Reduction formulae question?
Prove that
n∫cosh^n(x) dx = cosh^(n-1) x * sinhx + (n-1)∫cosh^(n-2) x dx
Many thanks.
Prove that
n∫cosh^n(x) dx = cosh^(n-1) x * sinhx + (n-1)∫cosh^(n-2) x dx
Many thanks.
Moise Gunen
Favorite Answer
I(x) = ∫cosh^(n-1) (x) dx =
∫cosh^(n-1) (x) d(sinh (x)) =
cosh^(n-1) (x) sinh (x) - (n-1) ∫ sinh(x) cosh^(n-2) (x) sinh (x) dx =
cosh^(n-1) (x) sinh (x) - (n-1) ∫ sinh^2(x) cosh^(n-2) (x)dx =
cosh^(n-1) (x) sinh (x) - (n-1) ∫ (cosh^2 (x) -1)cosh^(n-2) (x) dx =
cosh^(n-1) (x) sinh (x) + (n-1) ∫ cosh^(n-2) dx - (n-1) ∫ cosh^n (x) dx
You get
I(x) = cosh^(n-1) (x) sinh (x) + (n-1) ∫ cosh^(n-2) dx - (n-1)I(x)
Solve equation and get: (undefined I(x) )
nI(x) = cosh^(n-1) (x) sinh (x) + (n-1) ∫ cosh^(n-2) dx
QED