Mathematics

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


ANSWER

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


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
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