int_0^{2 pi} frac{x cos x}{1+cos x} d x= | vedclass

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(D) Let $I = \int_0^{2 \pi} \frac{x \cos x}{1+\cos x} d x$ ... $(i)$ 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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None of the above.

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(D) Let $I = \int_0^{2 \pi} \frac{x \cos x}{1+\cos x} d x$ ... $(i)$ Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get: $I = \int_0^{2 \pi} \frac{(2 \pi - x) \cos(2 \pi - x)}{1 + \cos(2 \pi - x)} d x$ Since $\cos(2 \pi - x) = \cos x$,we have: $I = \int_0^{2 \pi} \frac{(2 \pi - x) \cos x}{1 + \cos x} d x$ ... (ii) Adding $(i)$ and (ii): $2I = \int_0^{2 \pi} \frac{2 \pi \cos x}{1 + \cos x} d x = 2 \pi \int_0^{2 \pi} \frac{\cos x}{1 + \cos x} d x$ $I = \pi \int_0^{2 \pi} \frac{\cos x}{1 + \cos x} d x = 2 \pi \int_0^{\pi} \frac{\cos x}{1 + \cos x} d x$ $I = 2 \pi \int_0^{\pi} (1 - \frac{1}{1 + \cos x}) d x = 2 \pi \int_0^{\pi} (1 - \frac{1}{2 \cos^2(x/2)}) d x$ $I = 2 \pi \int_0^{\pi} (1 - \frac{1}{2} \sec^2(x/2)) d x$ $I = 2 \pi [x - 	an(x/2)]_0^{\pi}$ Evaluating at limits: $I = 2 \pi [(\pi - 	an(\pi/2)) - (0 - 	an(0))]$. Since $	an(\pi/2)$ is undefined,the integral diverges. Thus,the correct option is $D$.

$\int_0^{2 \pi} \frac{x \cos x}{1+\cos x} d x=$

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