Integration by Parts of x^5cos(x^3)

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Homework Statement



[tex]\int x^5cos(x^3) dx[/tex]

Homework Equations


[tex]\int uv' = uv - \int u'v[/tex]


The Attempt at a Solution


[tex]\int x^5cos(x^3) dx[/tex]

u = [tex]x^5[/tex]
du = [tex]5x^4[/tex]
v = [tex] \frac{sin(x^3)}{(3x^2)}[/tex]
dv = [tex]cos(x^3)[/tex]
[tex]\frac{(x^5)*(sin(x^3)}{(3x^2)} - \int\frac{5x^4*sin(x^3)}{(3x^2)} dx[/tex]

[tex]\frac{(x^3)*sin(x^3)}{3} - \int\frac{(5x^2)*sin(x^3)}{(3)} dx[/tex]

[tex]\frac{(x^3)*sin(x^3)}{3} - \frac{5}{3} \int x^2*sin(x^3) dx[/tex]

u = [tex]x^3[/tex]
du = [tex]3x^2 dx[/tex]
dx = [tex] \frac{du}{3x^2}[/tex]
[tex]\int x^2*sin(u) *\frac{du}{3x^2}[/tex]

[tex]\frac{1}{3} \int sin(u) du[/tex]

[tex]\frac{-1}{3} cos(u)[/tex]

[tex]\frac{-1}{3} cos(x^3)[/tex]

Revisting

[tex]\frac{(x^3)*sin(x^3)}{3} - \frac{5}{3} * \frac{-1}{3} cos(x^3)[/tex]

My answer:

[tex]\frac{(x^3)*sin(x^3)}{3} + \frac{5}{9} * cos(x^3) + C[/tex]

Correct Answer:

[tex]\frac{(x^3)*sin(x^3)}{3} + \frac{1}{3} * cos(x^3) + C[/tex]
 

Answers and Replies

  • #2
179
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no, your first u-dv doesn't work that way. think compostion chainrule and u-substitution.
 
  • #3
HallsofIvy
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The integral of [itex]cos(x^3)[/itex] is not [itex]sin(x^3)/3x^2[/itex]!

First write the integral as
[tex]\int x^3\left[x^2cos(x^3)\right]dx[/tex]
and let [itex]u= x^3[/itex].
 
  • #4
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May I ask what the derivative of sin(x^3)/3x^2 may be then?

Thanks, rewriting the function got me the right answer.
 
Last edited:
  • #5
HallsofIvy
Science Advisor
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I don't know why you would ask that. It might be helpful to recognize that, by the chain rule, the derivative of [itex]sin(x^3)[/itex] is [itex](2x^2)cos(x^3)[/itex].
 

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