Here is the problem as I have been given it:
A curve $C$ is given in Cartesian coordinates by $r(t) = (cos(sin(nt))cost,\; cos(sin(nt))sint,\; sin(sin(nt)))$, with $t$ between $0$ and $2$$\pi$ inclusive.
Show that $\int_C H.dr = 2\pi$, where $H(r) = (-y/(x^2 + y^2), \; x/(x^2 + y^2),\; 0)$ where C is traversed in the direction of increasing t.
Now my question: I know that I can do it simply by finding the derivative of $r$ w.r.t. $t$ and dotting with $H$, and it all simplifies nicely to a doable integral. But $H$ is very close to being $d((1/2)arctan(x^2 + y^2))$, except that the $x$ and $y$ are the wrong way around. So I've tried changing basis, by rotation of $\pi/2$ about the $z$-axis so that the new $y$-coordinate is the old $x$-coordinate, and the new $x$-coordinate is minus the old $y$-coordinate. Then, $\int_C H.dr = \int_C d((1/2)arctan(x^2 + y^2)) = [(1/2)arctan(x^2 + y^2)]$ with the limits put in (sub in for $t = 2\pi$ and $t = 0$ into $r$ in the new basis. But I can't get the integral to give an answer of $2\pi$. Instead, I keep getting zero.
Can someone try and make the method of changing basis work?
The integrand $H$ is the gradient vector of $$\arctan\Bigl(\frac yx\Bigr)\ ,$$ not $\frac12\arctan(x^2+y^2)$. However this is not valid when $x=0$ and so I don't think it can be used, at least not easily, to evaluate the integral.