Question

Integrate the function: \(tan^{-1}\sqrt{\frac{1-x}{1+x}}\)

Answer 1

\(I=tan^{-1}\sqrt{\frac{1-x}{1+x}}\)

\(Let\space x=cosθ⇒dx=-sinθdθ\)

\(I=\int tan^{-1}\sqrt{\frac{1-cosθ}{1+cosθ}}(-sinθdθ)\)

\(=-\int tan^{-1}\sqrt{\frac{2sin^{2}\frac{θ}{2}}{2cos^{2}\frac{θ}{2}}}sinθdθ\)

\(=-\int tan^{-1}tan\frac{θ}{2}.sinθdθ\)

\(=-\frac{1}{2}∫θ.sinθdθ\)

\(=-\frac{1}{2}[θ.(-cosθ)-∫1.(-cosθ)dθ]\)

\(=-\frac{1}{2}[-θcosθ+sinθ]\)

\(=+\frac{1}{2}θcosθ-\frac{1}{2}sinθ\)

\(=\frac{1}{2}cos^{-1}x.x-\frac{1}{2}\sqrt{1-x^{2}}+C\)

\(=\frac{x}{2}cos^{-1}x-\frac{1}{2}\sqrt{1-x^{2}}+C\)

\(=\frac{1}{2}(xcos^{-1}x-\sqrt{1-x^{2}})+C\)