int_{-pi}^pi frac{x sin x}{1+cos^2 x} dx = | vedclass

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

Question

(D) Let $I = \int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx$. Since $f(x) = \frac{x \sin x}{1+\cos^2 x}$ is an even function because $f(-x) = \frac{(-

Vedclass · Accepted Answer

$\frac{\pi^2}{2}$

Vedclass · Answer

(D) Let $I = \int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx$. Since $f(x) = \frac{x \sin x}{1+\cos^2 x}$ is an even function because $f(-x) = \frac{(-x) \sin(-x)}{1+\cos^2(-x)} = \frac{(-x)(-\sin x)}{1+\cos^2 x} = f(x)$,we can write: $I = 2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$: $I = 2 \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1+\cos^2(\pi-x)} dx = 2 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx$. $I = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - 2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$. $I = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - I$. $2I = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx \Rightarrow I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$. Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$; when $x=\pi, t=-1$. $I = \pi \int_1^{-1} \frac{-dt}{1+t^2} = \pi \int_{-1}^1 \frac{dt}{1+t^2} = \pi [	an^{-1} t]_{-1}^1$. $I = \pi (	an^{-1}(1) - 	an^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$.

$\int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx =$

Similar Questions

Let $f(x)$ be positive for all real $x$. If $I_1 = \int_{1-h}^{h} x f(x(1-x)) dx$ and $I_2 = \int_{1-h}^{h} f(x(1-x)) dx$,where $(2h-1) > 0$,then $\frac{I_1}{I_2}$ is

$\int_0^{2 \pi} \sin ^6 x \cos ^5 x \, dx$ is equal to

$\int_{-1}^1 (a x^3 + b x) dx = 0$ for

The value of $\int_1^3 \sqrt{3 + x^3} \,dx$ lies in the interval

If $h(a) = h(b)$,the value of the integral $\int_a^b {[f(g(h(x)))]^{-1} f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \, dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module