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

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

Question

(A) माना $I = \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$. गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर,हमें प्राप्त होता है: $I

Vedclass · Accepted Answer

$\frac{\pi^2}{4}$

Vedclass · Answer

(A) माना $I = \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$. गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर,हमें प्राप्त होता है: $I = \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1+\cos^2(\pi-x)} dx = \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx$. $I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$. $I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - I$. $2I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$. माना $u = \cos x$,तब $du = -\sin x dx$. जब $x=0, u=1$; जब $x=\pi, u=-1$. $2I = \pi \int_1^{-1} \frac{-du}{1+u^2} = \pi \int_{-1}^1 \frac{du}{1+u^2}$. $2I = \pi [	an^{-1} u]_{-1}^1 = \pi (	an^{-1}(1) - 	an^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$. $I = \frac{\pi^2}{4}$.

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

Similar Questions

$\int_{0}^{20 \pi}(|\sin x|+|\cos x|)^{2} d x$ का मान ज्ञात कीजिए।

$\int\limits_0^{\pi / 2n} \frac{dx}{1 + \tan^n(nx)} = $

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$ का मान . . . . . . है।

$\int_{0}^{\pi / 2} \log (\operatorname{cosec} x) d x$ का मान है

Vedclass Test Series

Exam Paper Generator

Online Exam Module