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Question Number 119920 by bramlexs22 last updated on 28/Oct/20

$$\int \sqrt{\frac{1-x}{1+x}}dx=?,xϵ(-1,1)$$

Answered by mathmax by abdo last updated on 28/Oct/20

$$\mathrm{I}=\int \sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\mathrm{cost}\Rightarrow$$$$\mathrm{I}=\int \sqrt{\frac{{2\sin }^2\left(\frac{\mathrm{t}}{2}\right)}{{2\cos }^2\left(\frac{\mathrm{t}}{2}\right)}}\mathrm{sint}\mathrm{dt}$$$$=\int \frac{\sin \left(\frac{\mathrm{t}}{2}\right)}{\cos \left(\frac{\mathrm{t}}{2}\right)}\times 2\sin \left(\frac{\mathrm{t}}{2}\right)\cos \left(\frac{\mathrm{t}}{2}\right)\mathrm{dt}$$$$=2\int {\sin }^2\left(\frac{\mathrm{t}}{2}\right)\mathrm{dt}=2\int \frac{1-\cos \left(\mathrm{t}\right)}{2}\mathrm{dt}=\int (1-\mathrm{cost})\mathrm{dt}$$$$=\mathrm{t}-\mathrm{sint}+\mathrm{c}=\mathrm{arcosx}-\sqrt{1-{\mathrm{x}}^2}+\mathrm{c}=\mathrm{arcsinx}-\sqrt{1-{\mathrm{x}}^2}+\mathrm{C}$$$$$$

Answered by 1549442205PVT last updated on 28/Oct/20

$$\int \sqrt{\frac{1-x}{1+x}}dx=?,xϵ(-1,1)$$$$\mathrm{Put}\mathrm{x}=\cos \phi \Rightarrow \mathrm{dx}=-\sin \phi \mathrm{d}\phi$$$$\mathrm{F}=-\int \sqrt{\frac{1-\cos \phi }{1+\cos \phi }}.\sin \phi \mathrm{d}\phi$$$$=-\int \sqrt{\frac{{2\sin }^2\frac{\phi }{2}}{{2\cos }^2\frac{\phi }{2}}}.\sin \phi \mathrm{d}\phi$$$$=-\int \tan \frac{\phi }{2}.2\sin \frac{\phi }{2}\cos \frac{\phi }{2}\mathrm{d}\phi$$$$=-2\int {\sin }^2\frac{\phi }{2}\mathrm{d}\phi =-\int (1-\cos \phi )\mathrm{d}\phi$$$$=-\phi +\sin \phi =-{\cos }^{-1}\left(\mathrm{x}\right)+\sqrt{1-{\mathrm{x}}^2}+\mathrm{C}$$

Answered by bobhans last updated on 28/Oct/20

$$\int \sqrt{\frac{1-x}{1+x}}dx=?,x\in (-1,1)$$$$K=\int \sqrt{\frac{1-x}{1+x}}dx[let\sqrt{1+x}=1+t]$$$$1+x=1+2t+t^2\Rightarrow dx=(2+2t)dt$$$$K=\int \frac{\sqrt{1-2t-t^2}}{1+t}(2+2t)dt=$$$$2\int \sqrt{2-{(1+t)}^2}dt=2\int \sqrt{{\left(\sqrt{2}\right)}^2-{(1+t)}^2}dt$$$$bysubstituting1+t=\sqrt{2}\sin w$$$$K=4\int {\cos }^{2}wdw=4\int (\frac{1}{2}+\frac{1}{2}\cos 2w)dw$$$$K=2w+\sin 2w+c$$$$K=2\mathrm{arc}\sin \left(\frac{1+t}{\sqrt{2}}\right)+2\left(\frac{1+t}{\sqrt{2}}\right)\left(\frac{\sqrt{1-2t-t^2}}{\sqrt{2}}\right)+c$$$$K=2\mathrm{arc}\sin \left(\sqrt{\frac{1+x}{2}}\right)+\sqrt{1-x^2}+c$$$$$$

Answered by Ar Brandon last updated on 28/Oct/20

$$\mathcal{I}=\int \sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\mathrm{dx}=\int \frac{1-\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}$$$$=\int \frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}-\int \frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}$$$$=\mathrm{Arcsin}\left(\mathrm{x}\right)+\sqrt{1-{\mathrm{x}}^2}+\mathcal{C}$$

Answered by Dwaipayan Shikari last updated on 28/Oct/20

$$\int \frac{1-x}{\sqrt{1-x^2}}dx$$$$=\int \frac{1}{\sqrt{1-x^2}}+\frac{1}{2}\int \frac{-2x}{\sqrt{1-x^2}}dx$$$$={sin}^{-1}x+\sqrt{1-x^2}+C$$

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