Show that $\int_0^{2 \pi} \dfrac{d \theta}{5+4 \sin\theta}=\dfrac{2 \pi}{3}.$

Question

By change of variable and letting $C_0$ be the unit circle about the origin, we have: \begin{align*} \int_0^{2 \pi} \dfrac{d \theta}{5+4 \sin\theta} &= \int_{C_0} \dfrac{dz/zi}{5+4(z-z^{-1})/2i}\\ & = \int_{C_0} \dfrac{dz}{2z^2+5iz-2}\\ & = \int_{C_0} \dfrac{dz}{(z+i/2)(z+2i)}\\ & = 2 \pi i (-i/2+2i)^{-1}\\ & = \dfrac{4 \pi}{3} \end{align*}

But wolframalpha gives $\dfrac{2 \pi}{3}.$ Is what I wrote valid?

Answer 1

$\oint_{|z|=1} \frac {1}{i(2z^2 + 5iz -2} \ dz\\ \oint_{|z|=1} \frac {1}{2(z +2i)(z + \frac 12i)} \ dz$

You dropped a factor of $\frac 12$

$2\pi i\frac {1}{2(-\frac 12 i + 2i)} = \frac {2\pi}{3}$