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Question Number 119930 by Lordose last updated on 28/Oct/20

Answered by mathmax by abdo last updated on 28/Oct/20

$$\mathrm{A}={\int }_0^1\arcsin \left(\tan \theta \right)\mathrm{d}\theta \mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}$$$$\arcsin \left(\tan \theta \right)=\mathrm{x}\Rightarrow \tan \theta =\mathrm{sinx}\Rightarrow \theta =\mathrm{artan}\left(\mathrm{sinx}\right)\Rightarrow$$$$\mathrm{A}={\int }_0^{\arcsin \left(\tan 1\right)}\mathrm{x}\times \frac{\mathrm{cosx}}{1+{\sin }^2\mathrm{x}}\mathrm{dx}\Rightarrow \mathrm{A}{=}_{\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}}{\int }_0^{\tan \left(\frac{\arcsin \left(\tan 1\right)}{2}\right)}\frac{2\mathrm{arctant}\times \frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}}{1+\frac{{4\mathrm{t}}^2}{{(1+{\mathrm{t}}^2)}^2}}\times \frac{2\mathrm{dt}}{1+{\mathrm{t}}^2}$$$$=4{\int }_0^{\mathrm{c}}\frac{(1-{\mathrm{t}}^2)\mathrm{arctant}}{{(1+{\mathrm{t}}^2)}^2(1+\frac{{4\mathrm{t}}^2}{{(1+{\mathrm{t}}^2)}^2})}\mathrm{dt}=4{\int }_0^{\mathrm{c}}\frac{(1-{\mathrm{t}}^2)\mathrm{arctant}}{{(1+{\mathrm{t}}^2)}^2+{4\mathrm{t}}^2}\mathrm{dt}$$$$=4{\int }_0^{\mathrm{c}}\frac{(1-{\mathrm{t}}^2)\arctan \left(\mathrm{t}\right)}{{\mathrm{t}}^4+{2\mathrm{t}}^2+1+{4\mathrm{t}}^2}\mathrm{dt}=4{\int }_0^{\mathrm{c}}\frac{(1-{\mathrm{t}}^2)\arctan \left(\mathrm{t}\right)}{{\mathrm{t}}^4+{6\mathrm{t}}^2+1}\mathrm{dt}$$$$\mathrm{let}\phi \left(\mathrm{a}\right)={\int }_0^{\mathrm{c}}\frac{(1-{\mathrm{t}}^2)\arctan \left(\mathrm{at}\right)}{{\mathrm{t}}^4+{6\mathrm{t}}^2+1}\mathrm{dt}(\mathrm{c}=\tan \left(\frac{\arcsin \left(\tan 1\right)}{2}\right))$$$${\phi }^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\mathrm{c}}\frac{\mathrm{t}(1-{\mathrm{t}}^2)}{(1+{\mathrm{a}}^2{\mathrm{t}}^2)({\mathrm{t}}^4+{6\mathrm{t}}^2+1)}\mathrm{dt}$$$${=}_{\mathrm{at}=\mathrm{z}}{\int }_0^{\mathrm{cz}}\frac{\frac{\mathrm{z}}{\mathrm{a}}(1-\frac{{\mathrm{z}}^2}{{\mathrm{a}}^2})}{(1+{\mathrm{z}}^2)(\frac{{\mathrm{z}}^4}{{\mathrm{a}}^4}+6\frac{{\mathrm{z}}^2}{{\mathrm{a}}^2}+1)}\frac{\mathrm{dz}}{\mathrm{a}}$$$$={\int }_0^{\mathrm{cz}}\frac{\mathrm{z}({\mathrm{a}}^2-{\mathrm{z}}^2)}{(1+{\mathrm{z}}^2)({\mathrm{z}}^4+{6\mathrm{a}}^2{\mathrm{z}}^2+{\mathrm{a}}^4)}\mathrm{dz}\mathrm{let}\mathrm{decompose}$$$$\mathrm{F}\left(\mathrm{z}\right)=\frac{\mathrm{z}({\mathrm{a}}^2-{\mathrm{z}}^2)}{({\mathrm{z}}^2+1)({\mathrm{z}}^4+{6\mathrm{a}}^2{\mathrm{z}}^2+{\mathrm{a}}^4)}$$$${\mathrm{z}}^4+{6\mathrm{a}}^2{\mathrm{z}}^2+{\mathrm{a}}^4=0\to {\mathrm{\Delta }}^{{}^{\prime }}={\left({3\mathrm{a}}^2\right)}^2-{\mathrm{a}}^4={8\mathrm{a}}^4\Rightarrow$$$${\mathrm{z}}_1^2=\frac{-{3\mathrm{a}}^2+2\sqrt{2}{\mathrm{a}}^2}{1}=(2\sqrt{2}-3\mathrm{a}){\mathrm{a}}^2\Rightarrow {\mathrm{z}}_1=\stackrel{-}{+}\mathrm{i}\sqrt{3\mathrm{a}-2\sqrt{2}}$$$${\mathrm{z}}_1^2=\frac{-{3\mathrm{a}}^2-2\sqrt{2}{\mathrm{a}}^2}{1}\Rightarrow {\mathrm{z}}_2=\stackrel{-}{+}\mathrm{i}\sqrt{3\mathrm{a}+2\sqrt{2}}.....\mathrm{be}\mathrm{continued}...$$$$$$

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