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Evaluate : $ \int_0^{\large\frac{\pi}{2}} (2\log\sin\: x-\log\sin 2x)dx$

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  • $m\log n=\log n^m$
  • $\log m-\log n=\log\large\frac{m}{n}$
  • $\sin 2x=2\sin x\cos x$
Step 1:
$I=\int_0^{\Large\frac{\pi}{2}}(2\log \sin x-\log \sin 2x)dx$
$\quad=\int_0^{\Large\frac{\pi}{2}}\log\big(\large\frac{\sin^2x}{\sin 2x}\big)$$dx$
$\log m-\log n=\log\large\frac{m}{n}$
$\quad=\int_0^{\Large\frac{\pi}{2}}\log \bigg(\large\frac{\sin^2x}{2\sin x\cos x}\bigg)$
$\quad=\int_0^{\Large\frac{\pi}{2}}\log\bigg(\large\frac{\tan x}{2}\bigg)dx$
$\quad=\int_0^{\Large\frac{\pi}{2}}\log(\tan x)-\log 2 dx$
$\quad=\int_0^{\Large\frac{\pi}{2}}\log(\tan x)-\int_0^{\Large\frac{\pi}{2}}\log 2 dx$
Step 2:
$I=I_1-\log 2\big[x\big]_0^{\Large\frac{\pi}{2}}$------(1)
$\;\;=I_1-(\large\frac{\pi}{2}-\normalsize 0)$$\log 2$
Where $I_1=\int_0^{\Large\frac{\pi}{2}}\log(\tan x)dx$
$\int_0^{\Large\frac{\pi}{2}}\log(\tan(\large\frac{\pi}{2}\normalsize -x))dx$
$I_1=\int_0^{\large\frac{\pi}{2}}\log(\cot x)dx$
Step 3:
$2I=\int_0^{\Large\frac{\pi}{2}}\log(\tan x)+\log(\cot x)]dx$
$\log m+\log n=\log (mn)$
$\Rightarrow \int_0^{\large\frac{\pi}{2}}\log(\tan x\cot x)dx$
On substituting the value of $I_1$ in equ(1) we get
$I=I_1-(\large\frac{\pi}{2}$$-0)\log 2$
$I=0-(\large\frac{\pi}{2}$$-0)\log 2$
$\;\;=\large\frac{-\pi}{2}$$\log 2$
answered Sep 24, 2013 by sreemathi.v
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