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$\int_0^{\pi / 2} \log _e(\sin 2 x) d x$
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1124 Upvotes
Verified Answer
The correct answer is:
$-\frac{\pi}{2} \log 2$
$I=\int_0^{\pi / 2} \log _e(\sin 2 x) d x$
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, $I=\frac{1}{2} \int_0^\pi \log _e \sin (t) d t=\int_0^{\pi / 2} \log _e \sin (t) d t$
$$
=-\frac{\pi}{2} \log _e 2
$$
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, $I=\frac{1}{2} \int_0^\pi \log _e \sin (t) d t=\int_0^{\pi / 2} \log _e \sin (t) d t$
$$
=-\frac{\pi}{2} \log _e 2
$$
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