= Solution
Evaluate the smooth one-form on unit velocities to obtain $f(x,v)=\theta_x(v)$. Its integral along a periodic orbit of the <geodesic flow> is exactly the one-form integral along the corresponding closed <geodesic>. Negative curvature makes this geodesic flow an <Anosov flow>; the geodesic flow of a connected closed negatively curved surface is also <topologically transitive>. Apply the smooth <Livsic theorem> to obtain $u\in C^\infty(SM)$ with
$$
Xu=\theta_x(v).
$$
If there is more than one connected component, apply this argument to each component separately.
The <vertical vector field of a surface unit tangent bundle> differentiates rotation of $v$, so
$$
VXu=\theta_x(iv),\qquad V^2Xu=-\theta_x(v).
$$
For each $x$, writing $\theta_x(v)=A\cos\vartheta+B\sin\vartheta$ shows explicitly that
$$
\int_0^{2\pi}\theta_x(v)^2\,d\vartheta
=\pi(A^2+B^2)
=\int_0^{2\pi}\theta_x(iv)^2\,d\vartheta.
$$
The fibre disintegration of <Liouville volume of a surface geodesic flow> is $d\mu=dA_g\,d\vartheta$. Integrating the displayed equality over the base therefore gives $\|VXu\|_{L^2}^2=\|Xu\|_{L^2}^2$. Substitution into the supplied <Pestov identity for a geodesic flow> yields
$$
\|XVu\|_{L^2}^2+\int_{SM}(-K)(Vu)^2\,d\mu=0.
$$
Both terms are nonnegative and $-K\ge k_0>0$. Thus $Vu=0$ everywhere. Since every unit-circle fibre is connected, $u$ is independent of velocity and descends to a smooth function $h$ on $M$: $u=h\circ\pi$. Its geodesic derivative is
$$
Xu(x,v)=dh_x(v).
$$
Comparison with $Xu=\theta_x(v)$, first on unit vectors and then on all vectors by linearity, proves
$$
\boxed{\theta=dh.}
$$
Thus the one-form is an <exact differential form>. This proves the degree-one <geodesic X-ray transform> conclusion directly from the given integral identity and the stated smooth coboundary theorem.
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