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Question Number 127772 by Bird last updated on 02/Jan/21

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

Answered by mathmax by abdo last updated on 02/Jan/21

$$\mathrm{I}={\int }_0^{2\pi }\frac{\mathrm{dx}}{{(\mathrm{cosx}+2\mathrm{sinx})}^2}\Rightarrow \mathrm{I}={\int }_0^{2\pi }\frac{\mathrm{dx}}{{\cos }^2\mathrm{x}+4\mathrm{sinx}\mathrm{cosx}+{4\sin }^2\mathrm{x}}$$$$={\int }_0^{2\pi }\frac{\mathrm{dx}}{1+4\mathrm{sinxcosx}+{3\sin }^2\mathrm{x}}={\int }_0^{2\pi }\frac{\mathrm{dx}}{1+2\sin \left(2\mathrm{x}\right)+3\frac{1-\cos \left(2\mathrm{x}\right)}{2}}$$$$={\int }_0^{2\pi }\frac{2\mathrm{dx}}{2+4\sin \left(2\mathrm{x}\right)+3-3\cos \left(2\mathrm{x}\right)}={\int }_0^{2\pi }\frac{2\mathrm{dx}}{5+4\sin \left(2\mathrm{x}\right)-3\cos \left(2\mathrm{x}\right)}$$$$\underset{2\mathrm{x}=\mathrm{t}}{=}{\int }_0^{4\pi }\frac{\mathrm{dt}}{5+4\mathrm{sint}-3\mathrm{cost}}={\int }_0^{2\pi }\frac{\mathrm{dt}}{4\mathrm{sint}-3\mathrm{cost}+5}+{\int }_{2\pi }^{4\pi }\frac{\mathrm{dt}}{4\mathrm{sint}-3\mathrm{cost}+5}(\to \mathrm{t}=2\pi +\mathrm{u})$$$$=2{\int }_0^{2\pi }\frac{\mathrm{dt}}{4\mathrm{sint}-3\mathrm{cost}+5}{=}_{{\mathrm{e}}^{\mathrm{it}}=\mathrm{z}}2{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{\mathrm{iz}(4\frac{\mathrm{z}-{\mathrm{z}}^{-1}}{2\mathrm{i}}-3\frac{\mathrm{z}+{\mathrm{z}}^{-1}}{2}+5)}$$$$=2{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{\mathrm{iz}(\frac{2}{\mathrm{i}}(\mathrm{z}-{\mathrm{z}}^{-1})-\frac{3}{2}(\mathrm{z}+{\mathrm{z}}^{-1})+5)}$$$$=-2\mathrm{i}{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{\frac{2}{\mathrm{i}}({\mathrm{z}}^2-1)-\frac{3}{2}({\mathrm{z}}^2+1)+5\mathrm{z}}$$$$=2{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{2({\mathrm{z}}^2-1)-\frac{3}{2}\mathrm{i}({\mathrm{z}}^2+1)+5\mathrm{iz}}=4{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{4({\mathrm{z}}^2-1)-{3\mathrm{iz}}^2-3\mathrm{i}+10\mathrm{iz}}$$$$=4{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{{4\mathrm{z}}^2-4-{3\mathrm{iz}}^2-3\mathrm{i}+10\mathrm{iz}}$$$$=4{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{(4-3\mathrm{i}){\mathrm{z}}^2+10\mathrm{iz}-4-3\mathrm{i}}$$$$\mathrm{poles}\mathrm{of}\mathrm{w}\left(\mathrm{z}\right)=\frac{1}{(4-3\mathrm{i}){\mathrm{z}}^2+10\mathrm{iz}-4-3\mathrm{i}}$$$${\mathrm{\Delta }}^{{}^{\prime }}={\left(5\mathrm{i}\right)}^2+(4-3\mathrm{i})(4+3\mathrm{i})=-25+16+9=0\to \mathrm{one}\mathrm{root}$$$${\mathrm{z}}_0=-\frac{{\mathrm{b}}^{{}^{\prime }}}{\mathrm{a}}=-\frac{5\mathrm{i}}{4-3\mathrm{i}}=-\frac{5\mathrm{i}(4+3\mathrm{i})}{25}=-\frac{4\mathrm{i}-15}{5}\Rightarrow$$$$\mathrm{w}\left(\mathrm{z}\right)=\frac{1}{(4-3\mathrm{i}){(\mathrm{z}-{\mathrm{z}}_0)}^2}\Rightarrow {\int }_{\mid \mathrm{z}\mid =1}\mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\mathrm{w},{\mathrm{z}}_0)$$$$\mid {\mathrm{z}}_0\mid =\frac{1}{5}\sqrt{4^2+{15}^2}>1\Rightarrow \mathrm{Res}(\mathrm{w},{\mathrm{z}}_0)=0\Rightarrow \mathrm{I}=0$$

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