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*cos(x)^3+2/3*x*(1/3*cos(x)^2*sin(x)+2/3*sin(x))+2/27*cos(x)^3+4/9*cos(x)+C
看样子是用分步积分算