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Question Number 109658 by 150505R last updated on 24/Aug/20

Answered by mathmax by abdo last updated on 25/Aug/20

$$\mathrm{A}={\int }_0^{\frac{\mathrm{e}}{\pi }}\frac{\arctan \left(\frac{\pi \mathrm{x}}{\mathrm{e}}\right)}{\pi \mathrm{x}+\mathrm{e}}\mathrm{dx}\mathrm{changement}\frac{\pi \mathrm{x}}{\mathrm{e}}=\mathrm{t}\mathrm{give}$$$$\mathrm{A}={\int }_0^1\frac{\arctan \left(\mathrm{t}\right)}{\mathrm{et}+\mathrm{e}}.\frac{\mathrm{e}}{\pi }\mathrm{dt}=\frac{1}{\pi }{\int }_0^1\frac{\arctan \left(\mathrm{t}\right)}{\mathrm{t}+1}\mathrm{dt}\Rightarrow \pi \mathrm{A}={\int }_0^1\frac{\arctan \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}$$$${=}_{\mathrm{by}\mathrm{parts}}{\left[\ln (1+\mathrm{t})\arctan \left(\mathrm{t}\right)\right]}_0^1-{\int }_0^1\frac{\ln (1+\mathrm{t})}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$=\frac{\pi }{4}\ln \left(2\right)-{\int }_0^1\frac{\ln (1+\mathrm{t})}{1+{\mathrm{t}}^2}\mathrm{dt}\mathrm{but}{\int }_0^1\frac{\ln (1+\mathrm{t})}{1+{\mathrm{t}}^2}\mathrm{dt}{=}_{\mathrm{t}=\tan \theta }$$$${\int }_0^{\frac{\pi }{4}}\frac{\ln (1+\tan \theta )}{1+{\tan }^2\theta }(1+{\tan }^2\theta )\mathrm{d}\theta ={\int }_0^{\frac{\pi }{4}}\ln (1+\tan \theta )\mathrm{d}\theta =\frac{\pi }{8}\ln \left(2\right)\left(\mathrm{proved}\right)$$$$\Rightarrow \pi \mathrm{A}=\frac{\pi }{4}\ln \left(2\right)-\frac{\pi }{8}\ln \left(2\right)=\frac{\pi }{8}\ln \left(2\right)\Rightarrow ★\mathrm{A}=\frac{\ln \left(2\right)}{8}★$$$$$$

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