# Derivation of relativistic E_kin

1. Oct 9, 2013

### tomwilliam2

In my textbook there is a derivation of relativistic kinetic energy starting from an integral of the force applied over the distance required to take the particle's speed from 0 to v.
There's one stage of the derivation I don't understand on mathematical grounds, which is going from:

$$E_k = \int_0^{v} u\ d\left( \frac{mu}{\sqrt{1-(u^2/c^2)}}\right )$$
To the next line, which is:
$$E_k = \left [ \frac{mu^2}{\sqrt{1-(u^2/c^2)}}\right]_0^v - \int_0^v \frac{mu}{\sqrt{1-(u^2/c^2)}}\ du$$
I guess they have used integration by parts, but how do you get the change of integrating variable? I only really know the rule $\int f'g dx = fg - \int fg' dx$ and can't seem to make it fit here.
Thanks in advance

2. Oct 9, 2013

### Simon Bridge

What was the original variable being integrated over?
What was substituted in it's place?

3. Oct 9, 2013

### phyzguy

Take:
$$f' = d (\frac{mu}{\sqrt{1-(u^2/c^2)}}), g = u$$

4. Oct 9, 2013

### tomwilliam2

Ah, thanks phyzguy...the penny has dropped.

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