Green Function Homework: Problem & Solutions

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Homework Statement



The problem is attached in the first picture, the provided solutions are in the second.

The Attempt at a Solution



I got to where they are, but aren't they missing an additional term of sin(t)*cos(∏)*f(∏) from the second integral in dx/dt ?
 

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Millennial said:

Ok! I worked it out finally. Will the leibniz rule work if i take 'cos(t)' and 'sin(t)' out of the integral such that the inside of the integral only becomes one variable: ∫-sin(ζ)f(ζ) and ∫-cos(ζ)f(ζ) ??


The answer seems to use the leibniz rule but they imply that the 'cos(t)' and 'sin(t)' were taken out of the integral..
 
cos(t) and sin(t) are treated as constants in the integral because the integral is with respect to zeta, and the integral itself is treated as a constant in differentiation because the derivative is taken with respect to t. The function inside inside the integral that involves t is differentiated as usual, and the integral is treated as a constant so it is left as-is.
 
Millennial said:
cos(t) and sin(t) are treated as constants in the integral because the integral is with respect to zeta, and the integral itself is treated as a constant in differentiation because the derivative is taken with respect to t. The function inside inside the integral that involves t is differentiated as usual, and the integral is treated as a constant so it is left as-is.

Sorry I'm not sure what you mean but here's the method that I used (attached in the picture)

In the picture its ∫ f(x,t) from u to v.

In the question of this thread, I am above to take 'sin(t)' and 'cos(t)' out of the integration so that the integration only has 1 variable.

My question is:

How do we still apply (or can we) leibniz's rule if this is the case? (Only 1 variable)
 

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Yes, you can, just take f(x,t)=f(t) (check the link I gave about differentiation under the integral sign.)