Finding the Primitive Function for y'(x)=xcos(x2) | Homework Help and Equations

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SUMMARY

The discussion focuses on finding the primitive function for the derivative y'(x) = x cos(x²). The correct primitive function is y(x) = (1/2) sin(x²), derived using substitution where u = x² and du = 2x dx. This method simplifies the integral ∫ x sin(x²) dx to (1/2) ∫ sin(u) du. The participants confirm that the constant factor of 1/2 is essential for accurately determining the primitive function.

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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?
 
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y(x)=1/2f(x)
 
A more "formal" way to do this would be to let u= x^2 so that du= 2xdx or (1/2)du= xdx. Then \int x sin(x^2)dx becomes (1/2)\int sin(u)du. Alternatively, write \int x sin(x^2)dx as (1/2)(2)\int x sin(x^2)dx= (1/2)\int sin(x^2)(2x)dx

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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conquest said:
y(x)=1/2f(x)

So, the answer should be f(x)=1/2 sin(x) ?
 
HallsofIvy said:
A more "formal" way to do this would be to let u= x^2 so that du= 2xdx or (1/2)du= xdx. Then \int x sin(x^2)dx becomes (1/2)\int sin(u)du[/tex]. Alternatively, write \int x sin(x^2)dx as (1/2)(2)\int x sin(x^2)dx= (1/2)\int sin(x^2)(2x)dx<br />
<br /> <br /> I&#039;m not completely understand everything you wrote there, and I feel a bit stupid now.
 
@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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