Question

x*(sinx)^3的积分原函数是什么

请写明详细过程，谢谢。

Answer 1

要用到分部积分。

因为∫(sinx)^3dx=∫((cosx)^2-1)dcosx=(cosx)^3/3-cosx

所以

∫x(sinx)^3dx=∫xd[(cosx)^3/3-cosx]

=x[(cosx)^3/3-cosx]-∫[(cosx)^3/3-cosx]dx

=x[(cosx)^3/3-cosx]+sinx -(1/3)∫(cosx)^3dx

=x[(cosx)^3/3-cosx]+sinx -(1/3)∫[1-(sinx)^2]dsinx

=x[(cosx)^3/3-cosx]+sinx -(1/3)[sinx-(sinx)^3/3]

=(2/3)sinx-xcosx+(1/3)x(cosx)^3+(1/9)(sinx)^3+C

不过这不是最简结果，最简单结果如下

Answer 2

Answer 3

原理:积分:u'v du=uv-积分:uv' dv

积分:x^2*(sinx)^3 dx
=积分:x^2*(sinx)*[1-(cosx)^2] dx
=积分:x^2*sinx dx + 积分:x^2*(sinx)*(cosx)^2 dx
=(-cosx)*x^2+积分:2x(cosx) dx +(1/3)(x^2)(cosx)^3-积分:(2/3)*x*(cosx)^3 dx
=(-cosx)*(x^2)+(2x)(sinx)-积分:2sinx dx + (1/3)(x^2)(cosx)^3-积分:(2/3)*x*(cosx)*[1-(sinx)^2] dx
=(-cosx)*(x^2)+(2x)(sinx)+2cosx+ (1/3)(x^2)(cosx)^3-积分:(2/3)*x*(cosx) dx + 积分:(2/3)*x*(cosx)*(sinx)^2 dx

=(-cosx)*(x^2)+(2x)(sinx)+2cosx+ (1/3)(x^2)(cosx)^3-(2/3)*x*(sinx)+积分:(2/3)(sinx) dx + (2/9)*x*(sinx)^3 - 积分:(2/9)*(sinx)^3 dx

=(-cosx)*(x^2)+(2x)(sinx)+2cosx+ (1/3)(x^2)(cosx)^3-(2/3)*x*(sinx)-(2/3)(cosx) + (2/9)*x*(sinx)^3 -积分:(2/9)*(sinx)[1-(cosx)^2] dx

=(-cosx)*(x^2)+(2x)(sinx)+2cosx+ (1/3)(x^2)(cosx)^3-(2/3)*x*(sinx)-(2/3)(cosx) + (2/9)*x*(sinx)^3 -积分:(2/9)(sinx) dx+积分:(2/9)(sinx)(cosx)^2 dx

=(-cosx)*(x^2)+(2x)(sinx)+2cosx+ (1/3)(x^2)(cosx)^3-(2/3)*x*(sinx)-(2/3)(cosx) + (2/9)*x*(sinx)^3 +(2/9)(cosx)-(2/27)(cosx)^3

=-(x^2)cosx+(4/3)xsinx+(14/9)cosx+(1/3)(x^2)(cosx)^3+(2/9)x(sinx)^3-(2/27)(cosx)^3+C