Mathematics

Prove $$\underset{0}{\overset{\frac{\pi}{2}}{\int}} \log (\sin \, x) dx = - \dfrac{\pi}{2} \log \, 2$$


SOLUTION
$$I=\int_{0}^{\frac{\pi}{2}}\log (\sin x)dx$$

$$I=\int_{0}^{\frac{\pi}{2}}\log (\cos x)dx$$

$$2I=\int_{0}^{\frac{\pi}{2}}[\log (\sin x)+\log (\cos x)]dx$$

$$2I=\int_{0}^{\frac{\pi}{2}}\log (\sin x \cos x)dx$$

$$2I=\int_{0}^{\frac{\pi}{2}}\log \left ( \dfrac{\sin 2x}{2} \right )dx$$

$$2I=\int_{0}^{\frac{\pi}{2}} \log (\sin 2x)dx-\int_{0}^{\frac{\pi}{2}} \log 2 dx$$

$$2I=I_1-\log 2 [x]_0^{\frac{\pi}{2}}$$

$$2I=I_1-\dfrac{\pi}{2}\log 2$$...(1)

$$I_1=\int_{0}^{\frac{\pi}{2}} \log (\sin 2x)dx$$

substitute $$2x=t\rightarrow dt=2dx\Rightarrow x=0-\frac{\pi}{2}\rightarrow t=0-\pi$$

$$\therefore I_1=\dfrac{1}{2}\int_{0}^{\pi}\log (\sin t)dt$$

$$\Rightarrow I_1=I$$...(2)

Placing (2)  in (1)

$$\Rightarrow 2I=I-\dfrac{\pi}{2}\log 2$$

$$\therefore I=-\dfrac{\pi}{2}\log 2$$
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Subjective Medium Published on 17th 09, 2020
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