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Question Number 144409 by liberty last updated on 25/Jun/21

$$\int \frac{\mathrm{dx}}{(\mathrm{x}+3)\sqrt{1-{\mathrm{x}}^2}}?$$

Answered by iloveisrael last updated on 25/Jun/21

$$\mathrm{let}\mathrm{x}=\sin \mathrm{t}$$$$\mathrm{I}=\int \frac{\cos \mathrm{t}}{(3+\sin \mathrm{t})\sqrt{1-\sin {}^2\mathrm{t}}}\mathrm{dt}$$$$\mathrm{I}=\int \frac{\mathrm{dt}}{3+\sin \mathrm{t}}$$$$\mathrm{let}\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{u}$$$$\Rightarrow \mathrm{dt}=\frac{2}{1+{\mathrm{u}}^2}\mathrm{du}$$$$\mathrm{I}=\int \left(\frac{2}{1+{\mathrm{u}}^2}\right)\left(\frac{1}{3+\frac{2\mathrm{u}}{1+{\mathrm{u}}^2}}\right)\mathrm{du}$$$$\mathrm{I}=\int \frac{2}{{3\mathrm{u}}^2+2\mathrm{u}+3}\mathrm{du}$$$$\mathrm{I}=\frac{2}{3}\int \frac{\mathrm{du}}{{\mathrm{u}}^2+\frac{2}{3}\mathrm{u}+1}$$$$\mathrm{I}=\frac{2}{3}\int \frac{\mathrm{du}}{{(\mathrm{u}+\frac{1}{3})}^2+\frac{8}{9}}$$$$\mathrm{I}=\frac{2}{3}\int \frac{\mathrm{du}}{{(\mathrm{u}+\frac{1}{3})}^2+{\left(\frac{2\sqrt{2}}{3}\right)}^2}$$

Answered by mathmax by abdo last updated on 25/Jun/21

$$\mathrm{\Psi }=\int \frac{\mathrm{dx}}{(\mathrm{x}+3)\sqrt{1-{\mathrm{x}}^2}}\mathrm{changement}\mathrm{x}=\mathrm{sint}\mathrm{give}$$$$\mathrm{\Psi }=\int \frac{\mathrm{cost}\mathrm{dt}}{(\mathrm{sint}+3)\mathrm{cost}}=\int \frac{\mathrm{dt}}{3+\mathrm{sint}}{=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{y}}\int \frac{2\mathrm{dy}}{(1+{\mathrm{y}}^2)(3+\frac{2\mathrm{y}}{1+{\mathrm{y}}^2})}$$$$=\int \frac{2\mathrm{dy}}{3+{3\mathrm{y}}^2+2\mathrm{y}}=\int \frac{2\mathrm{dy}}{{3\mathrm{y}}^2+2\mathrm{y}+3}$$$$=\frac{2}{3}\int \frac{\mathrm{dy}}{{\mathrm{y}}^2+\frac{2}{3}\mathrm{y}+1}=\frac{2}{3}\int \frac{\mathrm{dy}}{{\mathrm{y}}^2+\frac{2}{3}\mathrm{y}+\frac{1}{9}+1-\frac{1}{9}}$$$$=\frac{2}{3}\int \frac{\mathrm{dy}}{{(\mathrm{y}+\frac{1}{3})}^2+\frac{8}{9}}{=}_{\mathrm{y}+\frac{1}{3}=\frac{2\sqrt{2}}{3}\mathrm{z}}\frac{2}{3}\times \frac{9}{8}\int \frac{1}{{\mathrm{z}}^2+1}\times \frac{2\sqrt{2}}{3}\mathrm{dz}$$$$=\frac{\sqrt{2}}{2}\mathrm{arcctanz}+\mathrm{K}=\frac{\sqrt{2}}{2}\arctan \left(\frac{3\mathrm{y}+1}{2\sqrt{2}}\right)+\mathrm{K}$$$$=\frac{\sqrt{2}}{2}\arctan \left(\frac{3\tan \left(\frac{\mathrm{t}}{2}\right)+1}{2\sqrt{2}}\right)+\mathrm{K}$$$$\mathrm{\Psi }=\frac{\sqrt{2}}{2}\arctan \left(\frac{3\tan \left(\frac{\mathrm{arcsinx}}{2}\right)+1}{2\sqrt{2}}\right)+\mathrm{K}$$

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