∫积分上限1积分下限-1(x^2-1)^ndx(n为自然数)

Question

∫积分上限1积分下限-1 (x^2-1)^n dx(n为自然数)

Answer 1

In
=∫(-1->1) (x^2-1)^n dx
= [x(x^2-1)^n](-1,1) - 2∫(-1->1) x^2(x^2-1)^(n-1) dx
=- 2∫(-1->1) (x^2-1)^n dx - 2∫(-1->1) (x^2-1)^(n-1) dx
3In= -2I(n-1)
In= (-2/3)I(n-1)
In/I1 = (-2/3)^(n-1)
In =(-2/3)^(n-1) ∫(-1->1) (x^2-1) dx
= (-2/3)^(n-1) [ x^3/3-x](-1->1)
=(-2/3)^(n-1)[ 1/3-1 - (-1/3+1) ]
= 2.(-2/3)^(n)