Show that $\int_{0}^{\pi/2} \frac{\sin x~dx}{1+\sin 2x}=\frac{\coth^{-1} \sqrt{2}}{\sqrt{2}}$ The integral $$\int_{0}^{\pi/2} \frac{\sin x~dx}{1+\sin 2x}=\frac{\coth^{-1} \sqrt{2}}{\sqrt{2}}?$$ can be checked at Mathematica. The question is how to do it by hand?

$$I=\int_{0}^ {\pi/2}\frac{\sin x~dx}{1+\sin 2x} = \frac{1}{2} \int_{0}^{\pi/2} \frac{(\sin x+\cos x)-(\cos x-\sin x)}{(\sin x+ \cos x)^2} dx=\frac{1}{2}\left (\int_{0}^{\pi/2}  \frac{dx}{\sin x+\cos x}-\int_{1}^{1} \frac{dt}{t^2} \right).$$
$$\Rightarrow I=\frac{1}{2\sqrt{2}} \int_{0}^{\pi/2} \mbox{cosec} (x+\pi/4) dx=\frac{1}{2\sqrt{2}}\left . \ln(\tan(x/2+\pi/8)\right |_{0}^{\pi/2}
=\frac{1}{2\sqrt{2}} \ln\left( \frac{\tan (3\pi/8)}{\tan (\pi/8)} \right).$$
Note that $\tan(\pi/8)=\sqrt{2}-1=t$, then
$$ I= \frac{1}{2\sqrt{2}} \ln \left (\frac{3-t^2}{1-3t^2} \right )= \frac{1}{2\sqrt{2}} \ln (3+2\sqrt{2})=\frac{1}{\sqrt{2}} \ln (1+\sqrt{2})=
\frac{\mbox{coth}^{-1}\sqrt{2}}{\sqrt{2}}.$$
