complex-analysiscontour-integrationintegrationresidue-calculus
Evaluate $$\int_0^{2\pi} \frac{\sin^2\theta}{5+4\cos(\theta)}\mathrm d\theta$$
This is the final question on my review for my final exam tomorrow, and I will be honest and say that I have no clue how to begin problem. Any hints in the direction of how to solve this would be helpful.
Following from @Adam Hughes,
$-\dfrac{\pi}{2}[f'(0)+Res_2]$When I took the derivative of $f(z)=\dfrac{(z^2-1)^2}{2z^2+5z+2}$ and evaluated for 0, I got $-\dfrac{5}{4}$ Now, $Res_2$
I got this as the first part of this answer:
Start with $$ \begin{align} \int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x &=\frac12\int_0^\pi\log(\sin(x))\,\mathrm{d}x\\ &=\int_0^{\pi/2}\log(\sin(2x))\,\mathrm{d}x\\ &=\int_0^{\pi/2}\Big(\log(2)+\log(\sin(x))+\log(\cos(x))\Big)\,\mathrm{d}x\\ &=\frac\pi2\log(2)+2\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x\tag{1} \end{align} $$ Therefore, $$ \int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x=-\frac\pi2\log(2)\tag{2} $$
Thus, $$ \int_0^\pi\log(\sin(x))\,\mathrm{d}x=-\pi\log(2) $$
Using Gauss Mean Value
$\mathrm{Re}(\log(z))=\log(|z|)=\log\left(\sqrt{2-2\cos(x)}\right)$
$\hspace{4.5cm}$
$$
\begin{align}
\int_0^\pi\log(\sin(x))\,\mathrm{d}x
&=\int_0^\pi\log\left(\color{#C00000}{\frac12}\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x\\
&=\pi\color{#00A000}{\frac1{2\pi}\int_0^{2\pi}\log\left(\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x}\color{#C00000}{-\pi\log(2)}\\[6pt]
&=\pi\color{#00A000}{\log(1)}-\pi\log(2)\\[12pt]
&=-\pi\log(2)
\end{align}
$$
Your question is basically whether $$\int_0^{\pi}\frac{\sin 2\theta \,\mathrm{d}\theta}{5-3\cos\theta}=\int_{-1}^1\frac{2x\,\mathrm{d}x}{5-3x}= \tfrac{4}{9}(5\,\ln 2-3)$$ can be written as an integral about a closed curve of some algebraic function over the rationals. This statement is believed to be false; it is in fact an open problem to demonstrate as much (Kontsevich and Zagier, Problem 5).
The usual method (substituting $x=\cos\theta$) will have to do.
Best Answer
Using the standard trigonometric substitution $\theta=2\arctan t$ your integral boils down to: $$I=2\int_{-\infty}^{+\infty}\frac{4t^2}{(1+t^2)^2(9+t^2)}\,dt.$$ Now we just have to compute the residues in $t=i$ and $t=3i$ to get: $$ I = \frac{\pi}{4}.$$
As a side note, you should not use this site as a resource to solve your homeworks: apart from the debatable fairness, it is not designed for that task. When proposing a problem, it is strongly recommended to show your efforts - "I have no clue" or "my final exam is tomorrow" are really bad starters.