Solve the following integral $$ \int_0^{\frac{\pi}{2}}{\frac{x\ln \left( 1-\sin x \right)}{\sin x}dx} \\ I\left( a \right) =\int_0^{\frac{\pi}{2}}{\frac{x\ln \left( 1-a\sin x \right)}{\sin x}}dx\quad \implies \quad I'\left( a \right) =-\int_0^{\frac{\pi}{2}}{\frac{x}{\left( 1-a\sin x \right)}}dx$$
$$ \quad $$ $$ \int{\frac{1}{\left( 1-a\sin x \right)}dx}=\int{\frac{1}{\sin ^2\frac{x}{2}+\cos ^2\frac{x}{2}+a\sin \frac{x}{2}\cos \frac{x}{2}}dx} \\ =\int{\frac{1}{\sin ^2\frac{x}{2}+\cos ^2\frac{x}{2}+a\text{2}\sin \frac{x}{2}\cos \frac{x}{2}}dx}=2\int{\frac{d\left( \tan \frac{x}{2} \right)}{\left( a+\tan \frac{x}{2} \right) ^2+\left( \sqrt{1-a^2} \right) ^2}}$$
$$ \\ =2\frac{1}{\sqrt{1-a^2}}\arctan \frac{a+\tan \frac{x}{2}}{\sqrt{1-a^2}} \\ \int_0^{\frac{\pi}{2}}{\frac{x}{\left( 1-a\sin x \right)}}dx=\int_0^{\frac{\pi}{2}}{xd\left( 2\frac{1}{\sqrt{1-a^2}}\arctan \frac{a+\tan \frac{x}{2}}{\sqrt{1-a^2}} \right)}$$= $$ 2\frac{1}{\sqrt{1-a^2}}\arctan \frac{a+\tan \frac{x}{2}}{\sqrt{1-a^2}}x\Bigg\vert_{0}^{\frac{\pi}{2}}-\frac{2}{\sqrt{1-a^2}}\int_0^{\frac{\pi}{2}}{\arctan \frac{a+\tan \frac{x}{2}}{\sqrt{1-a^2}}dx}$$ \
