Mathematics

If $$\int {\sqrt {\dfrac{{1 - x}}{{1 + x}}} dx = \sqrt {1 - {x^2}} }  + f\left( x \right) + c,\,x \in [0,\,1)$$ where $$f\left( 0 \right) =  - \dfrac{\pi }{2}$$ then $$f\left( {\dfrac{1}{2}} \right)$$ is ______


ANSWER

$$ - \dfrac{\pi }{3}$$


SOLUTION
Solution -

$$ =\displaystyle \int \sqrt{\dfrac{1-x}{1+x}}\times \dfrac{\sqrt{1-x}}{\sqrt{1-x}}dx  $$

$$ = \displaystyle\int \dfrac{1-x}{\sqrt{1-x^{2}}}dx = \int \dfrac{dx}{\sqrt{1-x^{2}}}-\int \dfrac{x\,dx}{\sqrt{1-x^{2}}} $$

$$ =\displaystyle -cos^{-1}x-\int \dfrac{xdx}{\sqrt{1-x^{2}}} $$ 

put $$ 1-x^{2} = t $$

$$ -2\,x\,dx = dt $$

$$ =\displaystyle -cos^{-1}x+\dfrac{1}{2}\int \dfrac{dt}{\sqrt{t}}$$

$$ =\displaystyle -cos^{-1}x+\sqrt{1-x^{2}}+c $$

$$ = f(x) = -cos^{-1}x $$

$$ f(0) = \dfrac{-\pi }{2} $$   $$ f(\dfrac{1}{2} )= -cos^{-1}(\dfrac{1}{2}) = -\dfrac{\pi }{3} $$

A is correct 
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Single Correct Medium Published on 17th 09, 2020
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