Evaluate: $\int_{0}^{\frac{π}{2}} (2 \log \sin x - \log \sin 2 x) dx$

Solution

$I = \int_{0}^{\frac{π}{2}} (2 \log \sin x - \log \sin 2 x) dx I = \int_{0}^{\frac{π}{2}} (\log \frac{\sin^{2} x}{2 \sin x . \cos x} . dx) I = \int_{0}^{\frac{π}{2}} \log (\frac{\tan x}{2}) . dx . . . . . . . . . . . . . (i) Using property \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx$

We get,

$I = \int_{0}^{\frac{π}{2}} \log (\frac{\tan (\frac{π}{2} - x)}{2}) dx \Rightarrow I = \int_{0}^{\frac{π}{2}} \log (\frac{\cot x}{2}) dx . . . . . . . . (ii)$

Adding (i) and (ii)

$2 I = \int_{0}^{\frac{π}{2}} [\log (\frac{\tan x}{2}) + \log (\frac{\cot x}{2})] dx \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} \log [(\frac{\tan x}{2}) (\frac{\cot x}{2})] dx \Rightarrow I = \frac{1}{2} \int_{0}^{\frac{π}{2}} \log (\frac{1}{4}) dx \Rightarrow I = \frac{1}{2} \log (\frac{1}{4}) x (\frac{π}{2}) \Rightarrow I = \frac{1}{2} \log {(\frac{1}{4})}^{2} x (\frac{π}{2}) \Rightarrow I = \log (\frac{1}{2}) x (\frac{π}{2}) \Rightarrow I = \frac{π}{2} \log \frac{1}{2}$

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Evaluate: $\int_{0}^{\frac{π}{2}} (2 \log \sin x - \log \sin 2 x) dx$