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?
 A: Substitute $t=\tan\frac{x}{2}$ to rewrite your integral as $$\int_0^1\frac{4tdt}{(t^2-2t-1)^2}=\int_0^1\left(\frac{t^2+2t-1}{(t^2-2t-1)^2}-\frac{1}{t^2-2t-1}\right)dt\\=\left[-\frac{t+1}{t^2-2t-1}+\frac{1}{\sqrt{8}}\ln\frac{t+\sqrt{2}-1}{t-\sqrt{2}-1}\right]_0^1=\frac{1}{\sqrt{2}}\ln(\sqrt{2}+1).$$This agrees with your result, since$$\operatorname{arcoth}\sqrt{2}=\operatorname{artanh}\frac{1}{\sqrt{2}}=\frac12\ln\frac{\sqrt{2}+1}{\sqrt{2}-1}=\ln(\sqrt{2}+1).$$
A: $$I=\int_{0}^ {\pi/2}\frac{\sin x~dx}{1+\sin 2x} = \frac{1}{2} \int_{0}^{\pi/2} \frac{(\sin x+\cos x)-(\cos x-\sin x)}{(\sin x+ \cos x)^2} dx=\frac{1}{2}\left (\int_{0}^{\pi/2}  \frac{dx}{\sin x+\cos x}-\int_{1}^{1} \frac{dt}{t^2} \right).$$
$$\Rightarrow I=\frac{1}{2\sqrt{2}} \int_{0}^{\pi/2} \mbox{cosec} (x+\pi/4) dx=\frac{1}{2\sqrt{2}}\left . \ln(\tan(x/2+\pi/8)\right |_{0}^{\pi/2}
=\frac{1}{2\sqrt{2}} \ln\left( \frac{\tan (3\pi/8)}{\tan (\pi/8)} \right).$$
Note that $\tan(\pi/8)=\sqrt{2}-1=t$, then
$$ I= \frac{1}{2\sqrt{2}} \ln \left (\frac{3-t^2}{1-3t^2} \right )= \frac{1}{2\sqrt{2}} \ln (3+2\sqrt{2})=\frac{1}{\sqrt{2}} \ln (1+\sqrt{2})=
\frac{\mbox{coth}^{-1}\sqrt{2}}{\sqrt{2}}.$$
A: 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:
$$I = -\int_0^1 \frac{dt}{(t-1)^2+2} =- \int_{-1}^0 \frac{dt}{t^2+2} = \frac1{\sqrt{2}}\operatorname{Artanh}\left(\frac1{\sqrt{2}}\right)$$
A: Hint:
Use Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$.
to find $$2I=\int_0^{\pi/2}\dfrac{\sin x+\cos x}{1+\sin2x}dx$$
Set $\int(\cos x+\sin x)dx=u$
$u^2=1-\sin2x$
