Computing the integral $\displaystyle I=\int_{0}^{1}\sqrt{1-x^{\pi}}dx$

[Note] : You can use the beta function :

• Beta Function : $\displaystyle \mathfrak{B}(x,y):=\int_{0}^{1}(1-\mu)^{x-1}\mu^{y-1}\;\text{d}\mu=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

Let $\mu:=x^{\pi}$. Then : \begin{align*} x^{\pi}=\mu&\implies\text{d}\mu=\pi x^{\pi -1}\\ \\ &\iff\text{d}x=\frac{d\mu}{\pi\cdot\mu^{\displaystyle\frac{\pi-1}{\pi}}} =\frac{1}{\pi}\mu^{\displaystyle\frac{1-\pi}{\pi}}\text{d}\mu \end{align*} Therefore : \begin{align*} I&=\frac{1}{\pi}\int_{0}^{1}\sqrt{1-\mu}\cdot\mu^{\displaystyle\frac{1-\pi}{\pi}}\text{d}\mu\\ \\ &=\frac{1}{\pi}\int_{0}^{1}(1-\mu)^{\frac{3}{2}-1}\mu^{\frac{1}{\pi}-1}\;\text{d}\mu\\ \\ &=\frac{1}{\pi}\mathfrak{B}\left(\frac{3}{2},\frac{1}{\pi}\right)\\ \\ &=\frac{1}{\pi}\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{\pi}\right)}{\Gamma\left(\frac{3}{2}+\frac{1}{\pi}\right)}\\ \\ &\approx 0.846941\cdots \end{align*}