int_{0}^{pi} log (1+cos x) d x का मान ज्ञात क | vedclass

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(B) माना $I = \int_{0}^{\pi} \log (1+\cos x) d x \dots (i)$ गुणधर्म $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$ का उपयोग करने पर: $I = \int_{0}^

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) माना $I = \int_{0}^{\pi} \log (1+\cos x) d x \dots (i)$ गुणधर्म $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$ का उपयोग करने पर: $I = \int_{0}^{\pi} \log (1+\cos(\pi-x)) d x = \int_{0}^{\pi} \log (1-\cos x) d x \dots (ii)$ $(i)$ और $(ii)$ को जोड़ने पर: $2I = \int_{0}^{\pi} [\log(1+\cos x) + \log(1-\cos x)] d x$ $2I = \int_{0}^{\pi} \log(1-\cos^2 x) d x = \int_{0}^{\pi} \log(\sin^2 x) d x$ $2I = 2 \int_{0}^{\pi} \log(\sin x) d x$ $I = \int_{0}^{\pi} \log(\sin x) d x$ गुणधर्म $\int_{0}^{2a} f(x) d x = 2 \int_{0}^{a} f(x) d x$ का उपयोग करने पर यदि $f(2a-x) = f(x)$ हो: $I = 2 \int_{0}^{\pi/2} \log(\sin x) d x$ चूंकि $\int_{0}^{\pi/2} \log(\sin x) d x = -\frac{\pi}{2} \log 2$: $I = 2 	imes (-\frac{\pi}{2} \log 2) = -\pi \log 2 = \pi \log(\frac{1}{2})$

$\int_{0}^{\pi} \log (1+\cos x) d x$ का मान ज्ञात कीजिए।

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