# What approach should I use to solve integrals like this?

$$\int{\sqrt{1-{x^3}}}dx$$

I tried with $t=x^3$ but then I have the $3x^2$ dt that I can't get rid of.

• The result seems to be non-elementary (for example, could be expressed in an elliptic function), are you familiar with those? – mickep Jul 26 '16 at 14:43
• the result can be expressed by an elliptic function – Dr. Sonnhard Graubner Jul 26 '16 at 15:16
• Are you really interested in the general primitive (that is indeed given by an elliptic integral) or it is enough to compute the integral over some interval, by chance $(0,1)$? In such a case the answer is given by the Euler's beta function. – Jack D'Aurizio Jul 26 '16 at 16:27
• – Jack D'Aurizio Jul 26 '16 at 16:37
• @AlexPatient: Wikipedia is your friend: en.wikipedia.org/wiki/Elliptic_integral – Jack D'Aurizio Jul 26 '16 at 18:10

For any real number of $x$ ,

When $|x|\leq1$ ,

$\int\sqrt{1-x^3}~dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{3n}}{4^n(n!)^2(1-2n)}dx$

$=\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{3n+1}}{4^n(n!)^2(1-2n)(3n+1)}+C$

When $|x|\geq1$ ,

$\int\sqrt{1-x^3}~dx$

$=i\int\sqrt{x^3-1}~dx$

$=i\int x^\frac{3}{2}\sqrt{1-\dfrac{1}{x^3}}~dx$

$=i\int x^\frac{3}{2}\sum\limits_{n=0}^\infty\dfrac{(2n)!}{4^n(n!)^2(1-2n)x^{3n}}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{i(2n)!x^{\frac{3}{2}-3n}}{4^n(n!)^2(1-2n)}dx$

$=\sum\limits_{n=0}^\infty\dfrac{i(2n)!x^{\frac{5}{2}-3n}}{4^n(n!)^2(1-2n)\left(\dfrac{5}{2}-3n\right)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{i(2n)!}{2^{2n-1}(n!)^2(2n-1)(6n-5)x^{3n-\frac{5}{2}}}+C$

• Thank you very very much!!! – Healshine Jul 28 '16 at 14:00