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Question Number 27215 by abdo imad last updated on 03/Jan/18

$$findthevalueof{\int }_{-1}^1\frac{dx}{\sqrt{1-x^2}+\sqrt{1+x^2}}.$$

Answered by Giannibo last updated on 03/Jan/18

$$$$$$$$$$\underset{-1}{\stackrel{1}{\int }}\frac{(\sqrt{1-{\mathrm{x}}^2}-\sqrt{1+{\mathrm{x}}^2}}{(\sqrt{1-{\mathrm{x}}^2}+\sqrt{1-{\mathrm{x}}^2})(\sqrt{1-{\mathrm{x}}^2}-\sqrt{1+{\mathrm{x}}^2})}\mathrm{dx}$$$$\underset{-1}{\stackrel{1}{\int }}\frac{\sqrt{1-{\mathrm{x}}^2}-\sqrt{1+{\mathrm{x}}^2}}{2}\mathrm{dx}$$$$=\underset{\mathrm{u}\to 1}{\lim }(\underset{-\mathrm{u}}{\stackrel{\mathrm{u}}{\int }}\frac{\sqrt{1-{\mathrm{x}}^2}}{2}\mathrm{dx}-\underset{-\mathrm{u}}{\stackrel{\mathrm{u}}{\int }}\frac{\sqrt{1+{\mathrm{x}}^2}}{2}\mathrm{dx})$$$$=\underset{\mathrm{u}\to 1}{\lim }\frac{1}{4}({\sin }^{-1}\left(\mathrm{u}\right)+\mathrm{u}\sqrt{1-{\mathrm{u}}^2}-{\sin }^{-1}(-\mathrm{u})+\mathrm{u}\sqrt{1-{\mathrm{u}}^2}-(\ln (\mid \sqrt{{\mathrm{u}}^2+1}+\mathrm{u}\mid )+\mathrm{u}\sqrt{{\mathrm{u}}^2+1}-\ln (\mid \sqrt{{\mathrm{u}}^2+1}-\mathrm{u}\mid )+\mathrm{u}\sqrt{{\mathrm{u}}^2+1})$$$$=\frac{1}{4}(\frac{\pi }{2}+0+\frac{\pi }{2}-(\ln (\sqrt{2}+1)+\sqrt{2}-\ln (\sqrt{2}-1)+\sqrt{2}))$$$$=\frac{1}{4}(\pi -2\sqrt{2}-\ln \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right))$$

Commented by abdo imad last updated on 05/Jan/18

$${\int }_{-1}^1\frac{dx}{\sqrt{1-x^2}+\sqrt{1+x^2}}={\int }_{-1}^1\frac{\sqrt{1-x^2}-\sqrt{1+x^2}}{-2x^2}dx...!!$$

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