Mathematics

# $\displaystyle \int \sin^{-1}\left ( \frac{2x}{1+x^2} \right )dx$ is equal to

$\displaystyle 2(x \tan^{-1} x +\ln \left | \cos (\tan^{-1}x) \right |)+C$

##### SOLUTION
$\displaystyle I=\int \sin^{-1}\left ( \frac{2x}{1+x^{2}} \right )dx$
Let $x=\tan \theta$ $\implies \displaystyle dx=\sec ^{2}\theta d\theta$

$\therefore \displaystyle I=\int \sin^{-1}\left ( \frac{2\tan \theta }{1+\tan^{2}\theta} \right ) \sec^{2}\theta d\theta$

$=\displaystyle \int \sin^{-1}\left(\dfrac{2\frac{\sin \theta}{\cos \theta}}{1+\frac{\sin^{2}\theta}{\cos^{2}\theta}}\right)\sec^{2}\theta d\theta$

$=\displaystyle \int \sin^{-1}\left(\dfrac{2\frac{\sin \theta}{\cos \theta}}{\frac{\sin^{2}\theta+\cos^{2}\theta}{\cos^{2}\theta}}\right)\sec^{2}\theta d\theta$

$=\displaystyle \int \sin^{-1}(2\sin{\theta} \cos{\theta})\sec^{2}\theta d\theta$

$=\displaystyle \int \sin^{-1}(\sin{2\theta})\sec^{2}\theta d\theta$

$=\displaystyle 2\int \theta \sec ^{2}\theta \:d\theta$

$=\displaystyle 2(\theta \:\tan \:\theta +\ln \left | \cos \:\theta \right |)+C$

$=\displaystyle 2(x \tan^{-1} x +\ln \left | \cos (\tan^{-1}x) \right |)+C$

Ans: A

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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