Evaluate $\int_0^{2\pi} \frac{\sin^2\theta}{5+4\cos\theta}\,\mathrm d\theta$

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.

Recall that on the unit circle $\sin\theta ={1\over 2i}(z-z^{-1})$ and $\cos\theta = {1\over 2}(z+z^{-1})$, since $z=e^{i\theta}$ and $dz=iz\,d\theta$

Rewriting gives

$$-{1\over 4i}\int_{C}{z^2-2+z^{-2}\over 5z+2z^2+2}\,dz$$

Clearing out the denominators gives

$$-{\pi\over 2}\cdot {1\over 2\pi i}\int_C{(z^2-1)^2\over z^2(2z^2+5z+2)}\,dz$$

By using Cauchy's integral formula, we see this is

$$-{\pi\over 2}\cdot \left(f'(0)+\text{Res}_2\right)$$

where $f(z)={(z^2-1)^2\over 2z^2+5z+2}$ and Res$_2$ is the reside at the zero of $2z^2+5z+2$ inside the disc--there is exactly one by Rouché, since $|5z+2|\ge 3>2=|2z^2|$ on the boundary and $-{2\over 5}$ is a zero of the dominating function inside the disc. I'll leave the computations to you, since it seems you're trying to practice for the final.

Write $\sin(\theta)$ as $$ \sin(\theta) = \frac{z-z^{-1}}{2i} $$ and $\cos(\theta)$ as $$ \cos(\theta) =\frac{z+z^{-1}}{2} $$ where $z=e^{i\theta}$. Then $\frac{dz}{iz} = d\theta$.