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)

Add (i) & (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
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