Mathematics

Solve the integral $$\int \sqrt{\dfrac{1+x}{1-x}}dx$$.


SOLUTION

Put $$x=\cos 2\theta$$ on differentiating with respect to $$\theta$$ , and we get

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

put given function  and solve given function

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

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

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

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

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

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

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

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

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

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

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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