int_0^pi frac{x tan x}{sec x + cos x} ,dx = | vedclass

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

Question

(A) माना $I = \int_0^\pi \frac{x 	an x}{\sec x + \cos x} \,dx$. गुणधर्म $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$ का उपयोग करने पर: $I = \int_0^\pi

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) माना $I = \int_0^\pi \frac{x 	an x}{\sec x + \cos x} \,dx$. गुणधर्म $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$ का उपयोग करने पर: $I = \int_0^\pi \frac{(\pi - x) 	an(\pi - x)}{\sec(\pi - x) + \cos(\pi - x)} \,dx = \int_0^\pi \frac{(\pi - x)(-	an x)}{-\sec x - \cos x} \,dx = \int_0^\pi \frac{(\pi - x) 	an x}{\sec x + \cos x} \,dx$. $I$ के दोनों व्यंजकों को जोड़ने पर: $2I = \int_0^\pi \frac{\pi 	an x}{\sec x + \cos x} \,dx = \pi \int_0^\pi \frac{\sin x / \cos x}{1/\cos x + \cos x} \,dx = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \,dx$. माना $t = \cos x$,तब $dt = -\sin x \,dx$. जब $x=0, t=1$ और जब $x=\pi, t=-1$. $2I = \pi \int_1^{-1} \frac{-dt}{1 + t^2} = \pi \int_{-1}^1 \frac{dt}{1 + t^2} = \pi [	an^{-1} t]_{-1}^1$. $2I = \pi (	an^{-1}(1) - 	an^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$. अतः,$I = \frac{\pi^2}{4}$.

$\int_0^\pi \frac{x \tan x}{\sec x + \cos x} \,dx = $

Similar Questions

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) dx$ का मान ज्ञात कीजिए,जहाँ $f(x) = \sin |x| + \cos |x|$ और $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

समाकलन $\int_{0}^{1} x \cot^{-1}(1 - x^2 + x^4) dx$ का मान क्या है?

$\int_0^{\pi /2} \frac{x \sin x \cos x}{\cos^4 x + \sin^4 x} \, dx = $

$\int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x=$

यदि ${I_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,dx}$ है,तो $\mathop {\lim }\limits_{n \to \infty } \,n({I_n} + {I_{n - 2}})$ का मान ज्ञात कीजिए।

Vedclass Test Series

Exam Paper Generator

Online Exam Module