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) माना $I = \int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx$. चूँकि $f(x) = \frac{x \sin x}{1+\cos^2 x}$ एक सम फलन है,इसलिए: $I = 2 \int_0^\pi \frac{

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) माना $I = \int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx$. चूँकि $f(x) = \frac{x \sin x}{1+\cos^2 x}$ एक सम फलन है,इसलिए: $I = 2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$. गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर: $I = 2 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - I$. $2I = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx \Rightarrow I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$. माना $\cos x = t$,तब $-\sin x dx = dt$. $I = \pi \int_{-1}^1 \frac{dt}{1+t^2} = \pi [	an^{-1} t]_{-1}^1 = \pi (\frac{\pi}{4} + \frac{\pi}{4}) = \frac{\pi^2}{2}$.

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

Similar Questions

यदि $f(a + b - x) = f(x)$ है,तो $\int_{a}^{b} x \cdot f(a + b - x) \, dx = $

$\int_0^{\pi /2} \log(\sin x) \, dx = $

मान लीजिए $I = \int_{0}^{1} \frac{x^{3} \cos 3x}{2+x^{2}} dx$. तो

$\int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} dx$ का मान ज्ञात कीजिए।

समाकलन $I = \int_{1/2014}^{2014} \frac{\tan^{-1} x}{x} dx$ का मान है

Vedclass Test Series

Exam Paper Generator

Online Exam Module