The value of int_{0}^{pi} log (1+cos x) d x i | vedclass

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(B) Let $I = \int_{0}^{\pi} \log (1+\cos x) d x \dots (i)$ Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get: $I = \int_{0}^

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$\pi \log \frac{1}{2}$

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(B) Let $I = \int_{0}^{\pi} \log (1+\cos x) d x \dots (i)$ Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get: $I = \int_{0}^{\pi} \log (1+\cos(\pi-x)) d x = \int_{0}^{\pi} \log (1-\cos x) d x \dots (ii)$ Adding $(i)$ and $(ii)$: $2I = \int_{0}^{\pi} [\log(1+\cos x) + \log(1-\cos x)] d x$ $2I = \int_{0}^{\pi} \log(1-\cos^2 x) d x = \int_{0}^{\pi} \log(\sin^2 x) d x$ $2I = 2 \int_{0}^{\pi} \log(\sin x) d x$ $I = \int_{0}^{\pi} \log(\sin x) d x$ Using the property $\int_{0}^{2a} f(x) d x = 2 \int_{0}^{a} f(x) d x$ if $f(2a-x) = f(x)$: $I = 2 \int_{0}^{\pi/2} \log(\sin x) d x$ Since $\int_{0}^{\pi/2} \log(\sin x) d x = -\frac{\pi}{2} \log 2$: $I = 2 	imes (-\frac{\pi}{2} \log 2) = -\pi \log 2 = \pi \log(\frac{1}{2})$

The value of $\int_{0}^{\pi} \log (1+\cos x) d x$ is

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