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

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?

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In what way are you stuck? (We're just guessing if you don't tell us.) –  Eric Towers Mar 23 at 7:24
    
Random hint: $\cos|x| = \cos x$. –  Eric Towers Mar 23 at 7:25
    
numerator is $\pi +4x^3$ and denominator is $4-cosx+ \sqrt(3) sinx$. –  Kumar Mar 23 at 7:26
    
Try doing $u=x+\pi/3$. It's much better to have trigonometry simple and symmetric. –  orion Mar 23 at 7:34
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Random hint 2: Can you exploit the symmetry to reduce the integral? –  sos440 Mar 23 at 8:31

1 Answer 1

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

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

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