Evaluate: _0^ 2 (2 x - 2x) dx - Vidyadip

Question

$\text{Evaluate:} \int\limits_0^\frac{\pi}{2} (2\log \sin \text{x} - \log \sin 2\text{x}) \text{dx}$

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Answer

$\text{I} = \int\limits_0^\frac{\pi}{2} (2\log \sin \text{x} - \log \sin 2\text{x}) \text{dx} = \int\limits_0^\frac{\pi}{2}\log\bigg(\frac{\sin^{2}{x}}{2\sin x \cos x}\bigg)\text{dx}$
$= \int\limits_0^\frac{\pi}{2} \log \tan \text{x dx} - \int\limits_0^\frac{\pi}{2} \log 2 .\text{dx} = \text{I}_{1} - \text{I}_{2}$
$\text{I}_{1} = \int\limits_0^\frac{\pi}{2} \log \tan \text{x dx} = \int\limits_0^\frac{\pi}{2} \log \tan\bigg(\frac{\pi}{2} - \text{x}\bigg) \text{dx} = \int\limits_0^\frac{\pi}{2} \log \cot\text{x dx}$
$\Rightarrow \text{2 I}_{1} = 0 \Rightarrow \text{I}_{1} = 0$
$\text{I}_{2} = \log 2. \bigg[\text{x}\bigg]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} . \log 2$
$\therefore \text{I} = - \frac{\pi}{2} . \log 2$

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