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Primitive funktion help

  • Thread starter jakobs
  • Start date
  • #1
15
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Homework Statement


I have y'(x)= xcos(x2)
How do I get the primitive function for this?

Homework Equations





The Attempt at a Solution


I know that f'(x)=2xcos(x2) is f(x)=sin(x2)
How will removing that 2 in front of xcos affect the primitive function?
 

Answers and Replies

  • #2
133
4
y(x)=1/2f(x)
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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A more "formal" way to do this would be to let [itex]u= x^2[/itex] so that [itex]du= 2xdx[/itex] or [itex](1/2)du= xdx[/itex]. Then [itex]\int x sin(x^2)dx[/itex] becomes [itex](1/2)\int sin(u)du[/itex]. Alternatively, write [itex]\int x sin(x^2)dx[/itex] as [itex](1/2)(2)\int x sin(x^2)dx= (1/2)\int sin(x^2)(2x)dx[/itex]

Note that you can only move constants into and out of the integral like that. If that first "x" were not there, you could not do the integral.
 
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  • #4
15
0
y(x)=1/2f(x)
So, the answer should be f(x)=1/2 sin(x) ?
 
  • #5
15
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A more "formal" way to do this would be to let [itex]u= x^2[/itex] so that [itex]du= 2xdx[/itex] or [itex](1/2)du= xdx[/itex]. Then [itex]\int x sin(x^2)dx[/itex] becomes [itex](1/2)\int sin(u)du[/tex]. Alternatively, write [itex]\int x sin(x^2)dx[/itex] as [itex](1/2)(2)\int x sin(x^2)dx= (1/2)\int sin(x^2)(2x)dx[/itex]
I'm not completely understand everything you wrote there, and I feel a bit stupid now.
 
  • #6
133
4
@jakobs
well you actually put f(x)=sin(x²). So I meant y(x)= 1/2 sin(x²). Then indeed y'(x)=xcos(x²).

@Halls of Ivy
Agreed this would be the 'follow the recepy' way of doing it, but with a problem this straightforward it really isn't necessary.
 

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