Mathematics

# Evaluate:$\displaystyle\int_{0}^{\pi}\log (1-\cos x)\ dx$

##### SOLUTION
$I = \int_0^{\pi} \log (1-\cos x) dx$      ...(i)

$I = \int_0^{\pi} \log (1-\cos (\pi -x)) dx$

$I = \int_0^{\pi} \log (1 + \cos x) dx$    ...(ii)

$2I = \int_0^{\pi} [\log (1 - \cos x) + \log (1 + \cos x)]dx$

$2I = \int_0^{\pi} \log (1-\cos ^2x)dx$

$2I = \int_0^{\pi} \log \sin^2 x\, dx$

$\Rightarrow 2I = 2\int_0^{\pi} \log \sin x$

$I = \int _0^{\pi} \log \sin x \, dx$

$I = -\pi \log 2$............................(as $\int _0^{\pi} \log \sin x \, dx= -\pi \log 2$ )

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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