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Question Number 63509 by mathmax by abdo last updated on 05/Jul/19

$$calculate{\int }_{-1}^1(\sqrt{1+x^2}-\sqrt{1-x^2})dx$$

Commented by Prithwish sen last updated on 05/Jul/19

$$2{\int }_0^1(\sqrt{1+{\mathrm{x}}^2}-\sqrt{1-{\mathrm{x}}^2})\mathrm{dx}$$$$=2{[\frac{\mathrm{x}}{2}\sqrt{1+{\mathrm{x}}^2}+\frac{1}{2}\ln \mid \mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\mid -\frac{\mathrm{x}}{2}\sqrt{1-{\mathrm{x}}^2}-\frac{1}{2}{\sin }^{-1}\mathrm{x}]}_0^1$$$$=2[\frac{1}{2}\sqrt{2}+\frac{1}{2}\ln \mid 1+\sqrt{2}\mid -\frac{\pi }{4}]=(\sqrt{2}-\frac{\pi }{2})+\ln \mid 1+\sqrt{2}\mid$$$$\mathrm{please}\mathrm{check}.$$

Answered by MJS last updated on 05/Jul/19

$$-\underset{-1}{\stackrel{1}{\int }}\sqrt{1-x^2}dx=-\frac{\pi }{2}$$$$\underset{-1}{\stackrel{1}{\int }}\sqrt{1+x^2}dx=\frac{1}{2}{[x\sqrt{1+x^2}+\ln (x+\sqrt{1+x^2})]}_{-1}^1=\sqrt{2}+\ln (1+\sqrt{2})$$$$\underset{-1}{\stackrel{1}{\int }}(\sqrt{1+x^2}-\sqrt{1-x^2})dx=\sqrt{2}+\ln (1+\sqrt{2})-\frac{\pi }{2}$$

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