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Question Number 109378 by shahria14 last updated on 23/Aug/20

Answered by john santu last updated on 23/Aug/20

$$\frac{\approx JS\approx }{◼★◼}$$$$\sqrt{\frac{1+x}{1-x}.\frac{1+x}{1+x}}=\frac{1+x}{\sqrt{1-x^2}}$$$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \frac{1}{\sqrt{1-x^2}}dx+\int \frac{x}{\sqrt{1-x^2}}dx$$$$=\mathrm{arc}\sin \left(x\right)-\frac{1}{2}\int \frac{d(1-x^2)}{\sqrt{1-x^2}}$$$$=\mathrm{arc}\sin \left(x\right)-\frac{1}{2}.2\sqrt{1-x^2}+c$$$$=\mathrm{arc}\sin \left(x\right)-\sqrt{1-x^2}+c$$

Commented by shahria14 last updated on 23/Aug/20

thanks

Answered by 1549442205PVT last updated on 23/Aug/20

$$\mathrm{Put}\mathrm{x}=\cos \phi ,\phi \in [0;\pi ]\Rightarrow \mathrm{dx}=-\sin \phi \mathrm{d}\phi$$$$\int \sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\mathrm{dx}=\int \sqrt{\frac{1+\cos \phi }{1-\cos \phi }}(-\sin \phi )\mathrm{d}\phi$$$$=-\int \sqrt{\frac{{2\cos }^2\frac{\phi }{2}}{{2\sin }^2\frac{\phi }{2}}.}2\sin \frac{\phi }{2}\cos \frac{\phi }{2}\mathrm{d}\phi$$$$=-\int {2\cos }^2\frac{\phi }{2}\mathrm{d}\phi =-\int (1+\cos \phi )\mathrm{d}\phi$$$$=-\phi +\sin \phi =-{\cos }^{-1}\left(\mathrm{x}\right)+\sqrt{1-{\mathrm{x}}^2}+\mathrm{C}$$

Answered by Dwaipayan Shikari last updated on 23/Aug/20

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

Answered by mathmax by abdo last updated on 23/Aug/20

$$\mathrm{let}\mathrm{I}=\int \sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\cos \left(\theta \right)\Rightarrow$$$$\mathrm{I}=\int \sqrt{\frac{{2\cos }^2\left(\frac{\theta }{2}\right)}{{2\sin }^2\left(\frac{\theta }{2}\right)}}(-\sin \theta )\mathrm{d}\theta =-\int \frac{\cos \left(\frac{\theta }{2}\right)}{\mathrm{sn}\left(\frac{\theta }{2}\right)}2\cos \left(\frac{\theta }{2}\right)\sin \left(\frac{\theta }{2}\right)\mathrm{d}\theta$$$$=-\int {2\cos }^2\left(\frac{\theta }{2}\right)\mathrm{d}\theta =-\int (1+\cos \theta )\mathrm{d}\theta =-\theta -\sin \theta +\mathrm{C}$$$$=-\mathrm{arcosx}-\sqrt{1-{\mathrm{x}}^2}+\mathrm{C}$$$$$$

Commented by mathmax by abdo last updated on 23/Aug/20

$$\mathrm{another}\mathrm{method}\mathrm{I}=\int \sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}=\mathrm{t}$$$$\Rightarrow \frac{1+\mathrm{x}}{1-\mathrm{x}}={\mathrm{t}}^2\Rightarrow 1+\mathrm{x}={\mathrm{t}}^2-{\mathrm{t}}^2\mathrm{x}\Rightarrow (1+{\mathrm{t}}^2)\mathrm{x}={\mathrm{t}}^2-1\Rightarrow \mathrm{x}=\frac{{\mathrm{t}}^2-1}{{\mathrm{t}}^2+1}\Rightarrow$$$$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2\mathrm{t}({\mathrm{t}}^2+1)-2\mathrm{t}({\mathrm{t}}^2-1)}{{({\mathrm{t}}^2+1)}^2}=\frac{4\mathrm{t}}{{({\mathrm{t}}^2+1)}^2}\Rightarrow \mathrm{I}=\int \mathrm{t}\frac{4\mathrm{t}}{{({\mathrm{t}}^2+1)}^2}\mathrm{dt}$$$$=4\int \frac{{\mathrm{t}}^2+1-1}{{({\mathrm{t}}^2+1)}^2}\mathrm{dt}=4\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+1}-4\int \frac{\mathrm{dt}}{{({\mathrm{t}}^2+1)}^2}\mathrm{we}\mathrm{have}$$$$\int \frac{\mathrm{dt}}{1+{\mathrm{t}}^2}=\arctan \left(\mathrm{t}\right)+{\mathrm{c}}_0$$$$\int \frac{\mathrm{dt}}{{(1+{\mathrm{t}}^2)}^2}{=}_{\mathrm{t}=\tan \theta }\int \frac{1+{\tan }^2\theta }{{(1+{\tan }^2\theta )}^2}\mathrm{d}\theta =\int \frac{\mathrm{d}\theta }{1+{\tan }^2\theta }=\int {\cos }^2\theta \mathrm{d}\theta$$$$=\frac{1}{2}\int (1+\cos \left(2\theta \right))\mathrm{d}\theta =\frac{\theta }{2}+\frac{1}{4}\sin \left(2\theta \right)=\frac{1}{2}\arctan \left(\mathrm{t}\right)+\frac{1}{4}\times \frac{2\mathrm{t}}{1+{\mathrm{t}}^2}\Rightarrow$$$$\mathrm{I}=4\arctan \left(\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\right)-2\arctan \left(\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\right)+\frac{1}{2}\frac{\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}}{1+\frac{1+\mathrm{x}}{1-\mathrm{x}}}+\mathrm{C}$$$$=2\arctan \left(\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\right)+\frac{1}{4}(1-\mathrm{x})\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}+\mathrm{C}$$$$$$

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