求微分sin^2xcos^5xdx

来源：学生作业帮编辑：搜搜考试网作业帮分类：数学作业时间：2024/05/17 02:55:26

求微分sin^2x*cos^5x*dx

∫(cosx)^5·(sinx)²dx
=∫(cosx)^4·(sinx)²d(sinx)
=∫[(1-sinx)²]²(sinx)²d(sinx)
=∫(sin³x-2sin²x+sinx)² d(sinx)
=∫[(sinx)^6+4(sinx)^4+sin²x-4(sinx)^5-4sin³x+2(sinx)^4 ]d(sinx)
=1/7·(sinx)^7+4/5·(sinx)^5+1/3·sin³x-2/3·(sinx)^6-(sinx)^4+2/5·(sinx)^5+C
=1/7·(sinx)^7+6/5·(sinx)^5+1/3·sin³x-2/3·(sinx)^6-(sinx)^4+C
再问：答案应该是1/3(sinx)^3-2/5(sinx)^5+1/7(sinx)^7+C
再答： ∫(cosx)^5·(sinx)²dx =∫(cosx)^4·(sinx)²d(sinx) =∫(1-sin²x)²·(sinx)²d(sinx) =∫[(sinx)^6+2(sinx)^4+(sinx)^2]d(sinx) =1/3(sinx)^3-2/5(sinx)^5+1/7(sinx)^7+C

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