Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 28882 by abdo imad last updated on 31/Jan/18

$$find{\int }_{-\pi }^{\pi }\frac{2dt}{2+sint+cost}.$$

Commented by abdo imad last updated on 02/Feb/18

$$letusethech.e^{it}=z$$$$I={\int }_{\mid z\mid =1}\frac{2}{2+\frac{z-z^{-}}{2i}+\frac{z+z^{-1}}{2}}\frac{dz}{iz}$$$$I={\int }_{\mid z\mid =1}\frac{1}{iz(1+\frac{z-z^{-1}}{4i}+\frac{z+z^{-1}}{4})}dz$$$$I={\int }_{\mid z\mid =1}\frac{dz}{z(i+\frac{z-z^{-1}}{4}+\frac{i(z+z^{-1})}{4})}$$$$I={\int }_{\mid z\mid =1}\frac{4dz}{z(4i+z-z^{-1}+iz+{iz}^{-1})}$$$$I={\int }_{\mid z\mid =1}\frac{4dz}{4iz+z^2-1+{iz}^2+i}$$$$={\int }_{\mid z\mid =1}\frac{4dz}{(1+i)z^2+4iz-1+i}letintroducethecomplex$$$$functionw\left(z\right)=\frac{4}{(1+i)z^2+4iz-1+i}polesofw?$$$${\mathrm{\Delta }}^{{}^{\prime }}={\left(2i\right)}^2-(1+i)(-1+i)=-4-(-2)=-2={\left(i\sqrt{2}\right)}^2$$$$z_1=\frac{-2i+i\sqrt{2}}{1+i}and{z}_2=\frac{-2i-i\sqrt{2}}{1+i}$$$$\mid z_1\mid =\frac{\mid -2i+i\sqrt{2}\mid }{\mid 1+i\mid }=\frac{2-\sqrt{2}}{\sqrt{2}}=\sqrt{2}-1and\mid z_1\mid -1=\sqrt{2}-2<0$$$$so\mid z_1\mid <1$$

Commented by abdo imad last updated on 02/Feb/18

$$\mid z_2\mid =\frac{2+\sqrt{2}}{\sqrt{2}}=\sqrt{2}+1\Rightarrow \mid z_2\mid -1=\sqrt{2}>0\left(toeliminatefrmresidus\right)$$$${\int }_{\mid z\mid =1}w\left(z\right)dz=2i\pi Res(w,z_1)but$$$$w\left(z\right)=\frac{4}{(1+i)(z-z_1)(z-z_2)}\Rightarrow$$$$Res(w,z_1)=\frac{4}{(1+i)(z_1-z_2)}=\frac{4}{(1+i)\frac{2i\sqrt{2}}{1+i}}=\frac{2}{i\sqrt{2}}=\frac{\sqrt{2}}{i}$$$${\int }_{\mid z\mid =1}w\left(z\right)dz=2i\pi .\frac{\sqrt{2}}{i}=2\pi \sqrt{2}.\Rightarrow I=2\pi \sqrt{2}.$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com