问题标题：
【(sinx)9dx的不定积分怎么求？】
问题描述：
(sinx)9dx的不定积分怎么求？
沈熙回答：
Generalsolution：
∫sin^9(x)dx
=-∫sin^8(x)d(cosx)
=-∫(sin??x)??d(cosx)
=-∫(1-cos??x)??d(cosx)
=-∫(1-u??)??du
=-∫[1+4(-u??)+6(-u??)??+4(-u??)??+(-u??)??]du
=∫(-1+4u??-6u??+4u^6-u^8)du
=-u+(4/3)u??-(6/5)u^5+(4/7)u^7-(1/9)u^9+C
=-cosx+(4/3)cos??x-(6/5)cos^5(x)+(4/7)cos^7(x)-(1/9)cos^9(x)+C
Alternatives：
BytheReductionformulaI_(n)=∫[sinx]^ndx=-(1/n)cosx[sinx]^(n-1)+(n-1)/n*I_(n-2)
∫sin^9xdx
=-(1/9)sin^8xcosx+(8/9)∫sin^7xdx
=-(1/9)sin^8xcosx-(8/63)sin^6xcosx+(16/21)∫sin^5xdx
=-(1/9)sin^8xcosx-(8/63)sin^6xcosx-(16/105)sin??xcosx+(64/105)∫sin??xdx
=-(1/9)sin^8xcosx-(8/63)sin^6xcosx-(16/105)sin??xcosx-(64/315)sin??xcosx+(128/315)∫sinxdx
=-(1/9)sin^8xcosx-(8/63)sin^6xcosx-(16/105)sin??xcosx-(64/315)sin??xcosx-(128/315)cosx+C
杜尊桐回答：
等于-cosx*9x
沈晓冬回答：
f(x)=∫(sinx)^9dx=-∫(sinx)^8d(cosx)=-∫(1-cos^2x)^4dcosx
令cosx＝t
f(t)=-∫(1-t^2)^4dt=x-4x^3/3+6x^5/5-4x^7/7+x^9/9+C
带入t=cosx就可以了
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