Mathematics

Solve$$\displaystyle \int_{0}^{\pi /2}\left ( 2\log \sin x-\log \sin 2x \right )dx $$


ANSWER

$$-\dfrac {\pi}{2}\log 2$$


SOLUTION
$$\displaystyle I= \int_{0}^{\pi /2}\left ( 2\log \sin x-\log 2\sin x\cos x \right )dx$$

$$\displaystyle = \int_{0}^{\pi /2}\left [ 2\log \sin x-\left \{ \log 2+\log \sin x+\log \cos x \right \} \right ]dx$$

$$\displaystyle = \int_{0}^{\pi /2}\left [ \left ( \log \sin x-\log \cos x \right )-\log2  \right ]dx$$

$$\displaystyle = 0-\left [ x\log 2 \right ]_{0}^{\pi /2}= -\frac{\pi }{2}\log 2= \frac{\pi }{2}\log \frac{1}{2}$$

$$\because \int _{ 0 }^{ \pi /2 } \log  \sin  xdx=\int _{ 0 }^{ \pi /2 } \log  \sin { \left( \dfrac { \pi  }{ 2 } -x \right)  } dx=\int _{ 0 }^{ \pi /2 } \log  \cos  xdx$$ by integration property
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Single Correct Medium Published on 17th 09, 2020
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