参考下面的证明方法,注意积分函数不同
In=∫（sinx)＾ndx
=-∫（sinx)＾(n-1)dcosx
=-cosx*(sinx)＾(n-1)+∫cosxd[(sinx)＾(n-1)]
=-cosx*(sinx)＾(n-1)+(n-1)∫cosx*(sinx)＾(n-2)*cosxdx
=-cosx*(sinx)＾(n-1)+(n-1)∫(cosx)＾2*(sinx)＾(n-2)dx
=-cosx*(sinx)＾(n-1)+(n-1)∫[(sinx)＾(n-2)-(sinx)＾n]dx
=-cosx*(sinx)＾(n-1)+(n-1)l(n-2)-(n-1)ln
移向得
nln=-cosx*(sinx)＾(n-1)+(n-1) I(n-2)
即ln=-1/n(sinx)＾(n-1)cosx+(n-1)/n I(n-2)