What is the relationship between variational calculus and integration by parts?

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The discussion explains the relationship between variational calculus and integration by parts in the context of the action integral S and the Lagrangian density. The variation of S leads to an expression that requires integration by parts to manipulate the second term involving the derivative of the Lagrangian. By applying the chain rule, the participants clarify how the partial derivative can be rearranged, demonstrating that the derivative commutes with the variation operator. This manipulation ultimately reveals the connection between variational calculus and integration by parts. Understanding this relationship is crucial for grasping the underlying principles of the calculus of variations.
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I'm reading a book - and I've been stuck for a while on the same page. This is only a calculus question. We have the action:

S=\int d^4x \;\mathcal{L}

with the Lagrangian (density):

\mathcal{L}=\mathcal{L}(\phi,\dot{\phi},\nabla\phi)

We then vary S:

\delta S = \int d^4x \left[\frac{\partial \mathcal{L}}{\partial \phi}\delta\phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \delta(\partial_\mu \phi)\right]

which is all fine and dandy, but now the next line says:
= \int d^4 x \left[\frac{\partial \mathcal{L}}{\partial \phi}-\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial (\partial_\mu\phi)} \right) \right]\delta \phi + \partial_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \delta \phi\right)

Something like integration by parts must have befallen the second term...but I don't see it. I'm very inadequate in variational calculus, the mere sight of \delta throws me off.

So what happened there between the two lines? Thanks
 
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Indeed it is integration by parts. By the chain rule we have:

\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_ \mu\phi)}\delta\phi\right)=\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial( \partial_\mu\phi)}\right)\delta\phi+\frac{\partial\mathcal{L}}{ \partial(\partial _\mu\phi)}\partial_\mu(\delta\phi)

The only additional step you need then is to know that the partial derivative commutes with the ##\delta## so that the above equation can be rearranged to:

\frac{\partial\mathcal{L}}{\partial(\partial_ \mu\phi)}\delta(\partial_\mu\phi)=\partial_\mu\left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi\right)-\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial _\mu\phi)}\right)\delta\phi
 
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