intlimits_0^1 {frac{{{{tan }^{ - 1}}x}}{x},dx | vedclass

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Question

(C) Let $I = \int\limits_0^1 {\frac{{{{	an }^{ - 1}}x}}{x}\,dx} $. Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$. When $x = 0$,$\

Vedclass · Accepted Answer

$\frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\frac{x}{{\sin x}}\,dx} $

Vedclass · Answer

(C) Let $I = \int\limits_0^1 {\frac{{{{	an }^{ - 1}}x}}{x}\,dx} $. Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$. When $x = 0$,$	heta = 0$. When $x = 1$,$	heta = \frac{\pi}{4}$. $I = \int\limits_0^{\frac{\pi}{4}} {\frac{	heta}{{	an 	heta}} \cdot \sec^2 	heta \, d	heta} = \int\limits_0^{\frac{\pi}{4}} {\frac{	heta}{{\sin 	heta \cos 	heta}} \, d	heta} = \int\limits_0^{\frac{\pi}{4}} {\frac{{2	heta}}{{\sin 2	heta}} \, d	heta}$. Let $2	heta = y$,then $2 \, d	heta = dy$ or $d	heta = \frac{1}{2} dy$. When $	heta = 0$,$y = 0$. When $	heta = \frac{\pi}{4}$,$y = \frac{\pi}{2}$. $I = \int\limits_0^{\frac{\pi}{2}} {\frac{y}{{\sin y}} \cdot \frac{1}{2} \, dy} = \frac{1}{2} \int\limits_0^{\frac{\pi}{2}} {\frac{y}{{\sin y}} \, dy} = \frac{1}{2} \int\limits_0^{\frac{\pi}{2}} {\frac{x}{{\sin x}} \, dx}$.

$\int\limits_0^1 {\frac{{{{\tan }^{ - 1}}x}}{x}\,dx} = $

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