equation of motion for the scalar field via variational principle in general relativity

I would like to find the equation of motion for the scalar field $\phi$ by varying the following action in General Relativity.

Special Relativity: $$S = -\tfrac{1}{2}\int d^4\xi\, \eta^{ab} \partial_a \phi\partial_b\phi$$

General Relativity: $$S = -\tfrac{1}{2}\int d^4x \sqrt{g}\, g^{\mu\nu} \partial_{\mu} \phi\partial_\nu \phi$$

I was able to get the correct equation using the covariant derivatives. Since they are constant with respect to the metric partial integration works and one obtains $\Box\phi = 0$. But since for the scalar field the covariant and the partial derivates are the same I wanted to vary the action with the partial derivatives.

Note: $\phi \rightarrow \phi + \delta\phi$ \begin{align} \delta S &= -\tfrac{1}{2}\int d^4x \sqrt{g}\, (g^{\mu\nu} \partial_{\mu} \delta\phi\partial_\nu\phi + g^{\mu\nu} \partial_{\mu} \phi\partial_{\nu} \delta\phi ) \\ % &=-\int d^4x \sqrt{g}\, g^{\mu\nu} \partial_{\mu} \delta\phi\partial_\nu\delta\phi \\ % &=\int d^4x \, \partial_\nu(\sqrt{g}\,g^{\mu\nu}\partial_\mu\phi)\delta\phi \end{align} The third line one obtains after partial integration. But at this point I'm stuck. I know the following definition: $$\Box\phi = \tfrac{1}{\sqrt{g}}\partial_\mu(\sqrt{g}g^{\mu\nu}\partial_\nu\phi)$$ But I'm not able to convert my obtained solution into this. Thanks for your help!

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You already have your answer though. Given your result for $\delta S$, the equation you get is $\partial_\mu( \sqrt{g} g^{\mu\nu} \partial_\nu \phi) = 0$. This is the wave equation. Dividing by $\sqrt{g}$ (which can be done since the metric is positive-definite) and exchanging the indices $\mu$ and $\nu$ (note that the metric is symmetric), we produce $\Box \phi = 0$. –  Prahar Jan 21 '14 at 15:26
@Prahar Small comment: the metric of GR is not necessarily positive definite, it is however non-degenerate. –  joshphysics Jan 21 '14 at 18:09