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Question

Mathematics Question on integral

Integrate the function: sin−1(2x1+x2)sin^{-1}(\frac{2x}{1+x^2})

Answer

The correct answer is: 2xtan−1−log(1+x2)+C2xtan^{-1}-log(1+x^2)+C
Let x=tanθ  dx=sec2θ.dθx=tanθ\,\,dx=sec^2θ .dθ
∴sin−1(2x1+x2)=sin−1(2tanθ1+tan2θ)=sin−1(sin2θ)∴sin^{-1}(\frac{2x}{1+x^2})=sin^{-1}(\frac{2tanθ}{1+tan^2θ})=sin^{-1}(sin2θ)
∫sin−1(2x1+x2)dx=∫2θ.sec2θ dθ=2∫θ.sec2θ dθ∫sin^{-1}(\frac{2x}{1+x^2})dx=∫2θ.sec^2θ\,dθ=2∫θ.sec^2θ\,dθ
Integrating by parts,we obtain
2[θ.∫sec2θ dθ−∫[(ddθθ)∫sec2θ dθ]dθ2[θ.∫sec^2θ\,dθ-∫[(\frac{d}{dθ}θ)∫sec^2θ\,dθ]dθ
=2[θ.tanθ−∫tanθ dθ]=2[θ.tanθ-∫tanθ\,dθ]
=2[θ tanθ+log∣cosθ∣]+C=2[θ\,tanθ+log|cosθ|]+C
=2[xtan−1x+log∣11+x2∣]+C=2[xtan^{-1}x+log|\frac{1}{\sqrt{1+x^2}}|]+C
=2xtan−1x+2log(1+x2)−12+C=2xtan^{-1}x+2log(1+x^2)^{\frac{-1}{2}}+C
=2xtan−1x+2[−12log(1+x2)]+C=2xtan^{-1}x+2[\frac{-1}{2}log(1+x^2)]+C
=2xtan−1−log(1+x2)+C=2xtan^{-1}-log(1+x^2)+C