(sinx)^n的不定积分递推公式如何导出?是不定积分!

Question

(sin x)^n 的不定积分递推公式如何导出?是不定积分!

Answer 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.