Mathematics

# Evaluate: $\int\limits_0^\pi {\cos }(2x)$.$\log\sin x\;dx$

##### SOLUTION
Solution : -
$I = \int_{0}^{\pi} cos\,2x.\,log\,sin\,x\,dx$

consider $log\,sin\,x = I$         $cos\,2x = II$

then $I = log\,sin\,x \int cos\,2x\,dx-\int \dfrac{d}{dx}(log\,sin\,x).cos\,2x\,dx$

$= log\,.sin.x\dfrac{sin\,2x}{2}-\int \dfrac{cos\,x}{sin\,x}\times \dfrac{sin\,2x}{2}dx$

$= log\, sin\,x.\dfrac{sin\,2x}{2}-\dfrac{2}{2}\int cos^{2}x\,dx$

$= \dfrac{1}{2}log\, sin\,x.sin\,2x\dfrac{-1}{2}\int (cos\,2x+1)dx$

$= [\dfrac{1}{2}sin\,x.sin\,2x-\dfrac{1}{2}[\dfrac{sin\,2x}{2}+x]]_{0}^{\pi }$

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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