Contour intergal of a rational trigonometric function, can't find my mistake.

Question

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

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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_{z\to-\frac{1}{2}} \frac{d}{dz} \left( \frac{t(t+1/2)^2}{(t+2)^2(t+1/2)^2}\right) &= \lim_{z\to-\frac{1}{2}} \frac{d}{dz} \frac{t}{(t+2)^2} \\&= \lim_{z\to-\frac{1}{2}} \frac{(t+2)^2-2t(t+2)}{(t+2)^4} \\&= \lim_{z\to-\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 = i\pi\frac{20}{27}$

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

Answer 1

Your mistake is when using the residue theorem:

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

You forgot the factor $2$ in this formula.

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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)

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You also made a mistake in calculating the residue.

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

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

Answer 2

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 $2\pi i$.

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You could also use the geometric/binomial series to get the correct residual coefficient. Then, using $u=t+\frac12$, $$ \frac{t}{(t+2)^2(t+\frac12)^2}=\frac{u-\tfrac12}{(u+\frac32)^2u^2} =\frac49(u-\tfrac12)\sum_{k=0}^\infty (k+1)\left(-\frac23\right)^ku^{k-2} $$ so that the coefficient for the power $u^{-1}=(t+\frac12)^{-2}$ is $\frac49(1+\frac12\cdot2\cdot\frac23)=\frac{20}{27}$. The integration over the unit circle adds the factor $2\pi i$, so that in the final result $I=\frac{10\pi}{27}$.