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.

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    $\begingroup$ It shouts substitute sin(x) = u to me... $\endgroup$ – 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.

  • $\begingroup$ Please consider a hint rather than a solution. $\endgroup$ – Paul Jan 5 '17 at 9:55

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