Let $f$ be a continuous function. Prove that for every closed simple (smooth) curve $\gamma$: $\oint_\gamma f(\mathbf{x} \cdot \mathbf{x} ) \mathbf{x}\cdot d\mathbf{x} = 0$
I know that Integral of continuous function over a closed smooth curve is $0$ if and only if the value of the integral is independent of the curve, i.e. for every $2$ smooth closed curves that starts and ends at the same point $\gamma_1, \gamma_2$ we get that $\oint_{\gamma_1} f(\mathbf{x} \cdot \mathbf{x}) \mathbf{x} \cdot d\mathbf{x} = \oint_{\gamma_2} f(\mathbf{x} \cdot \mathbf{x} ) \mathbf{x} \cdot d\mathbf{x}$.(Mark it as condition (*))
Also, I know that condition (*) exists if and only if $f(x \cdot x ) x =\nabla h$ for some $h$.
So all I need is to prove that $f(\mathbf{x} \cdot \mathbf{x} ) \mathbf{x} $ is a gradient of some other function.
I define $g(u)=\frac{1}{2}\int_0^uf(t)dt$.
Therefore, am I correct that $\frac{\partial}{\partial \mathbf{x}}g(\mathbf{x}^2)= \frac{1}{2}f(\mathbf{x} \cdot \mathbf{x})\frac{\partial}{\partial \mathbf{x}}(\mathbf{x}^2)=f(\mathbf{x}^2)\mathbf{x}$? (and then $f = \nabla g(\mathbf{x}^2)...)$
And then I can use the two theorems mentioned above, and conclude that $\gamma$: $\oint_\gamma f(\mathbf{x} \cdot \mathbf{x} ) \mathbf{x} \cdot d\mathbf{x} = 0$?