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Question Number 120761 by bramlexs22 last updated on 02/Nov/20

$$\int {\tan }^{-1}\left(\sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\right)\mathrm{dx}?$$

Answered by liberty last updated on 02/Nov/20

$$\mathrm{We}\mathrm{put}\mathrm{x}=\cos \psi \Rightarrow \mathrm{dx}=-\sin \psi \mathrm{d}\psi$$$$\int {\tan }^{-1}\sqrt{\frac{1-\cos \psi }{1+\cos \psi }}(-\sin \psi \mathrm{d}\psi )=$$$$\int {\tan }^{-1}\sqrt{\frac{2\sin {}^2\left(\psi /2\right)}{2\cos {}^2\left(\psi /2\right)}}(-\sin \psi \mathrm{d}\psi )=$$$$\int {\tan }^{-1}\left(\tan \left(\psi /2\right)\right)(-\sin \psi \mathrm{d}\psi )=$$$$\int -\left(\frac{\psi }{2}\right)\sin \psi \mathrm{d}\psi =-\frac{1}{2}[-\psi \cos \psi +\sin \psi ]+\mathrm{c}$$$$=-\frac{1}{2}[-\mathrm{x}{\cos }^{-1}\left(\mathrm{x}\right)+\sqrt{1-{\mathrm{x}}^2}]+\mathrm{c}$$

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