The value of intlimits_{frac{-pi}{2}}^{frac{p | vedclass

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

Question

(C) Let $I = \int\limits_{-\pi/2}^{\pi/2} \frac{x^2}{1 + 	an x + \sec x} \, dx$ (since $\sqrt{1+	an^2 x} = |\sec x| = \sec x$ for $x \in [-\pi/2, \p

Vedclass · Accepted Answer

$\frac{\pi^3}{24}$

Vedclass · Answer

(C) Let $I = \int\limits_{-\pi/2}^{\pi/2} \frac{x^2}{1 + 	an x + \sec x} \, dx$ (since $\sqrt{1+	an^2 x} = |\sec x| = \sec x$ for $x \in [-\pi/2, \pi/2]$).Using the property $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} (f(x) + f(-x)) \, dx$,we have:$I = \int_{0}^{\pi/2} \left( \frac{x^2}{1 + 	an x + \sec x} + \frac{(-x)^2}{1 + 	an(-x) + \sec(-x)} ight) \, dx$$I = \int_{0}^{\pi/2} x^2 \left( \frac{1}{1 + 	an x + \sec x} + \frac{1}{1 - 	an x + \sec x} ight) \, dx$$I = \int_{0}^{\pi/2} x^2 \left( \frac{1 + \sec x - 	an x + 1 + \sec x + 	an x}{(1 + \sec x)^2 - 	an^2 x} ight) \, dx$$I = \int_{0}^{\pi/2} x^2 \left( \frac{2 + 2\sec x}{1 + 2\sec x + \sec^2 x - 	an^2 x} ight) \, dx$Since $1 + 	an^2 x = \sec^2 x$,the denominator becomes $1 + 2\sec x + 1 = 2 + 2\sec x$.$I = \int_{0}^{\pi/2} x^2 \left( \frac{2(1 + \sec x)}{2(1 + \sec x)} ight) \, dx = \int_{0}^{\pi/2} x^2 \, dx$$I = \left[ \frac{x^3}{3} ight]_{0}^{\pi/2} = \frac{(\pi/2)^3}{3} = \frac{\pi^3}{8 	imes 3} = \frac{\pi^3}{24}$.

The value of $\int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{x^2}{1 + \tan x + \sqrt{1 + \tan^2 x}} \, dx$ is

Similar Questions

If $f$ and $g$ are continuous functions in $[0, a]$ satisfying $f(x) = f(a - x)$ and $g(x) + g(a - x) = 4$,then $\int_{0}^{a} f(x) g(x) dx$ is equal to:

$\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x=$

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

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

$\int_0^1 x(1-x)^n dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module