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Evaluate the integral $ \int_0^{2\pi}\frac{1}{3-2\cos \theta +\sin\theta}\,\mathrm d\theta. $


This must be solved by using $d \theta = dz/(iz)$ and transforming $\sin, \cos$ to complex form, but I am stuck after transforming it. It is now $\frac{1}{(2i+1)z^2 - 6i z + (2i+1)}$ I don't know how to complete it now to find the singularities and solve by the residue theorem.

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    it is now 1/([2i+1]z^2 - 6i z + [2i+1] ) i dont know how to complete it now to find the singularities and solve by resduie2012-12-18

2 Answers 2

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With complex methods we have that $I=\int_{0}^{2\pi}\frac{dt}{3-2\cos t+\sin t}=\oint_{C}\frac{1}{3-z-z^{-1}+\frac{z-z^{-1}}{2i}}\frac{dz}{iz}=\oint_{C}\frac{1}{3zi-z^2i-i+\frac{z^2-1}2}dz$ where $C$ is the unit circle.

Consider $f(z)=\frac{1}{3zi-z^2i-i+\frac{z^2-1}2}=\frac{2}{z^2(1-2i)+6zi-2i-1}$ A trivial factorisation gives $f(z)=\frac{2}{(1-2i)(z+\frac{i}{1-2i})(z+\frac{5i}{1-2i})}$ You can then compute the residues and use the residue theorem

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    $\sin t=\frac{z-z^{-1}}{2i}$2012-12-18
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Hint: try using the substitution $u=\tan \left(\frac{\theta}{2}\right)$.

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    th$i$s must be solve by usi$n$g d theta = (dz/iz) and sin , cos will be in Z form , but i am stuck after transforming it to the complex !2012-12-18