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Question Number 112666 by bemath last updated on 09/Sep/20

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

Answered by Her_Majesty last updated on 09/Sep/20

$$byparts$$$$u^{\prime }=1\Rightarrow u=x$$$$v={tan}^{-1}\sqrt{\frac{1-x}{1+x}}\Rightarrow v^{\prime }=-\frac{1}{2\sqrt{1-x^2}}$$$${xtan}^{-1}\sqrt{\frac{1-x}{1+x}+}\frac{1}{2}\int \frac{x}{\sqrt{1-x^2}}dx=$$$$={xtan}^{-1}\sqrt{\frac{1-x}{1+x}}-\frac{\sqrt{1-x^2}}{2}+C$$

Answered by bobhans last updated on 09/Sep/20

$$\mathrm{solve}\int {\tan }^{-1}\left(\sqrt{\frac{1-\mathrm{x}}{\mathrm{x}+1}}\right)\mathrm{dx}$$$$\mathrm{by}\mathrm{substitute}\mathrm{method}$$$$\mathrm{let}\mathrm{x}=\cos \mathrm{z}\Rightarrow \mathrm{dx}=-\sin \mathrm{z}\mathrm{dz}$$$$\mathrm{I}=\int {\tan }^{-1}\left(\sqrt{\frac{1-\cos \mathrm{z}}{1+\cos \mathrm{z}}}\right)(-\sin \mathrm{z}\mathrm{dz})$$$$\mathrm{I}=-\int {\tan }^{-1}\left(\sqrt{\frac{2\sin {}^2\left(\frac{\mathrm{z}}{2}\right)}{2\cos {}^2\left(\frac{\mathrm{z}}{2}\right)}}\right)\left(\sin \mathrm{z}\mathrm{dz}\right)$$$$\mathrm{I}=-\int {\tan }^{-1}\left(\tan \left(\frac{\mathrm{z}}{2}\right)\right)\sin \mathrm{z}\mathrm{dz}$$$$\mathrm{I}=-\frac{1}{2}\int \mathrm{z}\sin \mathrm{z}\mathrm{dz}$$$$\mathrm{I}=-\frac{1}{2}(-\mathrm{zcos}\mathrm{z}+\sin \mathrm{z})+\mathrm{c}$$$$\mathrm{I}=\frac{1}{2}\mathrm{z}\cos \mathrm{z}-\frac{1}{2}\sin \mathrm{z}+\mathrm{c}$$$$\mathrm{I}=\frac{1}{2}{\mathrm{xcos}}^{-1}\left(\mathrm{x}\right)-\frac{1}{2}\sqrt{1-{\mathrm{x}}^2}+\mathrm{c}$$

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