Question

Integrate:
$\int {\tan }^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}},-\frac{\pi }{2}<x<\frac{\pi }{2}$

Answer 1

$\int {\tan }^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}}dx.\int {\tan }^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}\times \frac{1+\sin x}{1+\sin x}}dx\int {\tan }^{-1}\sqrt{\frac{{\left(1+\sin x\right)}^2}{1-{\sin }^{2}x}}dx.\int {\tan }^{-1}\sqrt{{\left(\frac{1+\sin x}{\cos x}\right)}^2}\int {\tan }^{-1}\left(\frac{1+\sin x}{\cos x}\right)dx.\int {\tan }^{-1}\left(\frac{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}\right)dx\int {\tan }^{-1}\frac{{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)}^2}{\left({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}\right)}dx\int {\tan }^{-1}\left(\frac{{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)}^2}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}dx\right)\left(A^2-B^2\right)=\left(A-B\right)\left(A+B\right)\int {\tan }^{-1}\left(\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right)dx\int {\tan }^{-1}\frac{\cos \frac{x}{2}\left(\tan \frac{x}{2}+1\right)}{\cos \frac{x}{2}\left(1-\tan \frac{x}{2}\right)}dx\int {\tan }^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)dx\int {\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}+\tan \frac{x}{2}}{1-\tan \frac{x}{2}.\tan \frac{\pi }{4}}\right)dx\left\{\begin{array}{c}weknow\\ \tan \frac{\pi }{4}=1\end{array}\right\}{\tan }^{-1}\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)dx.\{\begin{array}{c}Usingformulabe\\ \tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A.\tan B}\end{array}\int \left(\frac{\pi }{4}+\frac{x}{2}\right)dx\left(\frac{\pi }{4}x+\frac{x^2}{4}+c\right)Ans.$