Integrate the function in Exercise: ^ -1 ( 2x 1+x^2 ) - Vidyadip

Question

Integrate the function in Exercise:
$\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$

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Answer

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

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