Jtdylktuy

This is the integral :
$$I = int_{0}^{2pi} frac{dx}{(5+4cos x)^2} $$
Which, according to wolfram alpha, should evaluate to $frac{10pi}{27} $,
but the value i find is $frac{20pi}{27} $.

These are my calculations :

Using complex form of cosine i get $ cos x = frac{t^2+1}{2t}$
where $ t = e^{ix}$ and $ dx = frac{dt}{it}$.
If $partial{D} $ is the unit circle
$$ I = int_{partial{D}} frac{-i}{t(frac{5t+2t^2+2}{t})^2} dt =
-i int_{partial{D}}frac{t}{(t+2)^2(t+1/2)^2}dt = -i I_a$$

$I_a = pi i $ times the residue in -$frac{1}{2}$

The residue is equal to
begin{align}
lim_{zto-frac{1}{2}} frac{d}{dz} left( frac{t(t+1/2)^2}{(t+2)^2(t+1/2)^2}right)
&= lim_{zto-frac{1}{2}} frac{d}{dz} frac{t}{(t+2)^2}
\&= lim_{zto-frac{1}{2}} frac{(t+2)^2-2t(t+2)}{(t+2)^4}
\&= lim_{zto-frac{1}{2}} frac{2-t}{(t+2)^3}
= frac{2+1/2}{(-1/2+2)^3}
\&= frac{5/2}{27/9} = frac{20}{27}
end{align}

so $ I_a = ipifrac{20}{27}$

So our original integral $$I = -i I_a = -i^2 pifrac{20}{27} = frac{20pi}{27}$$

You made an error in calculating the factorization of the denominator. You should get $$5t+2t^2+2=(2+t)(1+2t)=2(t+2)(t+1/2),$$ where you missed to transfer the leading factor $2$ to the factorized expression. The correction leads to $I=-frac i4 I_a$ and the additional division by $4$ corrects the residuum that is computed with the correct factor $2pi i$.

You could also use the geometric/binomial series to get the correct residual coefficient. Then, using $u=t+frac12$,
$$
frac{t}{(t+2)^2(t+frac12)^2}=frac{u-tfrac12}{(u+frac32)^2u^2}
=frac49(u-tfrac12)sum_{k=0}^infty (k+1)left(-frac23right)^ku^{k-2}
$$
so that the coefficient for the power $u^{-1}=(t+frac12)^{-2}$ is $frac49(1+frac12cdot2cdotfrac23)=frac{20}{27}$. The integration over the unit circle adds the factor $2pi i$, so that in the final result $I=frac{10pi}{27}$.

Your mistake is when using the residue theorem:

$$oint_{C^+} f(z) dz = 2pi isum_{singularities} Res(f,singularity)$$

You forgot the factor $2$ in this formula.

Another mistake: $$5t + 2t^2 + 2 = 2(t+2)(t+1/2)$$

You again forgot a factor 2 (which will yield a factor $4$ eventually)

You also made a mistake in calculating the residue.

Note that $(-1/2 + 2)^3 = (3/2)^3 = 27/8$

Correcting these two mistakes, we easily find the correct answer.