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求微分sin^2x*cos^5x*dx
人气:451 ℃ 时间:2020-02-03 11:08:33
解答
∫(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)^5-4sin...答案应该是1/3(sinx)^3-2/5(sinx)^5+1/7(sinx)^7+C∫(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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