Please Help Integration!

  • Thread starter mckallin
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  • #1
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Please Help!! Integration!

can anyone help me solve the following integration? thanks a lot.

[tex]\int x^5 cos(x^3) dx[/tex]
 

Answers and Replies

  • #2
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try to solve it by parts
 
  • #3
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I have tried it, but it doesn't work, at least, for me.

If I make [tex] u=cos(x^3), dv=x^5 dx [/tex], the grade of x, which is in the [tex] cos(x^3) [/tex], won't be reduce.

If I make [tex] u=x^5, dv=cos(x^3) dx [/tex], I can't solve the [tex] \int cos(x^3) dx[/tex].

Could you give me some more advice? thanks
 
  • #4
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What would you like to have inside the integral in order to have

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

be solvable? Can you choose slightly different u and dv to accomplish that?
 
  • #5
From [itex](\sin x^3)'=3\,x^2\,\cos x^3[/itex] deduce that [itex]\cos x^3=\frac{1}{3\,x^2}\,(\sin x^3)'[/itex] and use that to integrate by parts.
 
  • #6
HallsofIvy
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Oh, for goodness sake! Make the substitution u= x3, then integrate by parts!

(I notice now that that is essentially what Rainbow Child said.)
 
  • #7
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From [itex](\sin x^3)'=3\,x^2\,\cos x^3[/itex] deduce that [itex]\cos x^3=\frac{1}{3\,x^2}\,(\sin x^3)'[/itex] and use that to integrate by parts.
How it can help?

Make the substitution u= x3, then integrate by parts!
How this can help?
 
  • #8
Hootenanny
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How this can help?
I don't think there's any harm in showing the substitution,

[tex]u = x^3 \Rightarrow \frac{du}{dx} = 3x^2 \Rightarrow dx = \frac{du}{3x^2}[/tex]

Hence when we make the substitution the integral becomes,

[tex]\int \frac{x^5}{3x^2}\cos(u)du = \frac{1}{3}\int u\cos(u)du[/tex]

Which is a simple integral to solve.
 
Last edited:

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