answer though by simply differentiating w.r.t. $x$.

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thanks a lot @fretty – dato datuashvili Feb 25 '12 at 13:12
@fretty: Very good point about checking. – André Nicolas Feb 25 '12 at 16:39

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