作业帮求x^2(cosx)^2sinx积分
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/10/05 03:35:08
求x^2(cosx)^2sinx积分
∫x^2(cosx)^2sinx dx
= (-1/3) ∫x^2 d(cosx)^3
= -(1/3) x^2 (cosx)^3 +(1/3) ∫2x(cosx)^3 dx
= -(1/3) x^2 (cosx)^3 +(1/3) ∫2x(1-(sinx)^2 ) dsinx
= -(1/3) x^2 (cosx)^3 +(2/3) ∫x dsinx - (2/9)∫ xd(sinx)^3
= -(1/3) x^2 (cosx)^3 +(2/3)[xinx - ∫sinxdx] - (2/9)x(sinx)^3 + (2/9)∫ (sinx)^3 dx
= -(1/3) x^2 (cosx)^3 +(2/3)[xinx +cosx] - (2/9)x(sinx)^3 - (2/9)∫ (1-(cosx)^2)dcosx
= -(1/3) x^2 (cosx)^3 +(2/3)[xinx +cosx] - (2/9)x(sinx)^3 - (2/9)[cosx -(cosx)^3/3)] + C
= (-1/3) ∫x^2 d(cosx)^3
= -(1/3) x^2 (cosx)^3 +(1/3) ∫2x(cosx)^3 dx
= -(1/3) x^2 (cosx)^3 +(1/3) ∫2x(1-(sinx)^2 ) dsinx
= -(1/3) x^2 (cosx)^3 +(2/3) ∫x dsinx - (2/9)∫ xd(sinx)^3
= -(1/3) x^2 (cosx)^3 +(2/3)[xinx - ∫sinxdx] - (2/9)x(sinx)^3 + (2/9)∫ (sinx)^3 dx
= -(1/3) x^2 (cosx)^3 +(2/3)[xinx +cosx] - (2/9)x(sinx)^3 - (2/9)∫ (1-(cosx)^2)dcosx
= -(1/3) x^2 (cosx)^3 +(2/3)[xinx +cosx] - (2/9)x(sinx)^3 - (2/9)[cosx -(cosx)^3/3)] + C
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