这个是用 部分积分+解方程+迭代 实现的.
解题须知:
∫ u dv = u v - ∫ v du
设:
J(n) = ∫ cos(t)^n dt
u = cos(t)^(n-1)
du = - (n-1) cos(t)^(n-2) sin(t) dt
v = sin(t)
dv = cos(t) dt
(上下限从略)
J(n) = ∫ cos(t)^n dt
J(n) = ∫ cos(t) cos(t)^(n-1) dt
J(n) = ∫ u dv
J(n) = u v - ∫ v du
J(n) = cos(t)^(n-1) sin(t) + ∫ sin(t) (n-1) cos(t)^(n-2) sin(t) dt
J(n) = 0 + (n-1) ∫ sin(t)^2 cos(t)^(n-2) dt ----------(注1)
J(n) = (n-1) ∫ (1 - cos(t)^2) cos(t)^(n-2) dt
J(n) = (n-1) ∫ (cos(t)^(n-2) - cos(t)^n) dt
J(n) = (n-1) (J(n-2) - J(n))
J(n) = (n-1) J(n-2) - (n-1) J(n)
n J(n) = (n-1) J(n-2)
J(n) = ((n-1) / n) J(n-2)
所以,
原式
= (2/3) J(6)
= (2/3) (5/6) J(4)
= (2/3) (5/6) (3/4) J(2)
= (2/3) (5/6) (3/4) (1/2) J(0)
= (2/3) (5/6) (3/4) (1/2) (π/2) ----------(注2)
= 5π / 48
注1:cos(t)^(n-1) sin(t) = 0 是因为要代入上下限并求其差,结果为0
注2:J(0) = ∫ cos(t)^0 dt = ∫ 1 dt = π/2 ,这个积分应该很容易求得.
