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
$2[θ.∫sec^2θ\,dθ-∫[(\frac{d}{dθ}θ)∫sec^2θ\,dθ]dθ$
$=2[θ.tanθ-∫tanθ\,dθ]$
$=2[θ\,tanθ+log|cosθ|]+C$
$=2[xtan^{-1}x+log|\frac{1}{\sqrt{1+x^2}}|]+C$
$=2xtan^{-1}x+2log(1+x^2)^{\frac{-1}{2}}+C$
$=2xtan^{-1}x+2[\frac{-1}{2}log(1+x^2)]+C$
$=2xtan^{-1}-log(1+x^2)+C$