3
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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 :-) $\endgroup$ – Claude Leibovici Oct 6 '14 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.