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

$\int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x=$

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

The value of the integral $\int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$ is equal to

The value of $\int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{x^2}{1 + \tan x + \sqrt{1 + \tan^2 x}} \, dx$ is

The value of $\int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx} $ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module