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Question Number 128775 by bramlexs22 last updated on 10/Jan/21

$$\int \frac{\mathrm{dx}}{{(1-\mathrm{x})}^2\sqrt{1-{\mathrm{x}}^2}}?$$

Answered by liberty last updated on 10/Jan/21

$$\mathrm{let}\phi =\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\Rightarrow \mathrm{x}=\frac{{\phi }^2-1}{{\phi }^2+1}\mathrm{and}\mathrm{dx}=\frac{4\phi }{{({\phi }^2+1)}^2}\mathrm{d}\phi$$$$\mathrm{Y}=\int \frac{{({\phi }^2+1)}^3}{8\phi }\times \frac{4\phi }{{({\phi }^2+1)}^2}\mathrm{d}\phi$$$$\mathrm{Y}=\frac{1}{2}\int ({\phi }^2+1)\mathrm{d}\phi =\frac{1}{2}(\frac{1}{3}{\phi }^3+\phi )+\mathrm{C}$$$$\mathrm{Y}=\frac{1}{6}(\frac{\mathrm{x}+1}{1-\mathrm{x}}+3)\sqrt{\frac{\mathrm{x}+1}{1-\mathrm{x}}}+\mathrm{C}$$$$\mathrm{Y}=\frac{2-\mathrm{x}}{3-3\mathrm{x}}\sqrt{\frac{\mathrm{x}+1}{1-\mathrm{x}}}+\mathrm{C}$$

Answered by mr W last updated on 10/Jan/21

$$x=\cos \theta$$$$\int \frac{-\sin \theta d\theta }{{(1-\cos \theta )}^2\sin \theta }$$$$=-\int \frac{d\theta }{{(1-\cos \theta )}^2}$$$$=-\int \frac{d\theta }{4{\sin }^4\frac{\theta }{2}}$$$$=-\int \frac{d\frac{\theta }{2}}{2{\sin }^4\frac{\theta }{2}}$$$$=\frac{3{\tan }^2\frac{\theta }{2}+1}{6{\tan }^3\frac{\theta }{2}}+C$$$$=\frac{3{\tan }^2\left(\frac{{\cos }^{-1}x}{2}\right)+1}{6{\tan }^3\left(\frac{{\cos }^{-1}x}{2}\right)}+C$$

Answered by mathmax by abdo last updated on 10/Jan/21

$$\int \frac{\mathrm{dx}}{{(1-\mathrm{x})}^2\sqrt{1-{\mathrm{x}}^2}}\mathrm{changement}\mathrm{x}=\mathrm{sint}\mathrm{give}$$$$\mathrm{I}=\int \frac{\mathrm{cosx}}{{(1-\mathrm{sint})}^2\mathrm{cosx}}\mathrm{dx}{=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{u}}\int \frac{2\mathrm{du}}{(1+{\mathrm{u}}^2){(1-\frac{2\mathrm{u}}{1+{\mathrm{u}}^2})}^2}$$$$=\int \frac{2\mathrm{du}}{({\mathrm{u}}^2+1){\left(\frac{{\mathrm{u}}^2+1-2\mathrm{u}}{{\mathrm{u}}^2+1}\right)}^2}=2\int \frac{{\mathrm{u}}^2+1}{{({\mathrm{u}}^2-2\mathrm{u}+1)}^2}\mathrm{du}$$$$=2\int \frac{{\mathrm{u}}^2-2\mathrm{u}+1+2\mathrm{u}}{{({\mathrm{u}}^2-2\mathrm{u}+1)}^2}\mathrm{du}=2\int \frac{\mathrm{du}}{({\mathrm{u}}^2-2\mathrm{u}+1)}+2\int \frac{2\mathrm{u}-2+2}{{({\mathrm{u}}^2-2\mathrm{u}+1)}^2}\mathrm{du}$$$$=2\int \frac{\mathrm{du}}{{(\mathrm{u}-1)}^2}+2\int \frac{2\mathrm{u}-2}{{({\mathrm{u}}^2-2\mathrm{u}+1)}^2}\mathrm{du}+4\int \frac{\mathrm{du}}{{(\mathrm{u}-1)}^4}$$$$=-\frac{2}{\mathrm{u}-1}-\frac{2}{{\mathrm{u}}^2-2\mathrm{u}+1}+4\times \frac{1}{-4+1}{(\mathrm{u}-1)}^{-4+1}+\mathrm{C}$$$$=\frac{2}{1-\mathrm{u}}-\frac{2}{{(1-\mathrm{u})}^2}+\frac{4}{3{(1-\mathrm{u})}^3}+\mathrm{C}$$$$=\frac{2}{1-\tan \left(\frac{\mathrm{t}}{2}\right)}-\frac{2}{{(1-\tan \left(\frac{\mathrm{t}}{2}\right))}^2}+\frac{4}{3(1-\tan \left(\frac{\mathrm{t}}{2}\right))}+\mathrm{C}$$$$=\frac{10}{3(1-\tan \left(\frac{\mathrm{arcsinx}}{2}\right))}-\frac{2}{{(1-\tan \left(\frac{\mathrm{arcsinx}}{2}\right))}^2}+\mathrm{C}$$

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