int_0^pi frac{x tan x}{sec x+tan x} d x is eq | vedclass

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

Question

(A) Let $I = \int_0^\pi \frac{x 	an x}{\sec x+	an x} d x \quad ...(i)$ Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,we get: $I = \

Vedclass · Accepted Answer

$\frac{\pi(\pi-2)}{2}$

Vedclass · Answer

(A) Let $I = \int_0^\pi \frac{x 	an x}{\sec x+	an x} d x \quad ...(i)$ Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,we get: $I = \int_0^\pi \frac{(\pi-x) 	an(\pi-x)}{\sec(\pi-x)+	an(\pi-x)} d x$ Since $	an(\pi-x) = -	an x$ and $\sec(\pi-x) = -\sec x$,we have: $I = \int_0^\pi \frac{(\pi-x)(-	an x)}{-\sec x-	an x} d x = \int_0^\pi \frac{(\pi-x) 	an x}{\sec x+	an x} d x \quad ...(ii)$ Adding $(i)$ and $(ii)$: $2I = \int_0^\pi \frac{x 	an x + (\pi-x) 	an x}{\sec x+	an x} d x = \int_0^\pi \frac{\pi 	an x}{\sec x+	an x} d x$ $I = \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\sin x} d x$ Multiply numerator and denominator by $(1-\sin x)$: $I = \frac{\pi}{2} \int_0^\pi \frac{\sin x(1-\sin x)}{1-\sin^2 x} d x = \frac{\pi}{2} \int_0^\pi \frac{\sin x - \sin^2 x}{\cos^2 x} d x$ $I = \frac{\pi}{2} \int_0^\pi (\sec x 	an x - 	an^2 x) d x = \frac{\pi}{2} \int_0^\pi (\sec x 	an x - (\sec^2 x - 1)) d x$ $I = \frac{\pi}{2} [\sec x - 	an x + x]_0^\pi$ $I = \frac{\pi}{2} [(\sec \pi - 	an \pi + \pi) - (\sec 0 - 	an 0 + 0)]$ $I = \frac{\pi}{2} [(-1 - 0 + \pi) - (1 - 0 + 0)] = \frac{\pi}{2} (\pi - 2)$

$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$ is equal to

Similar Questions

$\int_0^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x=$

Let $f(x)$ be integrable over $(a, b)$,where $b > a > 0$. If $I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} f(\tan \theta + \cot \theta) \sec^2 \theta \, d\theta$ and $I_2 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} f(\tan \theta + \cot \theta) \csc^2 \theta \, d\theta$,then the ratio $\frac{I_1}{I_2}$ is:

$\int_{-\pi}^{\pi} x^2 \sin x \, dx =$

Which of the following statements is incorrect for the function $g(\alpha)$ for $\alpha \in R$ such that $g(\alpha)=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\alpha} x}{\cos^{\alpha} x+\sin^{\alpha} x} dx$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module