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Question Number 101833 by bobhans last updated on 05/Jul/20

$$\int {}_{-1}^1\sqrt{\frac{1+x}{1-x}}dx?$$

Answered by Dwaipayan Shikari last updated on 05/Jul/20

$${\int }_{-1}^1\frac{1+x}{\sqrt{1-x^2}}dx={\int }_{-1}^1\frac{1}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}={\left[{sin}^{-1}x\right]}_{-1}^1-\frac{1}{2}{\int }_{-1}^1\frac{-2x}{\sqrt{1-x^2}}dx$$$$=\pi -\frac{1}{2}{\left[\sqrt{1-x^2}\right]}_{-1}^1=\pi +0=\pi$$$$$$

Answered by john santu last updated on 05/Jul/20

$$set\sqrt{\frac{1+x}{1-x}}=s\to dx=\frac{4s}{{(s^2+1)}^2}ds$$$$I=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{4s^2}{{(s^2+1)}^2}ds\left[byparts\right]$$$$\{\begin{array}{l}u=s\\ dv=\frac{4sds}{{(s^2+1)}^2}\end{array}$$$$I=-2.\mid \frac{s}{s^2+1}{\mid }_0^{\mathrm{\infty }}-\int -\frac{2ds}{s^2+1}$$$$I=0+{\left[2\arctan \left(s\right)\right]}_0^{\mathrm{\infty }}$$$$I=2\left(\frac{\pi }{2}\right)=\pi (JS⊛)$$

Answered by mathmax by abdo last updated on 05/Jul/20

$$\mathrm{A}={\int }_{-1}^1\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\mathrm{dx}\mathrm{changement}\mathrm{x}=\mathrm{cost}\mathrm{give}$$$$\mathrm{A}=-{\int }_0^{\pi }\sqrt{\frac{1+\mathrm{cost}}{1-\mathrm{cost}}}(-\mathrm{sint})\mathrm{dt}={\int }_0^{\pi }\sqrt{\frac{{2\cos }^2\left(\frac{\mathrm{t}}{2}\right)}{{2\sin }^2\left(\frac{\mathrm{t}}{2}\right)}}\mathrm{sint}\mathrm{dt}$$$$={\int }_0^{\pi }\frac{\cos \left(\frac{\mathrm{t}}{2}\right)}{\sin \left(\frac{\mathrm{t}}{2}\right)}\times 2\sin \left(\frac{\mathrm{t}}{2}\right)\cos \left(\frac{\mathrm{t}}{2}\right)\mathrm{dt}=2{\int }_0^{\pi }{\cos }^2\left(\frac{\mathrm{t}}{2}\right)\mathrm{dt}$$$$={\int }_0^{\pi }(1+\mathrm{cost})\mathrm{dt}=\pi +{\int }_0^{\pi }\mathrm{cost}\mathrm{dt}=\pi +{\left[\mathrm{sint}\right]}_0^{\pi }=\pi +0\Rightarrow \mathrm{A}=\pi$$

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