int_0^{pi / 2} log _e(sin 2 x) d x | vedclass

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Question

(D) Let $I = \int_0^{\pi / 2} \log _e(\sin 2 x) d x$. Substitute $2x = t$,so $dx = \frac{1}{2} dt$. When $x = 0$,$t = 0$. When $x = \frac{\pi}{2}$,$t

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int_0^{\pi / 2} \log _e(\sin 2 x) d x$. Substitute $2x = t$,so $dx = \frac{1}{2} dt$. When $x = 0$,$t = 0$. When $x = \frac{\pi}{2}$,$t = \pi$. Thus,$I = \frac{1}{2} \int_0^{\pi} \log _e(\sin t) dt$. Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$,we have $\int_0^{\pi} \log _e(\sin t) dt = 2 \int_0^{\pi/2} \log _e(\sin t) dt$. So,$I = \frac{1}{2} 	imes 2 \int_0^{\pi/2} \log _e(\sin t) dt = \int_0^{\pi/2} \log _e(\sin t) dt$. It is a standard result that $\int_0^{\pi/2} \log _e(\sin t) dt = -\frac{\pi}{2} \log _e 2$. Therefore,$I = -\frac{\pi}{2} \log _e 2$.

$\int_0^{\pi / 2} \log _e(\sin 2 x) d x$

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