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?

Using the substitution $x \mapsto \frac{\pi}2 - x$ we get
$$I = \int_0^{\frac{\pi}2} \frac{\sin x\,dx}{1+\sin2x} = \int_0^{\frac{\pi}2} \frac{\cos x\,dx}{1+\sin2x} = \frac12\int_0^{\frac{\pi}2} \frac{\sin x+\cos x}{1+\sin2x}\,dx = \frac12\int_0^{\frac{\pi}2} \frac{\sin x+\cos x}{(\sin x+\cos x)^2}\,dx$$
so we have
$$I = \frac12\int_0^{\frac{\pi}2} \frac{dx}{\sin x+\cos x} = \begin{bmatrix} t = \tan\frac{x}2  & dx = \frac{2dt}{1+t^2}\\ \sin x = \frac{2t}{1+t^2} & \cos x = \frac{1-t^2}{1+t^2}\end{bmatrix} = -\int_0^1\frac{dt}{t^2-2t-1} $$
We can calculate this as:
$$I = -\int_0^1 \frac{dt}{(t-1)^2+2} =- \int_{-1}^0 \frac{dt}{t^2+2} = \frac1{\sqrt{2}}\operatorname{Artanh}\left(\frac1{\sqrt{2}}\right)$$
