Mathematics

$\displaystyle\int^{1}_0\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)dx$.

SOLUTION
Let $x = tan\theta$
$dx = sec^2\theta d\theta$
when $x = 0, \theta = 0$
$x = 1 , \theta = \cfrac{\pi}{4}$
$\int_0^{\cfrac{\pi}{4}} sin^{-1}(\cfrac{2tan\theta}{1+tan^2\theta})sec^2\theta d\theta$
$\int_0^{\cfrac{\pi}{4}} sin^{-1}(sin2\theta) sec^2\theta d\theta$
$2\int_0^{\cfrac{\pi}{4}} \theta sec^2\theta d\theta$
Using byparts
$u = \theta , v = sec^2\theta$
$du = 1 , \int v dv = tan\theta$,we get
$[\theta tan\theta - \int tan\theta d\theta]^{\cfrac{\pi}{4}}_0$
$[\theta tan\theta - log(|cos\theta|)]^{\cfrac{\pi}{4}}_0$
$\cfrac{\pi}{4} +log\sqrt2$

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

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