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

[jakobs]

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?

 

[conquest]

y(x)=1/2f(x)

 

[HallsofIvy]

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.

 

[jakobs]

y(x)=1/2f(x)

So, the answer should be f(x)=1/2 sin(x) ?

 

[jakobs]

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 [itex](1/2)\int sin(u)du[/tex]. Alternatively, write $\int x sin(x^2)dx$ as $(1/2)(2)\int x sin(x^2)dx= (1/2)\int sin(x^2)(2x)dx$[/itex]

$<br /> I'm not completely understand everything you wrote there, and I feel a bit stupid now.$

 

[conquest]

@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.