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Question Number 111762 by mathmax by abdo last updated on 04/Sep/20

$$\mathrm{caoculate}{\int }_0^{\frac{\pi }{4}}\ln (1+2\mathrm{tanx})\mathrm{dx}$$

Answered by mathmax by abdo last updated on 12/Sep/20

$$\mathrm{A}={\int }_0^{\frac{\pi }{4}}\ln (1+2\mathrm{tanx})\mathrm{dx}\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{4}}\ln (1+\mathrm{atanx})\mathrm{dx}\mathrm{witha}>0$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{4}}\frac{\mathrm{tanx}}{1+\mathrm{atanx}}\mathrm{dx}{=}_{\mathrm{tanx}=\mathrm{t}}{\int }_0^1\frac{\mathrm{t}}{(1+at)(1+t^2)}dtletdecompose$$$$u\left(t\right)=\frac{t}{(at+1)(t^2+1)}\Rightarrow \mathrm{u}\left(\mathrm{t}\right)=\frac{\alpha }{\mathrm{at}+1}+\frac{\beta \mathrm{t}+\lambda }{{\mathrm{t}}^2+1}$$$$\alpha =\frac{-1}{\mathrm{a}(\frac{1}{{\mathrm{a}}^2}+1)}=\frac{-{\mathrm{a}}^2}{\mathrm{a}(1+{\mathrm{a}}^2)}=-\frac{\mathrm{a}}{{\mathrm{a}}^2+1}$$$${\lim }_{\mathrm{t}\to +\mathrm{\infty }}\mathrm{tu}\left(\mathrm{t}\right)=0=\frac{\alpha }{\mathrm{a}}+\beta \Rightarrow \beta =\frac{1}{{\mathrm{a}}^2+1}$$$$\mathrm{u}\left(0\right)=0=\alpha +\lambda \Rightarrow \lambda =\frac{\mathrm{a}}{{\mathrm{a}}^2+1}\Rightarrow \mathrm{u}\left(\mathrm{t}\right)=-\frac{\mathrm{a}}{({\mathrm{a}}^2+1)(\mathrm{at}+1)}$$$$+\frac{\frac{\mathrm{t}}{{\mathrm{a}}^2+1}+\frac{\mathrm{a}}{{\mathrm{a}}^2+1}}{{\mathrm{t}}^2+1}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-\frac{\mathrm{a}}{{\mathrm{a}}^2+1}{\int }_0^1\frac{\mathrm{dt}}{\mathrm{at}+1}+\frac{1}{{\mathrm{a}}^2+1}{\int }_0^1\frac{\mathrm{t}+\mathrm{a}}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$=-\frac{1}{{\mathrm{a}}^2+1}{\left[\ln (\mathrm{at}+1)\right]}_0^1+\frac{1}{2({\mathrm{a}}^2+1)}{\int }_0^1\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}\mathrm{dt}+\frac{\mathrm{a}}{{\mathrm{a}}^2+1}{\left[\mathrm{arctant}\right]}_0^1$$$$=-\frac{\ln (\mathrm{a}+1)}{{\mathrm{a}}^2+1}+\frac{\ln \left(2\right)}{2({\mathrm{a}}^2+1)}+\frac{\pi \mathrm{a}}{4({\mathrm{a}}^2+1)}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=-{\int }_0^{\mathrm{a}}\frac{\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}+\frac{\ln 2}{2}{\int }_0^{\mathrm{a}}\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}+\frac{\pi }{4}{\int }_0^{\mathrm{a}}\frac{\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}+\mathrm{c}$$$$=-{\int }_0^{\mathrm{a}}\frac{\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}+\frac{\ln 2}{2}\arctan \left(\mathrm{a}\right)+\frac{\pi }{8}\ln (1+{\mathrm{a}}^2)+\mathrm{c}$$$$\mathrm{c}=\mathrm{f}\left(0\right)=0\mathrm{and}\mathrm{A}=\mathrm{f}\left(2\right)=-{\int }_0^2\frac{\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}+\frac{\ln 2}{2}\arctan 2+\frac{\pi }{8}\ln 5$$$$\mathrm{rest}\mathrm{calculus}\mathrm{of}{\int }_0^2\frac{\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}....\mathrm{be}\mathrm{continued}...$$

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