int_0^1 {log sin left( {frac{pi }{2}x} right) | vedclass

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Question

(A) Let $I = \int_0^1 \log \sin \left( \frac{\pi }{2}x ight) dx$. Substitute $\frac{\pi }{2}x = 	heta$,then $dx = \frac{2}{\pi} d	heta$. When $x =

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$-\log 2$

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(A) Let $I = \int_0^1 \log \sin \left( \frac{\pi }{2}x ight) dx$. Substitute $\frac{\pi }{2}x = 	heta$,then $dx = \frac{2}{\pi} d	heta$. When $x = 0$,$	heta = 0$ and when $x = 1$,$	heta = \frac{\pi}{2}$. Thus,$I = \frac{2}{\pi} \int_0^{\pi/2} \log \sin 	heta \, d	heta$. Using the standard integral result $\int_0^{\pi/2} \log \sin 	heta \, d	heta = -\frac{\pi}{2} \log 2$,we get: $I = \frac{2}{\pi} \left( -\frac{\pi}{2} \log 2 ight) = -\log 2$.

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

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