Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$ \int_{-\pi/3}^{\pi/3} \frac{(\pi +4x^3)\,dx}{2-\cos(|x|+ \frac{\pi}{3})} $$

I have separated the integral into two parts, then expanded using $\cos(a+b)$ formula, after that I am lost. Can someone provide me hint?

share|cite|improve this question
In what way are you stuck? (We're just guessing if you don't tell us.) – Eric Towers Mar 23 '14 at 7:24
Random hint: $\cos|x| = \cos x$. – Eric Towers Mar 23 '14 at 7:25
numerator is $\pi +4x^3$ and denominator is $4-cosx+ \sqrt(3) sinx$. – Kumar Mar 23 '14 at 7:26
Try doing $u=x+\pi/3$. It's much better to have trigonometry simple and symmetric. – orion Mar 23 '14 at 7:34
Random hint 2: Can you exploit the symmetry to reduce the integral? – Sangchul Lee Mar 23 '14 at 8:31

You don't need to separate the integrand. First of all, the $x^3$ piece vanishes; this follows immediately from the fact that this is an odd integrand over an even interval. That leaves the first piece, which I can reduce to

$$2 \pi \int_{\pi/3}^{2 \pi/3} \frac{dx}{2-\cos{x}}$$

This may be evaluated using a substitution of the form $t=\tan{x/2}$; $dx=2 dt/(1+t^2)$. Then the integral is equal to

$$\begin{align}4 \pi \int_{1/\sqrt{3}}^{\sqrt{3}} dt \frac{1}{1+ 3 t^2} &= \frac{4 \pi}{3 \sqrt{3}} \left [\arctan{\frac{t}{\sqrt{3}}}\right ]_{1/\sqrt{3}}^{\sqrt{3}}\\ &= \frac{4 \pi}{3 \sqrt{3}} \left (\arctan{1}-\arctan{\frac13} \right )\\ &= \frac{4 \pi}{3 \sqrt{3}} \arctan{\frac12}\end{align}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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