Integrate ( x^5) sin ( x^3)
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Integrate ( x^5) sin ( x^3)

[From: ] [author: ] [Date: 12-02-14] [Hit: ]
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),......
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)

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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

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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.
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