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Question Number 108416 by mathdave last updated on 16/Aug/20

Answered by mathmax by abdo last updated on 16/Aug/20

$$\mathrm{I}=\int \ln (\sqrt{1+\mathrm{x}}+\sqrt{1-\mathrm{x}})\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\cos \left(2\mathrm{t}\right)\Rightarrow$$$$\mathrm{I}=\int \ln (\sqrt{{2\cos }^2\mathrm{t}}+\sqrt{{2\sin }^2\mathrm{t}})(-2\sin \left(2\mathrm{t}\right)\mathrm{dt}$$$$=-2\int \ln (\sqrt{2}(\mathrm{cost}+\mathrm{sint})\sin \left(2\mathrm{t}\right)\mathrm{dt}$$$$=-2\ln \left(\sqrt{2}\right)\int \sin \left(2\mathrm{t}\right)\mathrm{dt}-2\int \ln (\mathrm{cost}+\mathrm{sint})\sin \left(2\mathrm{t}\right)\mathrm{dt}$$$$=\frac{1}{2}\ln \left(2\right)\cos \left(2\mathrm{t}\right)-2\int \sin \left(2\mathrm{t}\right)\ln (\mathrm{cost}+\mathrm{sint})\mathrm{dt}\mathrm{by}\mathrm{psrts}$$$$\int \sin \left(2\mathrm{t}\right)\ln (\mathrm{cost}+\mathrm{sint})\mathrm{dt}=-\frac{1}{2}\cos \left(2\mathrm{t}\right)\ln (\mathrm{cost}+\mathrm{sint})$$$$+\frac{1}{2}\int \cos \left(2\mathrm{t}\right)\frac{\mathrm{cost}-\mathrm{sint}}{\mathrm{cost}+\mathrm{sint}}\mathrm{dt}$$$$\frac{1}{2}\int \cos \left(2\mathrm{t}\right)\frac{\mathrm{cost}-\mathrm{sint}}{\mathrm{cost}+\mathrm{sint}}\mathrm{dt}=\frac{1}{2}\int \cos \left(2\mathrm{t}\right)\times \frac{{(\mathrm{cost}-\mathrm{sint})}^2}{\cos \left(2\mathrm{t}\right)}\mathrm{dt}$$$$=\frac{1}{2}\int (1-2\mathrm{cost}\mathrm{sint})\mathrm{dt}=\frac{\mathrm{t}}{2}-\frac{1}{2}\int \sin \left(2\mathrm{t}\right)\mathrm{dt}$$$$=\frac{\mathrm{t}}{2}+\frac{1}{4}\cos \left(2\mathrm{t}\right)+\mathrm{c}=\frac{1}{4}\mathrm{arcosx}+\frac{\mathrm{x}}{4}\Rightarrow$$$$\mathrm{I}=\frac{\mathrm{x}}{2}\ln \left(2\right)+\mathrm{xln}(\sqrt{\frac{1+\mathrm{x}}{2}}+\sqrt{\frac{1-\mathrm{x}}{2}})-\frac{1}{2}\mathrm{arcosx}-\frac{\mathrm{x}}{2}+\mathrm{c}$$$$\mathrm{I}=\frac{\mathrm{x}}{2}\ln \left(2\right)+\mathrm{xln}(\sqrt{1+\mathrm{x}}+\sqrt{1-\mathrm{x}})+\mathrm{xln}\left(\frac{1}{\sqrt{2}}\right)-\frac{1}{2}\mathrm{arcosx}-\frac{\mathrm{x}}{2}+\mathrm{c}$$$$=\mathrm{xln}(\sqrt{1+\mathrm{x}}+\sqrt{1-\mathrm{x}})-\frac{1}{2}(\frac{\pi }{2}-\mathrm{arcsinx})-\frac{\mathrm{x}}{2}+\mathrm{c}$$$$★\mathrm{I}=\mathrm{xln}(\sqrt{1+\mathrm{x}}+\sqrt{1-\mathrm{x}})+\frac{\mathrm{arcsinx}}{2}-\frac{\mathrm{x}}{2}+\mathrm{C}★(\mathrm{C}=\mathrm{c}-\frac{\pi }{4})$$

Answered by Sarah85 last updated on 17/Aug/20

$$\int \ln (\sqrt{1+x}+\sqrt{1-x})dx$$$$\mathrm{integration}\mathrm{by}\mathrm{parts}\mathrm{leads}\mathrm{to}$$$$x\ln (\sqrt{1+x}+\sqrt{1-x})-\int \frac{x^2-1+\sqrt{1-x^2}}{2(x^2-1)}dx$$$$\mathrm{the}\mathrm{first}\mathrm{one}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\ln {(\sqrt{1+x}+\sqrt{1-x})}^x$$$$\mathrm{the}\mathrm{second}\mathrm{one}$$$$\int \frac{1}{2\sqrt{1-x^2}}dx-\int \frac{1}{2}dx=\frac{1}{2}\arcsin x-\frac{x}{2}+C$$

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