Evaluate \[ \int_0^{2\pi} \frac{\cos^2(3\theta)}{5-4\cos(2\theta)}\,d\theta. \]
Solution
1. For \(0<r<1\) and real \(\phi\),\[ \frac{1-r^2}{1-2r\cos\phi+r^2} = 1+2\sum_{k=1}^{\infty}r^k\cos(k\phi). \]
Since \(\abs r<1\), \[ \sum_{k=1}^{\infty}r^ke^{ik\phi} = \frac{re^{i\phi}}{1-re^{i\phi}}. \] Taking twice the real part and adding \(1\) gives \[ \begin{aligned} 1+2\sum_{k=1}^{\infty}r^k\cos(k\phi) &= 1+2\Re\left(\frac{re^{i\phi}}{1-re^{i\phi}}\right)\\ &= \frac{1-r^2}{1-2r\cos\phi+r^2}. \end{aligned} \] The series converges uniformly in \(\phi\) because \[ \sum_{k=1}^{\infty}2r^k<\infty. \]
2. One has the uniformly convergent expansion\[ \frac1{5-4\cos(2\theta)} = \frac13 \left( 1+2\sum_{k=1}^{\infty}2^{-k}\cos(2k\theta) \right). \]
1. Since\[ 5-4\cos(2\theta) = 4\left(1-2r\cos(2\theta)+r^2\right) \]and\[ 1-r^2=\frac34, \]division gives exactly the displayed formula.
3. One has\[ \int_0^{2\pi}\frac{d\theta}{5-4\cos(2\theta)} = \frac{2\pi}{3}. \]
Integrate the uniformly convergent series in step 2 term by term. For every \(k\geq1\), \[ \int_0^{2\pi}\cos(2k\theta)\,d\theta=0. \] Only the constant term remains, giving \[ \frac13\int_0^{2\pi}1\,d\theta = \frac{2\pi}{3}. \]
4. One has\[ \int_0^{2\pi} \frac{\cos(6\theta)}{5-4\cos(2\theta)} \,d\theta = \frac{\pi}{12}. \]
Multiply the series in step 2 by \(\cos(6\theta)\) and integrate term by term. Orthogonality of the cosine functions gives zero for every term except \(k=3\). Hence \[ \begin{aligned} \int_0^{2\pi} \frac{\cos(6\theta)}{5-4\cos(2\theta)} \,d\theta &= \frac13\cdot2\cdot2^{-3} \int_0^{2\pi}\cos^2(6\theta)\,d\theta\\ &= \frac13\cdot\frac14\cdot\pi\\ &= \frac{\pi}{12}. \end{aligned} \]
5. The required integral is\[ \boxed{\frac{3\pi}{8}}. \]
Using \[ \cos^2(3\theta) = \frac{1+\cos(6\theta)}2, \] steps 3 and 4 give \[ \begin{aligned} \int_0^{2\pi} \frac{\cos^2(3\theta)}{5-4\cos(2\theta)} \,d\theta &= \frac12 \left( \frac{2\pi}{3} + \frac{\pi}{12} \right)\\ &= \frac{3\pi}{8}. \end{aligned} \]
6. Q.E.D.
Step 5 is the required evaluation.