_0^ ^2 x d x= - Vidyadip

MCQ

$\int_0^\pi \log \sin ^2 x d x=$

✓
$2 \pi \log _e\left(\frac{1}{2}\right)$
B
$\pi \log _{ e } 2$
C
$\frac{\pi}{2} \log _e\left(\frac{1}{2}\right)$
D
$\pi \log _{ e }\left(\frac{1}{2}\right)$

✓

Answer

Correct option: A.

$2 \pi \log _e\left(\frac{1}{2}\right)$

(A)
$\int_0^\pi \log \sin ^2 x d x=\int_0^\pi 2 \log \sin x d x=\int_0^{2 \frac{\pi}{2}} \log \sin x d x$
$=2 \int_0^{\frac{\pi}{2}}[\log \sin x+\log \sin (\pi-x)] d x$
$\ldots .\left[\because \int_0^{2 a } f (x) d x=\int_0^{ a }[ f (x)+ f (2 a -x)] d x\right]$
$=4 \int_0^{\frac{\pi}{2}} \log \sin x d x$
$=4 \times\left(-\frac{\pi}{2} \log 2\right)=-2 \pi \log _{ e } 2=2 \pi \log _{ e }\left(\frac{1}{2}\right)$

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