For major players only: Integrate ∫ 1/√(1+x^4) dx
I broke this down using trig substitution (triangle: 1, x^2, √(x^4+1) to
∫ 1/√(2sin2ø) dø
If you could integrate in terms of ø or x it would be amazing.
For major players only: Integrate ∫ 1/√(1+x^4) dx
I broke this down using trig substitution (triangle: 1, x^2, √(x^4+1) to
∫ 1/√(2sin2ø) dø
If you could integrate in terms of ø or x it would be amazing.
The answer is a special function, called the "inverse hyperbolic lemniscatic sine"... The inverse lemniscatic sine equals $\displaystyle \int_0^x \frac{dt}{\sqrt{1-t^4}}$ (a generalization of the arcsinus, relating to the lemniscate curve like the sinus relates to the circle...), and its hyperbolic counterpart is the function you need: $\displaystyle \int_0^x\frac{dt}{\sqrt{1+t^4}}$. It is tightly connected to elliptic functions.
In a word: there is no expression for this anti-derivative in terms of "elementary functions"...