Someone asked this question and then deleted it just before I provided the answer which is,
S { (x^5) sin(x^3) }dx = (1/3){ sin(x^3) -(x^3) cos(x^3) }
Its a bit to envolved to show the details of the derevation but the key is to represent
sin(x^3) as (1/2j)[ e^(jx^3) - e(-jx^3)] , which will produce two integrals each of which can be integrated by parts , one using u=(1/3) X^3 and dv = 3jx^2 e^(jx^3) ,
and the second integral using u=(1/3)x^3 and dv= -3jx^2 e^(-jx^3) where j = (-1)^(1/2)
S { (x^5) sin(x^3) }dx = (1/3){ sin(x^3) -(x^3) cos(x^3) }
Its a bit to envolved to show the details of the derevation but the key is to represent
sin(x^3) as (1/2j)[ e^(jx^3) - e(-jx^3)] , which will produce two integrals each of which can be integrated by parts , one using u=(1/3) X^3 and dv = 3jx^2 e^(jx^3) ,
and the second integral using u=(1/3)x^3 and dv= -3jx^2 e^(-jx^3) where j = (-1)^(1/2)
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Your way is fine for you; mine feels easier to me:
Let w = x^3, dw = 3x^2 dx,
Integrate (1/3) w sin(w) dw.
u = (1/3) w, dv = sin(w) dw, v = -cos(w), du = (1/3) dw; we get
- (1/3) w cos(w) + (1/3) integral of cos(w) dw
= -(1/3) w cos(w) + (1/3) sin(w) = your answer
Let w = x^3, dw = 3x^2 dx,
Integrate (1/3) w sin(w) dw.
u = (1/3) w, dv = sin(w) dw, v = -cos(w), du = (1/3) dw; we get
- (1/3) w cos(w) + (1/3) integral of cos(w) dw
= -(1/3) w cos(w) + (1/3) sin(w) = your answer
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"Someone asked this question and then deleted it just before I provided the answer which is," This happens to all of us, I am guessing the answerer got an answer before you posted, saved it and deleted the question (for whatever reason they do it). While we are on topic though, I do not agree you did this in the most direct way, and your suggestion is not "key," (if I may suggest, that is).
The way this integral is set it up seems to encourage the substitution u = x^3, then du / 3 = x^2 dx, and x^5 = u du / 3
Thus the integral is (I1/3)u sin(u) du
which can be solved by one application of integration by parts to give -(1/3) u cos(u) + (1/3)sin(u) = - (1/3)x^3 cos(x^3) + (1/3)sin(x^3)
Note, that in the spirit of your method, you could also do this by writing the integral as Im [ x^5 sin(x^3)], the answer literally gets translated back to a real-valued basis by way of the imaginary operator. I think this pathway is more straightforward given it does not burden you to recognize sines and cosines in terms of complex exponentials after you integrate.
The way this integral is set it up seems to encourage the substitution u = x^3, then du / 3 = x^2 dx, and x^5 = u du / 3
Thus the integral is (I1/3)u sin(u) du
which can be solved by one application of integration by parts to give -(1/3) u cos(u) + (1/3)sin(u) = - (1/3)x^3 cos(x^3) + (1/3)sin(x^3)
Note, that in the spirit of your method, you could also do this by writing the integral as Im [ x^5 sin(x^3)], the answer literally gets translated back to a real-valued basis by way of the imaginary operator. I think this pathway is more straightforward given it does not burden you to recognize sines and cosines in terms of complex exponentials after you integrate.