# Primitive funktion help

• jakobs
In summary, the conversation discusses how to find the primitive function for y'(x)= xcos(x²). One method is to let u=x² and use the formula for the integral of sin(u). Another method is to write the integral as (1/2)(2)∫xsin(x²)dx and use the formula for the integral of sin(x²). The final solution is y(x)= 1/2 sin(x²).

## Homework Statement

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

## 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?

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

I'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.