\(\int \tan ^{-1}\left(\frac{1-\sin x}{1+\sin x}\right) d x=\)

Let \(\mathrm{I}=\int \tan ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) \mathrm{d} x\)
\(=\int \tan ^{-1}\left(\sqrt{\frac{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}-2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}}\right) \mathrm{d} x\)
\(=\int \tan ^{-1}\left(\sqrt{\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}}\right) d x\)
\(\begin{aligned} & =\int \tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) d x \\ & =\int \tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right) d x\end{aligned}\)
\(\begin{aligned} & =\int \tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right) \mathrm{d} x \\ & =\int\left(\frac{\pi}{4}-\frac{x}{2}\right) \mathrm{d} x \\ & =\frac{\pi}{4} x-\frac{x^2}{4}+\mathrm{c}\end{aligned}\)