Mathematics

# Evaluate: $I =\displaystyle \int \log \sin x dx$

##### SOLUTION
$=log\sin x\cdot \int dx - \int(\int dx)\cdot \dfrac{d}{dx}(log\sin x) dx + c$

$= log\sin x\cdot x - \int x \cdot \dfrac{1}{\sin x} \cdot \cos x dx + c$

$=xlog\sin x - \int x \cdot \dfrac{\cos x}{\sin x}dx + c$

$=xlog\sin x - \int x \cot x dx + c$

$=xlog\sin x - x \cdot \int \cot x dx - \int (\int \cot x dx) \dfrac{d}{dx}(x) dx +c +$

$=xlog \sin x - x \cdot log |\sin x| - \int log|\sin x|dx + c'$

$\because I = \int log |\sin x|$

$I = -I + c'$

$2I = c'$

$I = c''$

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

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