Question

$\int {\tan }^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}dx$.

Answer 1

$I=\int {\tan }^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}dx$

$=\int ta{n}^{-1}\sqrt{\frac{1+2sin\frac{x}{2}cox\frac{x}{2}}{1+2sin\frac{x}{2}cos\frac{x}{2}}}dx$ $[\because sin2x=2sinxcosx]$

$=\int ta{n}^{-1}\sqrt{\frac{co{s}^2\frac{x}{2}+si{n}^2\frac{x}{2}-2sin\frac{x}{2}cos\frac{x}{2}}{co{s}^2\frac{x}{2}+si{n}^2\frac{x}{2}+2sin\frac{x}{2}cos\frac{x}{2}}}dx$ $[\because co{s}^2\frac{x}{2}+si{n}^2\frac{x}{2}=1]$

$=ta{n}^{-1}{\sqrt{\frac{(cos\frac{x}{2}-sin\frac{x}{2}{)}^2}{cos\frac{x}{2}+sin\frac{x}{2}}}}^{2}dx$

$=\int ta{n}^{-1}|\frac{cos\frac{x}{2}-sin\frac{x}{2}}{cos\frac{x}{2}+sin\frac{x}{2}}|dx$

$=\int ta{n}^{-1}|\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}dx$

$=\int ta{n}^{-1}|\frac{tan\frac{\pi }{4}-tan\frac{x}{2}}{1+tan\frac{\pi }{4}tan\frac{x}{2}}|dx$$[\because tan(A-B)=\frac{tanA-tanB}{1+tanA+tanB}]$

$=\int ta{n}^{-1}tan(\frac{\pi }{4}-\frac{x}{2})dx$ $tan\frac{\pi }{4}=1$

$=\int (\frac{\pi }{4}-\frac{x}{2})dx$

$I=\frac{\pi }{4}\times \frac{-x^2}{4}+C$