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Question Number 144241 by mathmax by abdo last updated on 23/Jun/21

$$\mathrm{calculate}{\int }_0^{4\pi }\frac{\mathrm{dx}}{{(2+\mathrm{cosx})}^2}$$

Answered by Ar Brandon last updated on 05/Jul/21

$$\mathrm{I}={\int }_0^{4\pi }\frac{\mathrm{dx}}{{(2+\mathrm{cosx})}^2}=4{\int }_0^{\pi }\frac{\mathrm{dx}}{{(2+\mathrm{cosx})}^2},\mathrm{t}=\tan \frac{\mathrm{x}}{2}$$$$=4{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{(2+\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}^2}\cdot \frac{2}{1+{\mathrm{t}}^2}=8{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2+1}{{({\mathrm{t}}^2+3)}^2}\mathrm{dt}$$$$=8{\int }_0^{\mathrm{\infty }}\{\frac{1}{{\mathrm{t}}^2+3}-\frac{2}{{({\mathrm{t}}^2+3)}^2}\}\mathrm{dt}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+{\mathrm{a}}^2}=\frac{\pi }{\mathrm{a}}\Rightarrow \mathrm{f}{}^{\prime }\left(\mathrm{a}\right)=-{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{a}}{{({\mathrm{t}}^2+{\mathrm{a}}^2)}^2}=-\frac{\pi }{{\mathrm{a}}^2}$$$$\mathrm{I}=8(\frac{\pi }{\sqrt{3}}-\frac{\pi }{3\sqrt{3}})=8\left(\frac{2\pi }{3\sqrt{3}}\right)=\frac{16\sqrt{3}\pi }{9}$$

Answered by mathmax by abdo last updated on 25/Jun/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{4\pi }\frac{\mathrm{dx}}{\mathrm{a}+\mathrm{cosx}}\mathrm{with}\mathrm{a}>1\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_0^{4\pi }\frac{\mathrm{dx}}{{(\mathrm{a}+\mathrm{cosx})}^2}\Rightarrow$$$$-{\mathrm{f}}^{{}^{\prime }}\left(2\right)={\int }_0^{4\pi }\frac{\mathrm{dx}}{{(2+\mathrm{cosx})}^2}\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{2\pi }\frac{\mathrm{dx}}{\mathrm{a}+\mathrm{cosx}}+{\int }_{2\pi }^{4\pi }\frac{\mathrm{dx}}{\mathrm{a}+\mathrm{cosx}}(\to \mathrm{x}=2\pi +\mathrm{t})$$$$=2{\int }_0^{2\pi }\frac{\mathrm{dx}}{\mathrm{a}+\mathrm{cosx}}{=}_{{\mathrm{e}}^{\mathrm{ix}}=\mathrm{z}}2{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{\mathrm{iz}(\mathrm{a}+\frac{\mathrm{z}+{\mathrm{z}}^{-1}}{2})}$$$$=4{\int }_{\mid \mathrm{z}\mid =1}\frac{-\mathrm{idz}}{\mathrm{z}(2\mathrm{a}+\mathrm{z}+{\mathrm{z}}^{-1})}=-4\mathrm{i}{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{2\mathrm{az}+{\mathrm{z}}^2+1}$$$$\phi \left(\mathrm{z}\right)=\frac{1}{{\mathrm{z}}^2+2\mathrm{az}+1}$$$${\mathrm{\Delta }}^{{}^{\prime }}={\mathrm{a}}^2-1>0\Rightarrow {\mathrm{z}}_1=-\mathrm{a}+\sqrt{{\mathrm{a}}^2-1}\mathrm{and}{\mathrm{z}}_2=-\mathrm{a}-\sqrt{{\mathrm{a}}^2-1}$$$$\mid {\mathrm{z}}_1\mid -1=\mid \sqrt{{\mathrm{a}}^2-1}-\mathrm{a}\mid -1=\mathrm{a}-\sqrt{{\mathrm{a}}^2-1}-1\Rightarrow \phi \left(\mathrm{z}\right)=\frac{1}{(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)}$$$$=\mathrm{a}-1-\sqrt{{\mathrm{a}}^2-1}\mathrm{and}{(\mathrm{a}-1)}^2-{\mathrm{a}}^2+1={\mathrm{a}}^2-2\mathrm{a}+1-{\mathrm{a}}^2+1$$$$=2-2\mathrm{a}=2(1-\mathrm{a})<0\mathrm{and}\mid {\mathrm{z}}_2\mid -1=\mathrm{a}+\sqrt{{\mathrm{a}}^2-1}-1>0\Rightarrow$$$${\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,{\mathrm{z}}_1)=2\mathrm{i}\pi \times \frac{1}{{\mathrm{z}}_1-{\mathrm{z}}_2}=\frac{2\mathrm{i}\pi }{2\sqrt{{\mathrm{a}}^2-1}}=\frac{\mathrm{i}\pi }{\sqrt{{\mathrm{a}}^2-1}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=(-4\mathrm{i})\times \frac{\mathrm{i}\pi }{\sqrt{{\mathrm{a}}^2-1}}=\frac{4\pi }{\sqrt{{\mathrm{a}}^2-1}}=4\pi {({\mathrm{a}}^2-1)}^{-\frac{1}{2}}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-2\pi \left(2\mathrm{a}\right){({\mathrm{a}}^2-1)}^{-\frac{3}{2}}=\frac{-4\pi \mathrm{a}}{({\mathrm{a}}^2-1)\sqrt{{\mathrm{a}}^2-1}}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(2\right)=\frac{-8\pi }{3\sqrt{3}}\Rightarrow {\int }_0^{4\pi }\frac{\mathrm{dx}}{{(2+\mathrm{cosx})}^2}=\frac{8\pi }{3\sqrt{3}}$$

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