Show that $\int_{0}^{\pi/2} \frac{\sin x~dx}{1+\sin 2x}=\frac{\coth^{-1} \sqrt{2}}{\sqrt{2}}$

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$$\operatorname{arcoth}\sqrt{2}=\operatorname{artanh}\frac{1}{\sqrt{2}}=\frac12\ln\frac{\sqrt{2}+1}{\sqrt{2}-1}=\ln(\sqrt{2}+1).$$

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