数学求定积分的题 x^2 * (sinx)^3+tanx-1 的原函数是?打清楚一点,

∫(x^2 * (sinx)^3+tanx-1)dx
= - j /2∫x2*(ej3x-e-j3x)dx+∫(sinx/cosx)dx+x
又∫x2*ej3x dx
= - x2*ej3x/(3j)+2/(3j)*∫x*ej3xdx
= - x2*ej3x/(3j)-2*x*ej3x/9+9*∫ej3xdx
= - x2*ej3x/(3j)-2*x*ej3x/9+2ej3x/(27j)+C1
令x= -x,
则∫(-x2)*e-j3x d(-x)
== - x2*e-j3x/(3j)+2*x*e-j3x/9+2e-j3x/(27j)+C2
即 -∫x2*e-j3x dx
= - x2*e-j3x/(3j)+2*x*e-j3x/9+2e-j3x/(27j)+C2
∫(sinx/cosx)dx= -∫1/cosx d(cosx)= -ln｜cosx ｜+ C3
由以上得,
∫(x^2 * (sinx)^3+tanx-1)dx
= -j [-x2*(ej3x+ e-j3x)/(3j)- 2*x*(ej3x/9- e-j3x)+ 2(ej3x -e-j3x) /(27j)]/2 -ln｜cosx ｜+ x+C
= x2*(ej3x+ e-j3x)/6+j x*(ej3x- e-j3x) /9- (ej3x -e-j3x) /27- ln｜cosx ｜+ x+C
=x2*cos(3x)/3-2xsin(3x)/9-2jsin(3x)/27-ln｜cosx ｜+ x+C
仅供参考.