$\int_0^{\pi / 2} \log _e(\sin 2 x) d x$
$\int_0^{\pi / 2} \log _e(\sin 2 x) d x$
- $\pi \log 2$
- $-\pi \log 2$
- $\frac{\pi}{2} \log 2$
- $-\frac{\pi}{2} \log 2$
Solution
$I=\int_0^{\pi / 2} \log _e(\sin 2 x) d x$
Let
$
2 x=t
$
For lower limit at $x=0, t=0$
and upper limit at $x=\pi / 2, t=\pi$ and $d x=\frac{1}{2} d t$
So, $I=\frac{1}{2} \int_0^\pi \log _e \sin (t) d t=\int_0^{\pi / 2} \log _e \sin (t) d t$
$
=-\frac{\pi}{2} \log _e 2
$
Asked in: AP EAMCET 2018 (23 Apr Shift 2)
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