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Question Number 95844 by mathmax by abdo last updated on 28/May/20

$$\mathrm{cacuate}{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\ln (1+\mathrm{a}{\cos }^2\mathrm{t})\mathrm{dt}\mathrm{with}\mid \mathrm{a}\mid <1$$

Answered by mathmax by abdo last updated on 29/May/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\ln (1+{\mathrm{acos}}^2\mathrm{t})\mathrm{dt}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=2{\int }_0^{\frac{\pi }{4}}\ln (1+{\mathrm{acos}}^2\mathrm{t})\mathrm{dt}\Rightarrow$$ $${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=2{\int }_0^{\frac{\pi }{4}}\frac{{\cos }^2\mathrm{t}}{1+\mathrm{a}{\cos }^2\mathrm{t}}\mathrm{dt}=\frac{2}{\mathrm{a}}{\int }_0^{\frac{\pi }{4}}\frac{1+{\mathrm{acos}}^2\mathrm{t}-1}{1+{\mathrm{acos}}^2\mathrm{t}}\mathrm{dt}$$ $$=\frac{2}{\mathrm{a}}\times \frac{\pi }{4}-\frac{2}{\mathrm{a}}{\int }_0^{\frac{\pi }{4}}\frac{\mathrm{dt}}{1+{\mathrm{acos}}^2\mathrm{t}}=\frac{\pi }{2\mathrm{a}}-\frac{2}{\mathrm{a}}{\int }_0^{\frac{\pi }{4}}\frac{\mathrm{dt}}{1+{\mathrm{acos}}^2\mathrm{t}}\mathrm{we}\mathrm{have}$$ $${\int }_0^{\frac{\pi }{4}}\frac{\mathrm{dt}}{1+{\mathrm{acos}}^2\mathrm{t}}={\int }_0^{\frac{\pi }{4}}\frac{\mathrm{dt}}{1+\mathrm{a}\frac{1+\cos \left(2\mathrm{t}\right)}{2}}={\int }_0^{\frac{\pi }{4}}\frac{2\mathrm{dt}}{2+\mathrm{a}+\mathrm{acos}\left(2\mathrm{t}\right)}{=}_{2\mathrm{t}=\mathrm{u}}$$ $$={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{du}}{2+\mathrm{a}+\mathrm{acosu}}{=}_{\tan \left(\frac{\mathrm{u}}{2}\right)=\mathrm{z}}{\int }_0^1\frac{2\mathrm{dz}}{(1+{\mathrm{z}}^2)(2+\mathrm{a}+\mathrm{a}\frac{1-{\mathrm{z}}^2}{1+{\mathrm{z}}^2})}$$ $$={\int }_0^1\frac{2\mathrm{dz}}{(2+\mathrm{a})(1+{\mathrm{z}}^2)+\mathrm{a}-{\mathrm{az}}^2}={\int }_0^1\frac{2\mathrm{dz}}{2+\mathrm{a}+(2+\mathrm{a}-\mathrm{a}){\mathrm{z}}^2}={\int }_0^1\frac{2\mathrm{dz}}{2+\mathrm{a}+{2\mathrm{z}}^2}$$ $$=\frac{2}{2+\mathrm{a}}{\int }_0^1\frac{\mathrm{dz}}{1+\frac{2}{2+\mathrm{a}}{\mathrm{z}}^2}{=}_{\sqrt{\frac{2}{2+\mathrm{a}}}\mathrm{z}=\alpha }\frac{2}{2+\mathrm{a}}{\int }_0^{\sqrt{\frac{2}{2+\mathrm{a}}}}\frac{\sqrt{\frac{2+\mathrm{a}}{2}}\mathrm{d}\alpha }{1+{\alpha }^2}$$ $$=\frac{2}{\sqrt{2}}\times \frac{1}{\sqrt{2+\mathrm{a}}}{\left[\arctan \left(\alpha \right)\right]}_0^{\sqrt{\frac{2}{2+\mathrm{a}}}}=\frac{\sqrt{2}}{\sqrt{2+\mathrm{a}}}\arctan \left(\sqrt{\frac{2}{2+\mathrm{a}}}\right)\Rightarrow$$ $${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\frac{\pi }{2\mathrm{a}}-\frac{2}{\mathrm{a}}\times \frac{\sqrt{2}}{\sqrt{2+\mathrm{a}}}\arctan \left(\sqrt{\frac{2}{2+\mathrm{a}}}\right)\Rightarrow$$ $$\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{2}\ln \mid \mathrm{a}\mid -2\int \frac{1}{\mathrm{a}}\sqrt{\frac{2}{2+\mathrm{a}}}\arctan \left(\sqrt{\frac{2}{2+\mathrm{a}}}\right)\mathrm{da}+\mathrm{c}$$ $$\mathrm{changement}\sqrt{\frac{2}{2+\mathrm{a}}}=\mathrm{t}\mathrm{give}\frac{2}{2+\mathrm{a}}={\mathrm{t}}^2\Rightarrow \frac{2+\mathrm{a}}{2}=\frac{1}{{\mathrm{t}}^2}\Rightarrow 2+\mathrm{a}=\frac{2}{{\mathrm{t}}^2}\Rightarrow \mathrm{a}=\frac{2}{{\mathrm{t}}^2}-2\Rightarrow$$ $$\mathrm{da}=2\left(\frac{-2\mathrm{t}}{{\mathrm{t}}^4}\right)=\frac{-4}{{\mathrm{t}}^3}\Rightarrow$$ $$\int \frac{1}{\mathrm{a}}\sqrt{\frac{2}{2+\mathrm{a}}}\arctan \left(\sqrt{\frac{2}{2+\mathrm{a}}}\right)\mathrm{da}=\int \frac{1}{\frac{2}{{\mathrm{t}}^2}-2}\times \mathrm{t}\times \arctan \left(\mathrm{t}\right)\times \left(\frac{-4}{{\mathrm{t}}^3}\right)\mathrm{dt}$$ $$=-4\int \frac{1}{{\mathrm{t}}^2(\frac{2}{{\mathrm{t}}^2}-2)}\arctan \left(\mathrm{t}\right)\mathrm{dt}=-4\int \frac{\arctan \left(\mathrm{t}\right)}{2-{2\mathrm{t}}^2}\mathrm{dt}$$ $$=2\int \frac{\mathrm{arctant}}{{\mathrm{t}}^2-1}\mathrm{dt}=\int (\frac{1}{\mathrm{t}-1}-\frac{1}{\mathrm{t}+1})\arctan \left(\mathrm{t}\right)\mathrm{dt}$$ $$=\int \frac{\mathrm{arctant}}{\mathrm{t}-1}\mathrm{dt}-\int \frac{\mathrm{arctant}}{\mathrm{t}+1}\mathrm{dt}(\to \mathrm{t}=-\mathrm{u})$$ $$=\int \frac{\mathrm{arctant}}{\mathrm{t}-1}\mathrm{dt}-\int \frac{-\mathrm{arctanu}}{-\mathrm{u}+1}(-\mathrm{du})=\int \frac{\arctan \left(\mathrm{t}\right)}{\mathrm{t}-1}\mathrm{dt}+\int \frac{\mathrm{arctanu}}{\mathrm{u}-1}\mathrm{du}$$ $$=2\int \frac{\arctan \left(\mathrm{t}\right)}{\mathrm{t}-1}\mathrm{dt}....\mathrm{be}\mathrm{continued}....$$ $$$$

Commented bymaths mind last updated on 21/Jun/20

$$\int \frac{arctan\left(x\right)}{x-1}dx=arctan\left(x\right)ln\mid x-1\mid -\int \frac{ln(x-1)}{x^{(2}+1}dx$$ $$\int \frac{ln(x-1)}{x^2+1}dx=\frac{1}{2i}\int \frac{ln(x-1)(x+i-(x-i))}{(x+i)(x-i)}dx$$ $$=\frac{1}{2i}\int \frac{ln(x-1)}{x-i}-\frac{1}{2i}\int \frac{ln(x-1)}{(x+i)}dx$$ $$j=x-i\Rightarrow dj=dx$$ $$\int \frac{ln(j+i-1)}{j}dj=\int \frac{ln(i-1)ln(1-\frac{j}{1-i})}{\frac{j}{1-i}}d\left(\frac{j}{1-i}\right)$$ $$=-ln(i-1){Li}_2\left(\frac{j}{1-i}\right)$$ $$\int \frac{ln(x-1)}{x+i}dx=\int \frac{ln(y-i-1)}{y}dy=\int \frac{ln(-i-1)ln(1-\frac{y}{i+1})}{\frac{y}{i+1}}.d\left(\frac{y}{1+i}\right)$$ $$=-ln(-1-i){Li}_2\left(\frac{y}{i+1}\right)$$ $$weget-\frac{1}{2i}\{ln(i-1){Li}_2\left(\frac{x-i}{1-i}\right)-ln(-1-i){Li}_2\left(\frac{x+i}{i+1}\right)\}$$ $$$$

Answered by mathmax by abdo last updated on 29/May/20

$$\mathrm{let}\mathrm{determine}\mathrm{f}\left(\mathrm{a}\right)={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\ln (1+{\mathrm{acos}}^2\mathrm{x})\mathrm{dx}\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}$$ $$\mathrm{f}\left(\mathrm{a}\right)=2{\int }_0^{\frac{\pi }{4}}\ln (1+\mathrm{a}\times \frac{1+\cos \left(2\mathrm{x}\right)}{2})\mathrm{dx}=2{\int }_0^{\frac{\pi }{4}}\ln (2+\mathrm{a}+\mathrm{acos}\left(2\mathrm{x}\right))\mathrm{dx}$$ $$-2\times \frac{\pi }{4}\ln \left(2\right)+2{\int }_0^{\frac{\pi }{4}}\ln \{(2+\mathrm{a})(1+\frac{\mathrm{a}}{2+\mathrm{a}}\cos \left(2\mathrm{x}\right)\}\mathrm{dx}$$ $$=-\frac{\pi }{2}\ln \left(2\right)+\frac{\pi }{2}\ln (2+\mathrm{a})+2{\int }_0^{\frac{\pi }{4}}\ln (1+\frac{\mathrm{a}}{2+\mathrm{a}}\cos \left(2\mathrm{x}\right))\mathrm{dx}$$ $$\mathrm{we}\mathrm{have}{\ln }^{{}^{\prime }}(1+\mathrm{u})=\frac{1}{1+\mathrm{u}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}}\Rightarrow \ln (1+\mathrm{u})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}(\mathrm{c}=0)$$ $$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}{\mathrm{u}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow {\int }_0^{\frac{\pi }{4}}\ln (1+\frac{\mathrm{a}}{\mathrm{a}+2}\cos \left(2\mathrm{x}\right))\mathrm{dx}$$ $${=}_{2\mathrm{x}=\mathrm{t}}{\int }_0^{\frac{\pi }{2}}\ln (1+\frac{\mathrm{a}}{\mathrm{a}+2}\mathrm{cost})\frac{\mathrm{dt}}{2}=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\left(\frac{\mathrm{a}}{\mathrm{a}+2}\mathrm{cost}\right)}^{\mathrm{n}}\right)\mathrm{dt}$$ $$=\frac{1}{2}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\left(\frac{\mathrm{a}}{\mathrm{a}+2}\right)}^{\mathrm{n}}{\int }_0^{\frac{\pi }{2}}{\cos }^{\mathrm{n}}\mathrm{t}\mathrm{dt}=\frac{1}{2}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\mathrm{w}}_{\mathrm{n}}{\left(\frac{\mathrm{a}}{\mathrm{a}+2}\right)}^{\mathrm{n}}$$ $$\mathrm{with}{\mathrm{w}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{2}}{\cos }^{\mathrm{n}}\mathrm{t}\mathrm{dt}\left(\mathrm{wsllis}\mathrm{integral}\right)\Rightarrow$$ $$\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{2}(\ln (2+\mathrm{a})-\ln 2)+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\mathrm{w}}_{\mathrm{n}}{\left(\frac{\mathrm{a}}{\mathrm{a}+2}\right)}^{\mathrm{n}}$$ $$....\mathrm{be}\mathrm{continued}...$$

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