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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int_0^1 {{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} ight)\,dx$. Substitute $x = 	an 	heta$,so $dx = \sec^2 	heta \, d	heta$. When $x = 0$,$	heta = 0$. When $x = 1$,$	heta = \frac{\pi}{4}$. Since $\sin^{-1}(\frac{2	an 	heta}{1+	an^2 	heta}) = \sin^{-1}(\sin 2	heta) = 2	heta$ for $x \in [0, 1]$,the integral becomes: $I = \int_0^{\pi/4} 2	heta \sec^2 	heta \, d	heta$. Using integration by parts: $I = 2 [	heta 	an 	heta]_0^{\pi/4} - 2 \int_0^{\pi/4} 	an 	heta \, d	heta$. $I = 2 [\frac{\pi}{4} \cdot 1 - 0] - 2 [\ln |\sec 	heta|]_0^{\pi/4}$. $I = \frac{\pi}{2} - 2 \ln(\sec \frac{\pi}{4}) = \frac{\pi}{2} - 2 \ln(\sqrt{2})$. Thus,the correct option is $A$.

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

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