${\displaystyle \underset{\frac{\pi }{4}}{\overset{\frac{3\pi }{4}}{\int }}\frac{xdx}{1+\sin x}=}$

${\displaystyle {\int }_a^{b}f\left(x\right)dx}={\displaystyle {\int }_a^{b}f\left(a+b-x\right)dx\to }$

$I={\displaystyle \underset{\frac{\pi }{4}}{\overset{\frac{3\pi }{4}}{\int }}\frac{xdx}{1+\sin x},I}={\displaystyle \underset{\frac{\pi }{4}}{\overset{\frac{3\pi }{4}}{\int }}\frac{\left(\pi -x\right)}{1+\sin x}dx}$

$2I=\pi {\displaystyle \underset{\frac{\pi }{4}}{\overset{\frac{3\pi }{4}}{\int }}\frac{dx}{1+\sin x},t}=x-\frac{\pi }{2}$

     $=\pi {\displaystyle {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{dt}{1+\cos t}}=2\pi {\displaystyle {\int }_0^{\frac{\pi }{4}}\frac{dt}{1+\cos t}}$

      $=\pi {\displaystyle {\int }_0^{\frac{\pi }{4}}{\sec }^2\frac{t}{2}dt}=2{\pi \tan \frac{t}{2}|}_0^{\frac{\pi }{4}}$           (Using $1+\cos t=2\cos {}^2\frac{t}{2})$

      $=2\pi \tan \frac{\pi }{8}=2\pi \left(\sqrt{2}-1\right)\to I=\pi \left(\sqrt{2}-1\right)$