求I(n)=不定积分(sin^n(x)dx) 的递推式I(n)=f(I(n-1))

I = ∫(sinx)^ndx = ∫(sinx)^(n-2)(sinx)^2dx
= ∫(sinx)^(n-2)[1-(cosx)^2]dx
= ∫(sinx)^(n-2)dx - ∫(sinx)^(n-2)cosxdsinx
= I - [1/(n-1)]∫cosxd[(sinx)^(n-1)]
= I - [1/(n-1)](sinx)^(n-1)cosx
+ [1/(n-1)]∫[(sinx)^(n-1)](-sinx)dx
= I - [1/(n-1)](sinx)^(n-1)cosx -[1/(n-1)]I,
[n/(n-1)]I = - [1/(n-1)](sinx)^(n-1)cosx + I,
得递推公式 I = -(1/n)(sinx)^(n-1)cosx + [(n-1)/n] I.