the integral from 0-1 of √(1-x^4) dx = (Γ(1/4)^2)/(6√2π)

Here is how i tried to tackle it...

let t= x^4 , then dt/dx = 4x^3. And also x^3 = t ^(3/4).

Now the integral from 0-1 of (1-t) ^1/2 dt/4x^3

1/4 of the integral from 0-1 of x^(-3) (1-t)^1/2 dt

1/4 of the integral from 0-1 of. t^(-3/4) (1-t) ^1/2 dt

x-1 = -3/4 therefore x= 1/4

y-1 = 1/2 then. y= 3/2

which is 1/4 B(1/4 , 3/2) NOT equal to the LHS of the equation.

i also tried it by substituting x^2 = sin u and end up with a complete different answer of B(1/2, 3/2).

can some one pls point out what I missed.

Any help would be much appreciated