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Here is the integral I am trying to evaluate:

$$\int_0^{2\pi}\dfrac{\sin^2x}{5+4\cos x}\ dx$$

I am getting the final answer as $-{\pi}/8$ but the correct answer is ${\pi}/4$

Here are the steps I did: $$\begin{align} \dfrac{\sin^2x}{5+4\cos x} dx &= \\\\ &=\dfrac{\left(\dfrac{z-1/z}{2i}\right)^2}{5+4\dfrac{z+1/z}{2}} \dfrac{dz}{iz} \\\\ &=\dfrac{-1}{4i}\dfrac{{z^2+1/z^2-2}}{5z+2(z^2+1)} dz \\\\ &=\dfrac{-1}{4i}\dfrac{{z^4+1-2z^2}}{{z^2}(5z+2({z^2+1}))} dz \\\\ &=\dfrac{-1}{4i}\dfrac{{z^4+1-2z^2}}{{z^2}(2z+1)(z+2)} dz \end{align}$$

Clearly, singularities within $|z|=1$ are $0$, and $\frac{-1}{2}$. After applying Residue theorem, I am getting $\frac{-\pi}{8}$

Are my steps correct?

trancelocation
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XXX_fam_XXX
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  • Your steps so far are correct, but you need to show the rest of the approach in order for others to figure out why there's an additional $-\frac12$ factor. Note that $0$ is a pole of order $2$. – Zacky May 13 '20 at 10:30
  • I have rectified my mistake , I realized I wasn't handling numbers correctly(arithmetic). Was about to delete the question. Thanks for the response though. – XXX_fam_XXX May 13 '20 at 10:32
  • I'm glad if it worked out. Eventually this is also a duplicate of this question, see Adam Hughes's answer. – Zacky May 13 '20 at 10:55

1 Answers1

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So far the steps you showed are correct except for a little typo I corrected.

But you didn't show the calculation of the residues.

With $f(z) =\frac{-1}{4i}\frac{{z^4+1-2z^2}}{{z^2}(2z+1)(z+2)}$ I get

$$\operatorname{Res}_{z=0}f(z)= -\frac{5}{16}i \mbox{ and }\operatorname{Res}_{z=-\frac 12}f(z)= \frac{3}{16}i$$

Hence,

$$\int_0^{2\pi}\frac{\sin^2x}{5+4\cos x}\ dx = 2\pi i\left( -\frac{5}{16}i + \frac{3}{16}i\right) =\frac{\pi}{4}$$

Ali Shadhar
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trancelocation
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  • I realized the mistake a few minutes after posting the question(it was adding an extra 1/2 somehow), but I must thank you for taking the time to answer the question. It's my bad I didn't show the next steps. – XXX_fam_XXX May 13 '20 at 13:47