∫ sin∧n t dt等于多少?这是一个降幂公式.被积函数为sint的N次方.上标二分之派.下标为0

Question

∫ sin∧n t dt等于多少?
这是一个降幂公式.被积函数为sint的N次方.上标二分之派.下标为0

Answer 1

∫[0,π/2](sint)^ndt
=∫[0,π/2]sint^(n-1)d(-cost)
= -sint^(n-1)cost|[0,π/2]+∫[0,π/2]cost^2*(n-1)sint^(n-2)dt
=∫[0,π/2](1-sint^2)*(n-1)sint^(n-2)dt
=∫[0,π/2](n-1)sint^(n-2)dt - (n-1)∫[0,π/2](sint)^ndt
n∫[0,π/2]sint^ndt=(n-1)∫[0,π/2]sint^(n-2)dt
In=∫[0,π/2]sint^ndt
In=[(n-1)/n]In-2
∫[0,π/2]sintdt=-cost|[0,π/2]=1
∫[0,π/2]sint^2dt=∫[0,π/2](1-cos2t)/2 dt=π/4
I3=(2/3) I5=(2/3)(4/5) I(2n-1)=(2*3*..*(2n-2))/(1*3*5*..*(2n-1))
I6=(5/6)(π/4) I8=(5/6)(7/8)(π/4) (n>3) I2n=[(5*7*..*(2n-1))/(4*6*..*2n) ] π