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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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^

Vedclass · Accepted Answer

None of the above.

Vedclass · Answer

(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 $D$.

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

Similar Questions

If $I_n = \int_0^{\pi / 4} \tan^n x \, dx$,then $I_2+I_4, I_3+I_5, I_4+I_6, \ldots$ are in

The integral $\int_{-1}^{3} \left( \tan^{-1} \frac{x}{x^2+1} + \tan^{-1} \frac{x^2+1}{x} \right) dx = $

The integral value $\int_{-2}^{0} [x^3 + 3x^2 + 3x + 3 + (x + 1)\cos(x + 1)] \, dx$ is

$\int_{-8}^{8} \frac{x^{5}+x^{3}}{4-x^{2}} \, dx = $

The value of the integral $\frac{48}{\pi^{4}} \int_{0}^{\pi} \left(\frac{3 \pi x^{2}}{2} - x^{3}\right) \frac{\sin x}{1 + \cos^{2} x} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module