How would I evaluate this integral?

Question

I have an integral $$\int\sqrt{b^2-\frac{x^4}{a^4}} dx \quad a, b\in \mathbb{R}$$

And I have no idea how to solve it. The $x^4$ term is stumping me. I've tried the standard $u$ and trig substitution methods but to no avail. Any suggestions?

Answer 1

First of all, with you numbers, the integration bounds are exactly $$t_\pm=\pm \frac 3 5 \sqrt{\frac{31}{5}} \sim \pm 1.493987952$$ that is to say the limit of definition of the integrand.

$$I=\int\sqrt{b^2-\frac{x^4}{a^4}}\, dx$$

If you do not want to face nasty elliptic integrals, assuming $a>0$ and $b >0$, use $$\sqrt{b^2-\frac{x^4}{a^4}}=\sum_{n=0}^\infty (-1)^n a^{-4 n} b^{1-2 n} \binom{\frac{1}{2}}{n}x^{4n}$$ to make after integration $$I=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n}\frac {x^{4n+1}}{a^{4n}b^{2n-1}(4n+1) }$$

Now, for the integral $$J=\int_{-a \sqrt{b} }^{+a \sqrt{b} }\sqrt{b^2-\frac{x^4}{a^4}}\, dx=2 a b^{3/2}\sum_{n=0}^\infty (-1)^n \frac{\binom{\frac{1}{2}}{n}}{4 n+1}$$ $$\color{blue}{J= a\, b^{\frac 32}\,\sqrt{\pi }\,\frac{\Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{7}{4}\right)}}$$

More generally $$K=\int_{-ka \sqrt{b} }^{+ka \sqrt{b} }\sqrt{b^2-\frac{x^4}{a^4}}\, dx=2 a b^{3/2}\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n}\frac{k^{4n+1}}{4 n+1}$$ $$\color{blue}{K=2\, k\, a\, b^{\frac 32}\,\, _2F_1\left(-\frac{1}{2},\frac{1}{4};\frac{5}{4};k^4\right)}$$