int_0^1 frac{log x}{sqrt{1 - x^2}} , dx = | vedclass

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Question

(C) Let $I = \int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx$. Substitute $x = \sin 	heta$,then $dx = \cos 	heta \, d	heta$. When $x = 0$,$	heta = 0$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx$. Substitute $x = \sin 	heta$,then $dx = \cos 	heta \, d	heta$. When $x = 0$,$	heta = 0$. When $x = 1$,$	heta = \frac{\pi}{2}$. Substituting these into the integral: $I = \int_0^{\pi/2} \frac{\log(\sin 	heta)}{\sqrt{1 - \sin^2 	heta}} \cdot \cos 	heta \, d	heta$ $I = \int_0^{\pi/2} \frac{\log(\sin 	heta)}{\cos 	heta} \cdot \cos 	heta \, d	heta$ $I = \int_0^{\pi/2} \log(\sin 	heta) \, d	heta$ Using the standard definite integral result $\int_0^{\pi/2} \log(\sin 	heta) \, d	heta = - \frac{\pi}{2} \log 2$,we get: $I = - \frac{\pi}{2} \log 2$.

$\int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx = $

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