intlimits_0^{sqrt{3}} frac{1}{2} frac{d}{dx} | vedclass

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

Question

(B) Let $f(x) = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$.We know that $	an^{-1} \left( \frac{2x}{1-x^2} ight) = \begin{cases} 2	an^{-1}x & 	ext

Vedclass · Accepted Answer

$-\frac{\pi}{6}$

Vedclass · Answer

(B) Let $f(x) = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$.We know that $	an^{-1} \left( \frac{2x}{1-x^2} ight) = \begin{cases} 2	an^{-1}x & 	ext{if } |x|  1 \end{cases}$.The integral is $I = \int_0^{\sqrt{3}} \frac{1}{2} \frac{d}{dx} [f(x)] dx = \frac{1}{2} [f(x)]_0^{\sqrt{3}} = \frac{1}{2} [f(\sqrt{3}) - f(0)]$.For $x = \sqrt{3} > 1$,$f(\sqrt{3}) = 2	an^{-1}(\sqrt{3}) - \pi = 2(\frac{\pi}{3}) - \pi = \frac{2\pi}{3} - \pi = -\frac{\pi}{3}$.For $x = 0$,$f(0) = 2	an^{-1}(0) = 0$.Thus,$I = \frac{1}{2} [-\frac{\pi}{3} - 0] = -\frac{\pi}{6}$.

$\int\limits_0^{\sqrt{3}} \frac{1}{2} \frac{d}{dx} \left( \tan^{-1} \frac{2x}{1-x^2} \right) dx$ equals

Similar Questions

$\int_0^1 \frac{x}{(1-x)^{3/4}} dx = $

Suppose $g(x) = \int_0^x f(t) dt$,where $f$ is such that for $t \in [0, 1]$,$0 \le f(t) \le \frac{1}{2}$ and for $t \in [1, 2]$,$\frac{1}{2} \le f(t) \le 1$. Find the range of $g(2)$.

Evaluate the definite integral $\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$.

Evaluate the definite integral $\int_{-2}^{2.24} [x] \, dx$,where $[x]$ denotes the greatest integer function.

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^9 x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module