int_{-1}^{1} x tan^{-1} x , dx equals | vedclass

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

Question

(A) Let $I = \int_{-1}^{1} x 	an^{-1} x \, dx$. Since $f(x) = x 	an^{-1} x$ is an even function (because $f(-x) = (-x) 	an^{-1}(-x) = (-x)(-	an^{-

Vedclass · Accepted Answer

$\frac{\pi}{2} - 1$

Vedclass · Answer

(A) Let $I = \int_{-1}^{1} x 	an^{-1} x \, dx$. Since $f(x) = x 	an^{-1} x$ is an even function (because $f(-x) = (-x) 	an^{-1}(-x) = (-x)(-	an^{-1} x) = x 	an^{-1} x = f(x)$),we can write: $I = 2 \int_{0}^{1} x 	an^{-1} x \, dx$. Using integration by parts,let $u = 	an^{-1} x$ and $dv = x \, dx$. Then $du = \frac{1}{1+x^2} \, dx$ and $v = \frac{x^2}{2}$. $I = 2 \left[ \frac{x^2}{2} 	an^{-1} x \Big|_0^1 - \int_0^1 \frac{x^2}{2(1+x^2)} \, dx ight]$ $I = \left[ x^2 	an^{-1} x ight]_0^1 - \int_0^1 \frac{x^2}{1+x^2} \, dx$ $I = (1^2 	an^{-1} 1 - 0) - \int_0^1 \frac{x^2+1-1}{1+x^2} \, dx$ $I = \frac{\pi}{4} - \int_0^1 \left( 1 - \frac{1}{1+x^2} ight) \, dx$ $I = \frac{\pi}{4} - [x - 	an^{-1} x]_0^1$ $I = \frac{\pi}{4} - (1 - 	an^{-1} 1) = \frac{\pi}{4} - 1 + \frac{\pi}{4} = \frac{\pi}{2} - 1$.

$\int_{-1}^{1} x \tan^{-1} x \, dx$ equals

Similar Questions

The value of $\int_{0}^{\pi /2} \frac{2^{\sin x}}{2^{\sin x} + 2^{\cos x}} dx$ is

The value of $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \,dx$ is equal to

$\int_3^5(x-3)^3(5-x)^5 d x=$

If $f(x) = \int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt$ and $g(x) = \int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt$, then the value of $f(x) + g(x)$ is

If ${u_n} = \int_0^{\pi /4} {{\tan ^n}x\,dx,} $ then ${u_n} + {u_{n - 2}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module