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I want to prove that $\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$

My ideas, I don't know if they lead anywhere:

Let's substitute $\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$ and $z=e^{i\theta}$ right after:

$\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=-i\cdot\int_1^{-1}\frac{z^2+z^{-2}+\frac{1}{2}z^3+\frac{1}{2}z^{-3}}{5z+2z^2+2}dz$

This now gives me 4 new integrals, for example

$\displaystyle-i\int_1^{-1}\frac{z^2}{2z^2+5z+2}dz$, $\displaystyle-i\int_1^{-1}\frac{1}{2z^4+5z^3+2z^2}dz$ and so on.

But since I haven't been able to solve any of the new integrals, I'm a little lost.

Edit: Can't I do a partial fractions decomposition of all the 4 integrals and solve them seperately?

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If you wish to exploit the residue theorem, then first exploit the fact that the integral is even. In addition, use Euler's formula to write $2\cos(2\theta)+\cos(3\theta)=\text{Re}(2e^{i2\theta}+e^{i3\theta})$.

Then, we have

$$\begin{align} \int_0^\pi \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,d\theta&=\frac12\text{Re}\left(\oint_{|z|=1}\frac{2z^2+z^3}{5+2(z+z^{-1})}\,\frac{1}{iz}\,dz\right)\\\\ &=\frac12\text{Re}\left(\frac1i\oint_{|z|=1}\frac{z^2(2+z)}{(2z+1)(z+2)}\,dz\right)\\\\ &=\frac12\text{Re}\left(\frac1i\oint_{|z|=1}\frac{z^2}{2z+1}\,dz\right)\\\\ &=\frac{\pi}{8} \end{align}$$

And we are done!

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  • $\begingroup$ Isn't that residue equal to $2\pi i \left( \lim_{z \to \frac{-1}{2}} (2z + 1) \frac{z^2}{2z + 1} \right)= 2\pi i \cdot \frac{1}{4} = \frac{i\pi}{2}$? Which implies a final answer of $\frac{\pi}{4}$. Something's gone awry I think. $\endgroup$ – AlkaKadri Jul 19 '17 at 23:22
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    $\begingroup$ @AlkaKadri The residue is $\lim_{z\to -1/2}\frac{(z-1/2)z^2}{2z+1}=\frac18$. So, it is correct. ;-)) $\endgroup$ – Mark Viola Jul 19 '17 at 23:23
  • $\begingroup$ Good stuff, thanks Mark! A valuable lesson there on how not to evaluate residues (thank goodness for the internet, I get to hide my shame ;P). $\endgroup$ – AlkaKadri Jul 19 '17 at 23:27
  • $\begingroup$ @AlkaKadri Thank you for the nice comment! You're welcome. Miy pleasure. And no shame at all. $\endgroup$ – Mark Viola Jul 19 '17 at 23:33
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HINT: show that your integrand is equal to $${\frac {4\, \left( \cos \left( x \right) \right) ^{2}-2+4\, \left( \cos \left( x \right) \right) ^{3}-3\,\cos \left( x \right) }{5+4\, \cos \left( x \right) }} $$ and then use the Weierstrass substitution

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Looks like there should be a partial fraction decomposition:

$\frac{z^2+z^{-2}+\frac{1}{2}z^3+\frac{1}{2}z^{-3}}{5z+2z^2+2} = Az + B + \frac C{z} + \frac D{z^2} + \frac E{z^3} + \frac {F}{z+2} + \frac {G}{2z + 1}$

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As Dr. Sonnhard Graubner answered, expand $\cos(2x)$ and $\cos(3x)$ using the classical formulas and use the tangent half-angle substitution to get $$I=\int\frac{\cos(2x)+\cos(3x)}{5+4\cos(x}\,dx=\int\frac{2 \left(t^6+5 t^4-25 t^2+3\right)}{\left(t^2+1\right)^3 \left(t^2+9\right)}dt$$ Using partial fraction decomposition $$\frac{2 \left(t^6+5 t^4-25 t^2+3\right)}{\left(t^2+1\right)^3 \left(t^2+9\right)}=\frac{3}{8 \left(t^2+9\right)}+\frac{13}{8 \left(t^2+1\right)}-\frac{9}{\left(t^2+1\right)^2}+\frac{8}{\left(t^2+1\right)^3 }$$ Integration now to get $$I=\frac{1}{8} \left(\frac{4 t-12 t^3}{\left(t^2+1\right)^2}+\tan ^{-1}\left(\frac{t}{3}\right)+\tan ^{-1}(t)\right)$$ and the bounds are $0$ and $\infty$.

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