Use the residue theorem to calculate
$$\int_{0}^{2\pi} \frac {27} {(5+4\sin\theta)^2} d\theta $$
I know
$$ \operatorname{Res}_{z_0} f = \frac 1 {2\pi i} \int_\gamma d\theta f(\theta) $$
My question is how do I plug in that function in this formula?
Thanks!
Best Answer
The trick for this type of question is $$\int_0^{2\pi} \frac{d\theta}{(a+b\sin\theta)^2}= \displaystyle \frac{2\pi ab}{(a^2-b^2)^{3/2}} $$
Here $a=5$ and $b=4.$
Therefore,
$$\int_{0}^{2\pi} \frac {27} {(5+4\sin\theta)^2} d\theta = \frac{27 \cdot 2\pi \cdot 5\cdot 4}{(25-16)^{3/2}}=40\pi.$$