How to solve $\int{\sqrt{1-x^8}}dx$ I have tried to solve it with substitution, but I always have extra x variable that I can`t get rid off.
$$\int{\sqrt{1-x^8}}dx$$
Can you help me to solve this integral, how can I start? Thanks!
 A: As already mentioned, the indefinite integral is not expressible in terms of elementary functions.
However, $\displaystyle\int_0^1\sqrt[m]{1-x^n}dx=\int_0^1\sqrt[n]{1-x^m}dx=\frac{\Big(\frac1m\Big)!\cdot\Big(\frac1n\Big)!}{\Big(\frac1m+\frac1n\Big)!}$ . See beta function for more details.
A: Why not to start with the Taylor expansion of $\sqrt{1-y}$ and then replace $y$ by $x^8$ and integrate the infinite series.
I did it for the integration between $0$ and $1$. The exact result is $$\frac{\sqrt{\pi } \Gamma \left(\frac{1}{8}\right)}{16 \Gamma
   \left(\frac{13}{8}\right)} \simeq 0.930874$$ Adding one term at the time, the approximated values of the integral are $0.944444$, $0.937092$, $0.934592$, $0.933408$, $0.932741$, $0.932322$, $0.932040$, $0.931838$, $0.931689$, $0.931574$
A: If you are willing to use the Incomplete Beta Function, then the answer is
$$
\begin{align}
\int_0^a\sqrt{1-x^8}\,\mathrm{d}x
&=\frac18\int_0^{a^8}(1-u)^{1/2}u^{-7/8}\,\mathrm{d}u\\
&=\frac18\mathrm{B}(a^8;1/8,3/2)
\end{align}
$$
Of course, setting $a=1$ yields
$$
\int_0^1\sqrt{1-x^8}\,\mathrm{d}x=\frac18\mathrm{B}(1/8,3/2)
$$
where $\mathrm{B}(x,y)$ is the standard Beta function.
