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$$\int_{0}^{\pi/2}\frac{\sin t}{1+\sqrt{\sin 2t}}\,dt $$ i used properties of definite integrals to reduce it upto this $$I=\frac{1}{2}\int_{0}^{\pi/2}\frac{\sin t+\cos t}{1+\sqrt{\sin 2t}}\,dt$$ I'm not sure how to approach from here , it was asked in AMM ,problem 11961

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  • $\begingroup$ This one was also posted here some days ago. $\endgroup$ – mickep Jun 19 '17 at 10:49
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$$\begin{align}2I&=\int_{0}^{\pi/2}\frac{\sin t+\cos t}{1+\sqrt{\sin 2t}}\,dt\\&=\int_{0}^{\pi/2}\frac{\sin t+\cos t}{1+\sqrt{1-(\sin t-\cos t)^2}}\,dt\\& =\underbrace{ \int_{-\pi/2}^{\pi/2}\frac{\cos u}{1+\cos u}\,du}_{\sin t-\cos t=\sin u }\\&=2\int_0^{\pi/2}\left(1-\dfrac{1}{1+\cos u}\right)\,du\end{align}$$ Solving this you'll get $$I=\frac{\pi}{2}-1$$

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