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
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