Mathematics

# Solve: $\displaystyle \int^{\pi/2}_{0}(2log \sin x - log \sin 2x)dx$

$\frac{\pi}{2}log\frac{1}{2}$

##### SOLUTION
Consider, $I=$$\displaystyle \int^{\pi/2}_{0}(2\log \sin x - log \sin 2x)dx I=$$\displaystyle \int^{\pi/2}_{0}\log\left(\dfrac{ \sin^2 x}{2\sin x.\cos x}\right)dx$

$I=$$\displaystyle \int^{\pi/2}_{0}\log \tan xdx$$+\displaystyle \int^{\pi/2}_{0}\log \dfrac{1}{2}dx$$=\displaystyle \int^{\pi/2}_{0}\log \tan xdx$$+\dfrac{\pi}{2}\log \dfrac{1}{2}dx$........(1).

$I=$$\displaystyle \int^{\pi/2}_{0}\log \tan \left(\dfrac{\pi}{2}-x\right)dx+\dfrac{\pi}{2}\log\dfrac{1}{2} [ Using properties of definite integral] I=$$\displaystyle \int^{\pi/2}_{0}\log \cot xdx+\dfrac{\pi}{2}\log \dfrac{1}{2}$ .......(2).

Now adding (1) and (2) we get,

$2I=$$\displaystyle \int^{\pi/2}_{0}\log 1 dx+\pi \log \dfrac{1}{2}$

$2I=\pi \log \dfrac{1}{2}$

$I=\dfrac{\pi}{2}\log \dfrac{1}{2}$.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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