Mathematics

# $\displaystyle \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ \left( \dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } \right) } dx }$

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

##### SOLUTION
$\int_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{x}{{1 - {x^2}}}} \right)} \,dx$

Putting   $x = \sin \theta$    $dx = \cos \theta \,d\theta$

$= \int_0^{\dfrac{\pi }{2}} {{{\tan }^{ - 1}}} \left( {\dfrac{{\sin \theta }}{{\sqrt {1-{{\sin }^2}\theta } }}} \right)\cos \theta \,d\theta$

$= \int_0^{\dfrac{\pi }{2}} {{{\tan }^{ - 1}}} \left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)\cos \theta \,d\theta$

$= \int_0^{\dfrac{\pi }{2}} {{{\tan }^{ - 1}}} \left( {\tan \theta } \right)\cos \theta \,d\theta$

$= \int_0^{\dfrac{\pi }{2}} {\,\,\theta } \,\,\cos \theta \,d\theta$

$= {\left( {\theta \sin \theta } \right)^{\dfrac{\pi }{2}}} - \int_0^{\dfrac{\pi }{2}} {1.\sin \theta \,d\theta }$

$= {\left[ {\theta \,\,\sin \theta + \cos \theta } \right]_0}^{\dfrac{\pi }{2}}$

$= \left( {\dfrac{\pi }{2} + 0} \right) - \left( 1 \right)$

$= \dfrac{\pi }{2} - 1$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

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