If intlimits_0^1 frac{ln x}{sqrt{1 - x^2}} dx | vedclass

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(B) Let $I = \int\limits_0^1 \frac{\ln x}{\sqrt{1 - x^2}} dx$. Put $x = \sin 	heta$,then $dx = \cos 	heta d	heta$. As $x 	o 0, 	heta 	o 0$ and a

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$1/2$

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(B) Let $I = \int\limits_0^1 \frac{\ln x}{\sqrt{1 - x^2}} dx$. Put $x = \sin 	heta$,then $dx = \cos 	heta d	heta$. As $x 	o 0, 	heta 	o 0$ and as $x 	o 1, 	heta 	o \pi/2$. $I = \int\limits_0^{\pi/2} \ln(\sin 	heta) d	heta = -\frac{\pi}{2} \ln 2$. Now consider the $RHS$: $J = \int\limits_0^\pi \ln(1 + \cos x) dx$. Using the property $\int\limits_0^a f(x) dx = \int\limits_0^a f(a-x) dx$,we have $J = \int\limits_0^\pi \ln(1 - \cos x) dx$. $2J = \int\limits_0^\pi \ln(1 - \cos^2 x) dx = \int\limits_0^\pi \ln(\sin^2 x) dx = 2 \int\limits_0^\pi \ln(\sin x) dx$. Since $\int\limits_0^\pi \ln(\sin x) dx = 2 \int\limits_0^{\pi/2} \ln(\sin x) dx = 2(-\frac{\pi}{2} \ln 2) = -\pi \ln 2$,we get $2J = 2(-\pi \ln 2) = -2\pi \ln 2$,so $J = -\pi \ln 2$. Equating $I = kJ$,we have $-\frac{\pi}{2} \ln 2 = k(-\pi \ln 2)$,which gives $k = 1/2$.

If $\int\limits_0^1 \frac{\ln x}{\sqrt{1 - x^2}} dx = k \int\limits_0^\pi \ln(1 + \cos x) dx$,then the value of $k$ is:

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