The problem is to evaluate in closed-form:


An estimate for the integral is $1.76663875028545$.

I have tried all the usual techniques, including by parts and various $u$-subs, but they all seem to make the integral more contrived. The Weierstrass substitution $x=\tan{\left(\theta/2\right)}$ 'simplifies' the integrand to one with only polynomials, with a nice factorization as well:


However, I do not know how to continue from here, primarily because the exponent $\frac{2}{3}$ in the denominator still remains.

In addition, both Maple and WolframAlpha could not find a closed-form answer (though that does not mean that one does not exist).

Any ideas? Thanks!

  • $\begingroup$ Have you tried using $f(a+b-x)=f(x)$? $\endgroup$
    – Teh Rod
    Commented Aug 2, 2017 at 15:13
  • $\begingroup$ @TehRod : Do you mean that we can make the upper bound $\pi/4$ instead of $\pi/2$? How else can one apply this? $\endgroup$
    – Ant
    Commented Aug 2, 2017 at 15:18
  • $\begingroup$ I don't know, that's just the first thing that came to mind when I saw this. Especially since there does not appear to be a closed form for the antiderivative $\endgroup$
    – Teh Rod
    Commented Aug 2, 2017 at 15:19
  • $\begingroup$ Better numerical estimation: $1.766638750285906$ $\endgroup$
    – Enrico M.
    Commented Aug 2, 2017 at 15:36

1 Answer 1


First, enforcing the substitution $\theta=\arctan(x)$ reveals

$$\int_0^{\pi/2}\frac{1}{\left(\cos^3(\theta)+\sin^3(\theta)\right)^{2/3}}\,d\theta=\int_0^\infty \frac{1}{\left(1+x^3\right)^{2/3}}\,dx$$

Next, we let $y=x^3$ to obtain

$$\begin{align} \int_0^\infty \frac{1}{\left(1+x^3\right)^{2/3}}\,dx&=\frac13\int_0^\infty \frac{y^{-2/3}}{(1+y)^{2/3}}\,dy\\\\ &=\frac13 B\left(1/3,1/3\right)\tag 1\\\\ &=\frac13 \frac{\Gamma^2(1/3)}{\Gamma(2/3)}\tag 2 \end{align}$$

where $B(x,y)$ is the Beta function and $\Gamma(x)$ is the Gamma function. In going from $(1)$ to $(2)$ we used the relationship $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$.

  • $\begingroup$ I was going to post an answer, but yours is bloody awesome! +1 $\endgroup$
    – Enrico M.
    Commented Aug 2, 2017 at 15:38
  • $\begingroup$ @HenryTuring Wow! Much appreciated. $\endgroup$
    – Mark Viola
    Commented Aug 2, 2017 at 15:38
  • $\begingroup$ (+1) Very nice! I knew about the Beta function beforehand but didn't think to apply it here. Thanks! $\endgroup$
    – Ant
    Commented Aug 2, 2017 at 15:40
  • 2
    $\begingroup$ @JohnChessant, you should accept the answer I guess. $\endgroup$ Commented Aug 2, 2017 at 20:29

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