$I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{\sin x}{1+\sin2x}dx$

$=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{\sin x }{(\sin x+\cos x)^2}dx$

$=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{ \dfrac{1}{2}(\sin x+\cos x)-\dfrac{1}{2}(\cos x-\sin x)}{(\sin x+\cos x)^2}dx$

$=\underbrace{\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{1}{2}.\dfrac{1}{\sin x+\cos x}dx}_{I_1} - \underbrace{\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{1}{2}.\dfrac{\cos x-\sin x}{(\sin x+\cos x)^2}dx}_{I_2}$

+) $I_1$:

$I_1=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{1}{2}.\dfrac{1}{\sin x+\cos x}dx$

$=\dfrac{1}{2\sqrt2}\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{1}{\sin\left(x+\dfrac{\pi}{4}\right)}d\left(x+\dfrac{\pi}{4}\right)$

Xét $\displaystyle\int \dfrac{dx}{\sin x}$

$=\displaystyle\int \dfrac{dx}{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}$

$=\displaystyle\int\dfrac{ \dfrac{1}{\cos^2\dfrac{x}{2}}}{ 2\tan\dfrac{x}{2}}$ (*)

Đặt $u=\tan\dfrac{x}{2}\Rightarrow du=\dfrac{1}{2}.\dfrac{1}{\cos^2\dfrac{x}{2}}dx$

(*) $=\displaystyle\int \dfrac{2du}{2u}=\ln|u|+C=\ln| \tan\dfrac{x}{2}|+C$

Đo đó $I_1=\dfrac{1}{2\sqrt2}. \ln|\tan\left(0,5x+\dfrac{\pi}{8}\right)| |_0^{\frac{\pi}{2}}$

$=\dfrac{1}{2\sqrt2}.\ln\dfrac{\tan\dfrac{3\pi}{8}}{\tan\dfrac{\pi}{8}}$

+) $I_2$:

Đặt $t=\sin x+\cos x$

$\Rightarrow dt=(\cos x-\sin x)dx$

Đổi cận: $x=0\to t=1$; $x=\dfrac{\pi}{2}\to t=1$

$\Rightarrow I_2=0$

Ta có: 

$\tan\dfrac{3\pi}{4}= -1=\tan\left(2.\dfrac{3\pi}{8}\right)= \dfrac{2\tan\dfrac{3\pi}{8} }{1-\tan^2\dfrac{3\pi}{8}}$ 

$\Rightarrow \tan\dfrac{3\pi}{8}= 1+\sqrt2$ 

$\tan\dfrac{\pi}{4}=1=\tan\left(2.\dfrac{\pi}{8}\right)=\dfrac{ 2\tan\dfrac{\pi}{8}}{1-\tan^2\dfrac{\pi}{8}}$

$\Rightarrow \tan\dfrac{\pi}{8}= -1+\sqrt2$ 

Vậy $I=I_1+I_2= \dfrac{\sqrt2}{4}\ln\dfrac{\tan\dfrac{3\pi}{8}}{\tan\dfrac{\pi}{8}}=\dfrac{\sqrt2}{4}\ln (3+2\sqrt2)=\dfrac{\sqrt2}{2}\ln(1+\sqrt2)$