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Q22  Integrate the functions \sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )

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\sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )

\int \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx
let x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta
 

\\=\int\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int2\theta \sec^2\theta d\theta\\
Taking \theta as a first function and \sec^2\theta as a second function, by using by parts method

\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]+C\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]+C\\ =2x\tan^{-1}x+2\log (1+x^2)^{-1/2}\\ =2x\tan^{-1}x-\log(1+x^2)+C

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manish

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