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Question Number 148437 by aliibrahim1 last updated on 28/Jul/21

Answered by mathmax by abdo last updated on 28/Jul/21

$$\mathrm{I}={\int }_0^1\frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}}+\sqrt{1-\mathrm{x}}}\Rightarrow \mathrm{I}={\int }_0^1\frac{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}{1+\mathrm{x}-1+\mathrm{x}}\mathrm{dx}$$$$={\int }_0^1\frac{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}{2\mathrm{x}}\mathrm{dx}=\frac{1}{2}{\lim }_{\xi \to 0}{\int }_{\xi }^1\frac{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{we}{\mathrm{haveA}}_{\xi }={\int }_{\xi }^1\frac{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}{\mathrm{x}}\mathrm{dx}={\int }_{\xi }^1\frac{\sqrt{1+\mathrm{x}}}{\mathrm{x}}\mathrm{dx}-{\int }_{\xi }^1\frac{\sqrt{1-\mathrm{x}}}{\mathrm{x}}\mathrm{dx}$$$$={\mathrm{H}}_{\xi }-{\mathrm{K}}_{\xi }$$$${\mathrm{H}}_{\xi }{=}_{\sqrt{1+\mathrm{x}}=\mathrm{t}\to 1+\mathrm{x}={\mathrm{t}}^2}{\int }_{\sqrt{1+\xi }}^{\sqrt{2}}\frac{\mathrm{t}}{{\mathrm{t}}^2-1}\left(2\mathrm{t}\right)\mathrm{dt}$$$$=2{\int }_{\sqrt{1+\xi }}^{\sqrt{2}}\frac{{\mathrm{t}}^2-1+1}{{\mathrm{t}}^2-1}\mathrm{dt}=2(\sqrt{2}-\sqrt{1+\xi })+2{\int }_{\sqrt{1+\xi }}^{\sqrt{2}}\frac{\mathrm{dt}}{(\mathrm{t}-1)(\mathrm{t}+1)}$$$$=2(\sqrt{2}-\sqrt{1+\xi })+{\int }_{\sqrt{1+\xi }}^{\sqrt{2}}(\frac{1}{\mathrm{t}-1}-\frac{1}{\mathrm{t}+1})\mathrm{dt}$$$$=2(\sqrt{2}-\sqrt{1+\xi })+{[\log \mid \frac{\mathrm{t}-1}{\mathrm{t}+1}\mid ]}_{\sqrt{1+\xi }}^{\sqrt{2}}$$$$=2(\sqrt{2}-\sqrt{1+\xi })+\log \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)-\log \left(\frac{\sqrt{1+\xi }-1}{\sqrt{1+\xi }+1}\right)$$$${\mathrm{K}}_{\xi }={\int }_{\xi }^1\frac{\sqrt{1-\mathrm{x}}}{\mathrm{x}}\mathrm{dx}{=}_{\sqrt{1-\mathrm{x}}=\mathrm{t}\to \mathrm{x}=1-{\mathrm{t}}^2}{\int }_{\sqrt{1-\xi }}^0\frac{\mathrm{t}}{1-{\mathrm{t}}^2}(-2\mathrm{t})\mathrm{dt}$$$$=2{\int }_0^{\sqrt{1-\xi }}\frac{{\mathrm{t}}^2}{1-{\mathrm{t}}^2}\mathrm{dt}=-2{\int }_0^{\sqrt{1-\xi }}\frac{{\mathrm{t}}^2-1+1}{{\mathrm{t}}^2-1}\mathrm{dt}$$$$=-2\sqrt{1-\xi }-2{\int }_0^{\sqrt{1-\xi }}\frac{\mathrm{dt}}{{\mathrm{t}}^2-1}=-2\sqrt{1-\xi }-{\int }_0^{\sqrt{1-\xi }}(\frac{1}{\mathrm{t}-1}-\frac{1}{\mathrm{t}+1})\mathrm{dt}$$$$=-2\sqrt{1-\xi }-{[\log \mid \frac{\mathrm{t}-1}{\mathrm{t}+1}\mid ]}_0^{\sqrt{1-\xi }}$$$$=-2\sqrt{1-\xi }-\log \left(\frac{1-\sqrt{1-\xi }}{\sqrt{1-\xi }+1}\right)\Rightarrow$$$${\mathrm{A}}_{\xi }=2(\sqrt{2}-1)+\log \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)-\log \left(\frac{\sqrt{1+\xi }-1}{\sqrt{1+\xi }+1}\right)$$$$2\sqrt{1-\xi }+\log \left(\frac{1-\sqrt{1-\xi }}{1+\sqrt{1-\xi }}\right)\mathrm{we}\mathrm{have}$$$$\log \left(\frac{1-\sqrt{1-\xi }}{1+\sqrt{1-\xi }}\right)-\log \left(\frac{\sqrt{1+\xi }-1}{\sqrt{1+\xi }+1}\right)$$$$\log \left(\frac{1-\sqrt{1-\xi }}{1+\sqrt{1-\xi }}\times \frac{\sqrt{1+\xi }+1}{\sqrt{1+\xi }-1}\right)$$$$1-\sqrt{1-\xi }\sim 1-(1-\frac{\xi }{2})=\frac{\xi }{2}$$$$\sqrt{1+\xi }-1\sim 1+\frac{\xi }{2}-1=\frac{\xi }{2}\Rightarrow \log (....)\sim \log \left(\frac{\sqrt{1+\xi }+1}{1+\sqrt{1-\xi }}\right)\to 0(\xi \to 0)\Rightarrow$$$${\lim }_{\xi \to 0}{\mathrm{A}}_{\xi }=2\sqrt{2}-2+\log (\sqrt{2}-1)-\log (\sqrt{2}+1)+2$$$$=2\sqrt{2}+\log (\sqrt{2}-1)-\log (\sqrt{2}+1)\Rightarrow$$$$\mathrm{I}=\sqrt{2}+\frac{1}{2}\log (\sqrt{2}-1)-\frac{1}{2}\log (\sqrt{2}+1)$$

Commented by aliibrahim1 last updated on 29/Jul/21

$$thxsir$$$$$$$$$$

Answered by mathmax by abdo last updated on 28/Jul/21

$$\mathrm{la}\mathrm{plus}\mathrm{part}\mathrm{pense}\mathrm{que}\mathrm{cette}\mathrm{integrale}\mathrm{est}\mathrm{simple}\mathrm{mais}\mathrm{en}\mathrm{realite}$$$$\mathrm{il}\mathrm{n}\mathrm{est}\mathrm{pas}\mathrm{simple}...$$

Commented by puissant last updated on 29/Jul/21

$$\mathrm{prof}\mathrm{en}\mathrm{posant}\mathrm{x}=\cos \left(2\theta \right)ext\text{c}c\mathrm{ca}\mathrm{devient}$$$$\mathrm{beaucoup}\mathrm{simple}..\mathrm{en}\mathrm{fait}{\mathrm{c}}^{\prime }\mathrm{est}\mathrm{une}$$$$\mathrm{autre}\mathrm{methode}...$$

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