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 (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: . It is tightly connected to elliptic functions.
In a word: there is no expression for this anti-derivative in terms of "elementary functions"...