_ - 1 ^1 x ^ - 1 x ,dx equals - Vidyadip

MCQ

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

✓
$\left( {\frac{\pi }{2} - 1} \right)$
B
$\left( {\frac{\pi }{2} + 1} \right)$
C
$(\pi - 1)$
D
$0$

✓

Answer

Correct option: A.

$\left( {\frac{\pi }{2} - 1} \right)$

a
(a) $I = \int_{ - 1}^1 {x{{\tan }^{ - 1}}x\,dx = 2} \int_0^1 {x{{\tan }^{ - 1}}x\,dx} $

$\because \,\,\,x{{\tan }^{-1}}x$ is an even function

$I = [2\frac{{{x^2}}}{2}{\tan ^{ - 1}}x]_0^1 - 2\int_0^1 {\frac{1}{2}\frac{{{x^2}}}{{1 + {x^2}}}dx} $

$I = [{x^2}{\tan ^{ - 1}}x]_0^1 - \int_0^1 {\frac{{{x^2} + 1 - 1}}{{1 + {x^2}}}dx} $

$I = [{x^2}{\tan ^{ - 1}}x]_0^1 - [x]_0^1 + [{\tan ^{ - 1}}x]_0^1$

==> $I = \frac{\pi }{4} - 1 + \frac{\pi }{4} = \frac{\pi }{2} - 1$.

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