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Question Number 118575 by bramlexs22 last updated on 18/Oct/20

$$\int \cos \left({2\cot }^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx?$$

Answered by benjo_mathlover last updated on 18/Oct/20

$$set{\cot }^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)=\varphi \Rightarrow \sqrt{\frac{1-x}{1+x}}=\cot \varphi$$$$\Rightarrow \cot {}^2\varphi =\frac{1-x}{1+x},\tan {}^2\varphi =\frac{1+x}{1-x}$$$$\Rightarrow \sec {}^2\varphi =1+\frac{1+x}{1-x}=\frac{2}{1-x}$$$$\Rightarrow \cos {}^2\varphi =\frac{1-x}{2}\wedge \cos 2\varphi =2\cos {}^2\varphi -1$$$$\Rightarrow \cos 2\varphi =2\left(\frac{1-x}{2}\right)-1=-x$$$$Thus\int \cos \left({2\cot }^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right)dx=$$$$\int (-x)dx=-\frac{x^2}{2}+c$$

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