What is the generalized solution to $$M = \int_0^1 (1-x^m)^n \,dx$$ (where it can be expressed in terms of n and m) and m and n are rational numbers if considering m and n as natural numbers makes calculations easier use them.
Are there methods which solves M without the use of Beta function?
[Here, Beta function being referred to as $\beta$(m,n) = $\int_0^1 x^{m-1}(1-x)^{n-1}$ dx = $\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$]
Well, from the binomial theorem $$(1-x^m)^n=\sum_{k=0}^{n}{n\choose k}(-1)^kx^{km}$$ thus $$\int_0^1(1-x^m)^n\,dx=\int_0^1\sum_{k=0}^{n}{n\choose k}(-1)^kx^{km}\,dx=\sum_{k=0}^{n}{n\choose k}(-1)^k\frac{x^{km+1}}{km+1}\bigg|^1_0=\sum_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{km+1}$$ which according to WA equals $\frac{\frac{1}{m}!n!}{(\frac1m+n)!}$.
But since these are generalized factorials, these make use of the $\Gamma$ function, so this is equivalent to your $\beta$ function solution.
You can use this as closed form $$\int_0^{1}(1-x^m)^n\mathrm{d}x=\frac{\Gamma\left(1+\frac{1}{m}\right)\Gamma\left(1+n\right)}{\Gamma\left(1+n+\frac{1}{m}\right)}$$
