Equation of Motion: Deriving from Principle of Least Action

[thepopasmurf]

I'm going through Landau/Lifgarbagez's book II of theoretical physics. In it they have a derivation of the equation of motion from the principle of least action, however I don't understand one step.

Homework Statement

Derive the equation of motion:
\frac{d^2x^i}{ds^2}+\Gamma^i_{kj} \frac{dx^k}{ds} \frac{dx^j}{ds}=0
Using the principle of least action:
\delta S=-mc\delta\int ds=0

Homework Equations

\Gamma_{i,kj}=\frac{1}{2}\left(\frac{\partial g_{ik}}{\partial x^j}+\frac{\partial g_{ij}}{\partial x^k}-\frac{\partial g_{kj}}{\partial x^i}\right)

The Attempt at a Solution

\delta ds^2=2ds\delta ds = \delta(g_{ik}dx^i dx^k)=dx^i dx^k \frac{\partial g_{ik}}{\partial x^j}\delta x^j + 2g_{ik}dx^i d\delta x^k

Therefore

\delta S = -mc\int\left\{\frac{1}{2}\frac{dx^i}{ds}\frac{dx^k}{ds}\frac{\partial g_{ik}}{\partial x^j}\delta x^j + g_{ik}\frac{dx^i}{ds}\frac{d\delta x^k}{ds}\right\}

which equals

\delta S = -mc \int \left\{\frac{1}{2} \frac{dx^i}{ds} \frac{dx^k}{ds} \frac{\partial g_{ik}}{\partial x^j}\delta x^j - \frac{d}{ds}\left\{ g_{ik} \frac{x^i}{ds}\right\} \delta x^k \right\} ds

The step I don't understand is going from the second last line to the last line.
Thanks

 

[dextercioby]

Are you sure this is ok ? You missed the d there. Ok, he takes out d/ds for the whole term (this gives 0 upon integration, because of the limit/boundary conditions), then by partial integration he gets that term with -. This is a standard trick when deriving Euler-Lagrange eqns from the action functional.

 

[thepopasmurf]

Thanks, that has made it clear for me.