int_{-1}^1 frac{log 2 - log(1+x)}{sqrt{1-x^2} | vedclass

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

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$2 \pi \log 2$

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(D) Let $I = \int_{-1}^1 \frac{\log 2 - \log(1+x)}{\sqrt{1-x^2}} dx$. Substitute $x = \cos 	heta$, then $dx = -\sin 	heta d	heta$. When $x = -1$, $	heta = \pi$. When $x = 1$, $	heta = 0$. $I = \int_{\pi}^0 \frac{\log 2 - \log(1+\cos 	heta)}{\sqrt{1-\cos^2 	heta}} (-\sin 	heta) d	heta = \int_0^{\pi} \frac{\log 2 - \log(2 \cos^2(	heta/2))}{\sin 	heta} \sin 	heta d	heta$. $I = \int_0^{\pi} (\log 2 - \log 2 - 2 \log(\cos(	heta/2))) d	heta = -2 \int_0^{\pi} \log(\cos(	heta/2)) d	heta$. Let $u = 	heta/2$, then $d	heta = 2du$. $I = -2 \int_0^{\pi/2} \log(\cos u) (2 du) = -4 \int_0^{\pi/2} \log(\cos u) du$. Using the standard integral $\int_0^{\pi/2} \log(\cos u) du = -\frac{\pi}{2} \log 2$. $I = -4 	imes (-\frac{\pi}{2} \log 2) = 2 \pi \log 2$.

$\int_{-1}^1 \frac{\log 2 - \log(1+x)}{\sqrt{1-x^2}} dx =$

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