Question

设In=∫sinnxdx,证明： In= -1/n(sinn-1xcosx)+(n-1)/n〔In-2〕 设中是sinx

Answer 1

可以这样证明：

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)

命题得证