Evaluate the following integrals: ^ -1 ( 2x 1+x^2 )dx - Vidyadip

Question

Evaluate the following integrals:

$\int\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{dx}$

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Answer

Let $\text{x}=\tan\theta\Rightarrow\text{dx}=\sec^2\theta\text{d}\theta$
$\therefore\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)=\sin^{-1}(\sin2\theta)=\theta$
$\Rightarrow\int\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{dx}=\int2\theta\cdot\sec^2\theta\text{d}\theta=2\int\theta\cdot\sec^2\theta\text{d}\theta$
Integrating by parts, we obtain
$2\Big[\theta\cdot\int\sec^2\theta\text{d}\theta-\int\Big\{\Big(\frac{\text{d}}{\text{d}\theta}\theta\Big)\int\sec^2\theta\text{d}\theta\Big\}\text{d}\theta\Big]$
$=2\big[\theta\cdot\tan\theta-\int\tan\theta\text{d}\theta\big]$
$=2\big[\theta\tan\theta+\log|\cos\theta|\big]+\text{C}$
$=2\Big[\text{x}\tan^{-1}\text{x}+\log\Big|\frac{1}{\sqrt{1+\text{x}^2}}\Big|\Big]+\text{C}$
$=2\text{x}\tan^{-1}\text{x}+2\log(1+\text{x}^2)^{-\frac{1}{2}}+\text{C}$
$=2\text{x}\tan^{-1}\text{x}+2\Big[-\frac{1}{2}\log(1+\text{x}^2)\Big]+\text{C}$
$=2\text{x}\tan^{-1}\text{x}-\log(1+\text{x}^2)+\text{C}$

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