Please explain how this integration is done

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This discussion centers on the mathematical derivation of the right-hand side (RHS) of equation (2) from Lord Rayleigh's paper regarding pressure generated during bubble collapse in a liquid. The key technique involves substituting \(\frac{UR^2}{r^2}\) for \(u\) in the integral as indicated in equation (1). This substitution leads to a successful integration, clarifying the initial query posed by the user Stephen. The interaction concludes with Stephen acknowledging the resolution of his question.

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bksree
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Hi
Attached is an extract of a paper by Lord Rayleigh on pressure generated during collapse of a bubble in a liquid. Will someone please explain how the RHS of equation (2) in the attachment is obtained ?

TIA
 

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Use eq (1) to Substitute [itex]\frac{UR^2}{r^2}[/itex] for [itex]u[/itex] in the integral. Doesn't the integration work out after that?
 
Stephen
Thank you. Got it now. On hindsight, what a silly question to ask.

Thanks again
 

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