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
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