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Calculate the following integral: $$\int_{-\pi}^{\pi}\frac{1-\cos \frac{t}{4}}{5-2\cos t}dt$$

Any suggestions please? I do not really know where to start!

Mark
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1 Answers1

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My method is a little long, that's why I'm not writing the whole thing:

  • First as $\cos x=\cos(-x)$ $$I=\int_{-\pi}^{\pi}\frac{1-\cos \frac{t}{4}}{5-2\cos t}dt=2\int_{0}^{\pi}\frac{1-\cos \frac{t}{4}}{5-2\cos t}dt$$
  • Second, express every term in $u=\cos (t/4)$: $$I=8\int_1^{1/\sqrt2}\frac{du}{(16u^4-3-16u^2)\sqrt{u+1}}$$
  • Third use $v^2=\sqrt{u+1}$ to get: $$I=16\int_{\sqrt2}^{\sqrt{1+1/\sqrt2}}\frac{dv}{16(v^2-1)^4-16(v^2-1)^2-3}$$

  • Then use partial fraction(The v looks like u here): enter image description here

  • Now you can do the rest, ask if you need help integrating any term above(some of them go to $\ln,\arctan$ or some can be manipulated by something like algebraic twins).

RE60K
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  • You have been very brave to attack this monster. I wonder if there is not a problem between $First$ and $Second$ with the change of variable. Numerically, the integral which ends $First$ is $\approx 0.103891$, while the one which ends $Second$ is $\approx 0.584576$. Is there a problem with the first change of variable ? Cheers :-) – Claude Leibovici Oct 06 '14 at 07:50