int_0^1 sin ^{-1}left(frac{2 x}{1+x^2}right) | vedclass

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(C) Let $I = \int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}ight) d x$. Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta d	heta$. When $x = 0, 	he

Vedclass · Accepted Answer

$\frac{\pi}{2}-\log 2$

Vedclass · Answer

(C) Let $I = \int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}ight) d x$. Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta d	heta$. When $x = 0, 	heta = 0$ and when $x = 1, 	heta = \frac{\pi}{4}$. Using the identity $\sin^{-1}(\sin 2	heta) = 2	heta$ for $0 \le 	heta \le \frac{\pi}{4}$: $I = \int_0^{\frac{\pi}{4}} 2	heta \sec^2 	heta d	heta = 2 \int_0^{\frac{\pi}{4}} 	heta \sec^2 	heta d	heta$. Using integration by parts: $\int u dv = uv - \int v du$,where $u = 	heta$ and $dv = \sec^2 	heta d	heta$. $I = 2 \left[ 	heta 	an 	heta - \int 	an 	heta d	heta ight]_0^{\frac{\pi}{4}} = 2 [	heta 	an 	heta - \log |\sec 	heta|]_0^{\frac{\pi}{4}}$. $I = 2 [(\frac{\pi}{4} 	an \frac{\pi}{4} - \log \sec \frac{\pi}{4}) - (0 - \log \sec 0)]$. $I = 2 [\frac{\pi}{4}(1) - \log \sqrt{2} - 0] = 2 [\frac{\pi}{4} - \frac{1}{2} \log 2] = \frac{\pi}{2} - \log 2$.

$\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=$

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