∫ sin∧n t dt等于多少?

1个回答

  • ∫[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