# definite integration of inverse function

We have to evaluate the following integration $$\int_0^{\pi/2}\sin 2x\arctan(\sin x)dx.$$

In this question I thought of using integration by parts .

But stuck in that.

• It shouts substitute sin(x) = u to me... – Paul Jan 5 '17 at 9:29

Substituting $u=\sin x$ gives us $$I =\int_{0}^{1} 2u\arctan u du$$ Integrating by parts, keeping $f=\arctan u$ and $g'=u$, we get $$I =(u^2\arctan u-u + \arctan u)|_{0}^{1}$$ giving us the answer as $$\boxed {\frac {\pi}{2}-1}$$ Hope it helps.