1. ## An Integral #2

Here is the integral :
$\displaystyle \int_0^1{\sqrt[4]{1-x^4}}dx$

thanks very much.....

2. Originally Posted by Xingyuan
Here is the integral :
$\displaystyle \int_0^1{\sqrt[4]{1-x^4}}dx$

thanks very much.....
let $\displaystyle x=\sqrt{\sin t}.$ then $\displaystyle \int_0^1 \sqrt[4]{1-x^4} \ dx=\frac{1}{2} \int_0^{\frac{\pi}{2}}(\sin t)^{-\frac{1}{2}} (\cos t)^{\frac{3}{2}} \ dt$

$\displaystyle =\frac{1}{4}B\left(\frac{1}{4}, \frac{5}{4}\right)=\frac{\Gamma(\frac{1}{4})\Gamma (\frac{5}{4})}{4\Gamma(\frac{3}{2})}=\frac{1}{8 \sqrt{\pi}}\left(\Gamma\left(\frac{1}{4} \right)\right)^2. \ \ \ \square$

3. another approach, probably easier, is to put $\displaystyle x^4=t.$ then $\displaystyle dx=\frac{1}{4}t^{-\frac{3}{4}}dt.$ thus:

$\displaystyle \int_0^1 \sqrt[4]{1-x^4} \ dx = \frac{1}{4}\int_0^1 t^{-\frac{3}{4}}(1-t)^{\frac{1}{4}} \ dt=\frac{1}{4}B \left(\frac{1}{4},\frac{5}{4} \right) = ...$

the rest is the same as what i did in my previous post.
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for those viewers who might not be familiar with Beta function:

by definition: $\displaystyle B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1} \ dt, \ \ \text{Re}(x) > 0, \ \text{Re}(y) > 0.$

then it can be shown that: $\displaystyle B(x,y)=2\int_0^{\frac{\pi}{2}} (\sin t)^{2x-1}(\cos t)^{2y-1} \ dt=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}.$

see here for more details.