Question

Integrate the function: \(sin^{-1}(\frac{2x}{1+x^2})\)

Answer 1

The correct answer is: \(2xtan^{-1}-log(1+x^2)+C\)
Let \(x=tanθ\,\,dx=sec^2θ .dθ\)
\(∴sin^{-1}(\frac{2x}{1+x^2})=sin^{-1}(\frac{2tanθ}{1+tan^2θ})=sin^{-1}(sin2θ)\)
\(∫sin^{-1}(\frac{2x}{1+x^2})dx=∫2θ.sec^2θ\,dθ=2∫θ.sec^2θ\,dθ\)
Integrating by parts,we obtain
\(2[θ.∫sec^2θ\,dθ-∫[(\frac{d}{dθ}θ)∫sec^2θ\,dθ]dθ\)
\(=2[θ.tanθ-∫tanθ\,dθ]\)
\(=2[θ\,tanθ+log|cosθ|]+C\)
\(=2[xtan^{-1}x+log|\frac{1}{\sqrt{1+x^2}}|]+C\)
\(=2xtan^{-1}x+2log(1+x^2)^{\frac{-1}{2}}+C\)
\(=2xtan^{-1}x+2[\frac{-1}{2}log(1+x^2)]+C\)
\(=2xtan^{-1}-log(1+x^2)+C\)