Mathematics

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


SOLUTION

Put $$x=\cos 2\theta $$    …….  (1)

On differentiation with respect to $$\theta $$ , and we get

$$dx=-2\sin 2\theta d\theta $$

put given function  and solve given function

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

$$ =\int{\sqrt{\dfrac{1-\cos 2\theta }{1+\cos 2\theta }}\left( -2\sin 2\theta d\theta  \right)} $$

$$ =-2\int{\sqrt{\dfrac{1-\left( 1-2{{\sin }^{2}}\theta  \right)}{1+\left( 2{{\cos }^{2}}\theta -1 \right)}}\sin 2\theta d\theta } $$

$$ =-2\int{\sqrt{\dfrac{2{{\sin }^{2}}\theta }{2{{\cos }^{2}}\theta }}}\sin 2\theta d\theta  $$

$$ =-2\int{\sqrt{{{\tan }^{2}}\theta }}\sin 2\theta d\theta  $$

$$ =-2\int{\tan \theta }\sin 2\theta d\theta  $$

$$ =-2\int{\dfrac{\sin \theta }{\cos \theta }2\sin \theta \cos \theta d\theta } $$

$$ =-4\int{{{\sin }^{2}}\theta d\theta } $$

$$ =-4\int{\left( \dfrac{1-\cos 2\theta }{2} \right)}d\theta  $$

$$ =-2\int{1d\theta +2\int{\cos 2\theta d\theta }} $$

On integration, we get

$$ =-2\theta +2\dfrac{\sin 2\theta }{2}+C $$

$$ =-2\theta +\sin 2\theta +C $$

By equation (1),

$$\theta =\dfrac{1}{2}{{\cos }^{-1}}x$$

Put here and we get,

$$ =-2\times \dfrac{1}{2}{{\cos }^{-1}}x+\sin 2\left( \dfrac{1}{2}{{\cos }^{-1}}x \right)+C $$

$$ =-{{\cos }^{-1}}x+\sin {{\cos }^{-1}}x+C $$

$$ =-{{\cos }^{-1}}x+\sin {{\sin }^{-1}}\sqrt{1-{{x}^{2}}}+C $$

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

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

It is complete solution.

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