Integrate the function: \(tan^{-1}\sqrt{\frac{1-x}{1+x}}\)
Solution and Explanation
\(I=tan^{-1}\sqrt{\frac{1-x}{1+x}}\)
\(Let\space x=cosθ⇒dx=-sinθdθ\)
\(I=\int tan^{-1}\sqrt{\frac{1-cosθ}{1+cosθ}}(-sinθdθ)\)
\(=-\int tan^{-1}\sqrt{\frac{2sin^{2}\frac{θ}{2}}{2cos^{2}\frac{θ}{2}}}sinθdθ\)
\(=-\int tan^{-1}tan\frac{θ}{2}.sinθdθ\)
\(=-\frac{1}{2}∫θ.sinθdθ\)
\(=-\frac{1}{2}[θ.(-cosθ)-∫1.(-cosθ)dθ]\)
\(=-\frac{1}{2}[-θcosθ+sinθ]\)
\(=+\frac{1}{2}θcosθ-\frac{1}{2}sinθ\)
\(=\frac{1}{2}cos^{-1}x.x-\frac{1}{2}\sqrt{1-x^{2}}+C\)
\(=\frac{x}{2}cos^{-1}x-\frac{1}{2}\sqrt{1-x^{2}}+C\)
\(=\frac{1}{2}(xcos^{-1}x-\sqrt{1-x^{2}})+C\)
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