$\int_{0}^{2\pi} \frac{\cos{3\theta}}{5-4\cos{\theta}}\,d{\theta}$ by using complex/real integration

Question

How to evaluate $\int_{0}^{2\pi} \frac{\cos{3\theta}}{5-4\cos{\theta}}\,d{\theta}$ by using complex integration?

I assume $z=e^{i{\theta}}$, $\frac{1}{iz}dz=d{\theta}$, $$\cos{\theta}=\frac{z+z^{-1}}{2} \quad\mbox{and}\quad \cos3{\theta}=\frac{(z+z^{-1})(z^2+z^{-2}-1)}{2}.$$ Hence $$\frac{1}{2i}\oint_{|z|=1} \frac{{(z+z^{-1})(z^2+z^{-2}-1)}}{(5-4\frac{z+z^{-1}}{2})z}\,dz$$ and I'm stuck in this. Could you give me a hints or solution?

Answer 1

You are on the right track. Now you should use the Residue theorem and find the residues of the integrand at 0 and at $1/2$ (the poles inside the unit circle. The final result should be $\frac{\pi}{12}$. $$\int_{0}^{2\pi} \frac{\cos 3\theta }{5-4\cos{\theta}}\,d{\theta}= \frac{i}{4}\int_{|z|=1}\frac{z^6+1 }{z^3(z-\frac{1}{2})(z-2)}\,dz=-\frac{\pi}{2}\left(\mbox{Res}(f,0)+\mbox{Res}(f,1/2)\right)$$ where $f$ is the integrand function.

Answer 2

Complex Method

We better use $\cos 3x=Re(e^{3xi})=Re(z^3)$ to make the life easier.

Let $z=e^{xi}$, then the integral is transformed to a contour integration with a unit circle in anti-clockwise direction.

\begin{aligned} I&=\operatorname{Re} \oint_{|z |=1}\frac{z^3}{5-\frac{4}{2}\left(z+\frac{1}{z}\right)} \cdot \frac{d z}{i z}\\ & = \operatorname{Re}\left(\frac{-1}{i} \oint_{|z |=1} \frac{z^3}{(2 z-1)(z-2)} d z\right) \\ & =\operatorname{Re}\left(-2 \pi\left(\frac{1}{2} \cdot \frac{\frac{1}{8}}{-\frac{3}{2}}\right)\right) \\ & =\frac{\pi}{12} \end{aligned}

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Real Method

$\begin{aligned} \displaystyle \int_{0}^{2 \pi} {\frac{\cos(3 x)}{5 - 4 \cos(x)}} dx & = \int_{0}^{2 \pi} {\frac{4 \cos^{3}(x) - 3 \cos(x)}{5 - 4 \cos(x)}} dx \\ &= \int _ { 0 } ^ { 2 \pi } \left[ - \cos ^ { 2 } x - \frac { 5 } { 4 } \cos x - \frac { 13 } { 16 } + \frac { 65 } { 16 ( 5 - 4 \cos x ) } \right] d x\\ & = \left[ - \frac { x } { 2 } - \frac { \sin 2 x } { 4 } - \frac { 5 } { 4 } \sin x - \frac { 13 } { 16 } x \right] _ { 0 } ^ { 2 \pi } +\frac { 65 } { 16 } \int _ { 0 } ^ { 2 \pi } \frac { d x } { 5 - 4 \cos x }\\&= - \frac { 21 \pi } { 8 } + \frac { 65 } { 16 } \int _ { 0 } ^ { 2 \pi } \frac { d x } { 5 - 4 \cos x } \end{aligned} \tag*{} $ $\displaystyle \textrm{Let }t=\tan \frac{x}{2}, \textrm{ then }d t = \frac { 1 } { 2 } ( 1 + t ^ { 2 } ) d x$ $\displaystyle \begin{aligned} \int _ { 0 } ^ { 2 \pi } \frac { d x } { 5 - 4 \cos x } & =2 \int _ { 0 } ^ { \pi } \frac { d x } { 5 - 4 \cos x } \\ &=2 \int_0^{\infty} \frac{1}{5-4 \cdot \frac{1-t^2}{1+t^2}} \cdot \frac{2 d t}{1+t^2}\\& = 4 \int _ { 0 } ^ { \infty } \frac { d t } { ( 3 t ) ^ { 2 } + 1 } \\&= \frac { 4 } { 3 }\left [ \tan ^ { - 1 } ( 3 t )\right ] _ { 0 } ^ { \infty } \\ &= \frac { 4 } { 3 } \cdot \frac { \pi } { 2 } \\ &= \frac { \pi } { 6 } \end{aligned} \tag*{} $ $\displaystyle \therefore I = - \frac { 21 \pi } { 8 } + \frac { 65 } { 16 } \cdot \frac { 2 \pi } { 3 }=\boxed{\frac{\pi }{12}} \tag*{} $