The integral was: $$\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\frac{\pi + 4x^6}{1-\sin(|x|+\frac{\pi}{6})}$$

What I did was to identify that its an even function and write it as: $$2\int_{0}^{\frac{\pi}{6}}\frac{\pi + 4x^6}{1-\sin(x+\frac{\pi}{6})}$$

Then I wrote the denominator in terms of $\cos$ as $$\sin(x+\frac{\pi}{6})=\cos(\frac{\pi}{3}-x)$$ and then I applied the identity $1-\cos x=2\sin^2(\frac{x}{2})$ to finally get :


Now the first half is easy to get but how do I integrate the second one? Can someone suggest any steps or perhaps an alternative way?

  • 1
    $\begingroup$ I guess there's not a beautiful closed form for your integral. I tried to use mathematica to compute and it gives the answer $$\frac{10}{27} i \pi ^4 \text{Li}_2\left(\sqrt[6]{-1}\right)+\frac{80}{9} \pi ^3 \text{Li}_3\left(\sqrt[6]{-1}\right)-160 i \pi ^2 \text{Li}_4\left(\sqrt[6]{-1}\right)+\pi \left(4-1920 \text{Li}_5\left(\sqrt[6]{-1}\right)\right)+11520 i \left(\text{Li}_6\left(\sqrt[6]{-1}\right)-\text{Li}_6\left(\frac{1}{2} \left(i \sqrt{3}+1\right)\right)\right)+\frac{\pi ^6 \left(\sqrt{3}+(2+i)\right)}{5832}+\frac{1}{81} \pi ^5 \log \left(1-\sqrt[6]{-1}\right)$$ $\endgroup$
    – FFjet
    Sep 11, 2019 at 5:25

1 Answer 1


Looking at the monster given by Mathematica in comments, consider $$I=\int_{0}^{\frac{\pi}{6}}\frac{\pi + 4x^6}{1-\sin(x+\frac{\pi}{6})}\,dx=\pi\int_{0}^{\frac{\pi}{6}}\frac{dx}{1-\sin(x+\frac{\pi}{6})}+4\int_{0}^{\frac{\pi}{6}}\frac{x^6}{1-\sin(x+\frac{\pi}{6})}\,dx$$ The first one is simple.

For the second one, why not to try series expansion around $x=0$ $$\frac{1}{1-\sin(x+\frac{\pi}{6})}=2+2 \sqrt{3} x+5 x^2+\frac{11 x^3}{\sqrt{3}}+\frac{91 x^4}{12}+\frac{301 x^5}{20 \sqrt{3}}+O\left(x^6\right)$$ which, integrated termwise would give $$\int_{0}^{\frac{\pi}{6}}\frac{x^6}{1-\sin(x+\frac{\pi}{6})}\,dx\approx 0.00902$$ while the numerical integration would give $\approx 0.00937$


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