# Work-energy theorem integral

1. Jul 31, 2014

### DiracPool

I'm having trouble with an integral involved in deriving the work-energy theorem

1. The problem statement, all variables and given/known data

I'm trying to get from ∫mv/√(1-v^2/c^2)dv to -mc^2(1-v^2/c^2).

2. Relevant equations

3. The attempt at a solution

I start out by putting gamma on top to yield: ∫mv(1-v^2/c^2)^-1/2, then I square everything to get rid of the square root term and end up with: ∫m^2 v^2 - m^2 v^2 (v^2/c^2)dv, and then to get rid of the fractions I end up with: ∫c^2 m^2 v^2 - m^2 v^4dv, which, when I try to integrate, gets me no where close to the answer. What am I doing wrong? Should I be doing a U-substitution or chain rule?

2. Jul 31, 2014

### Simon Bridge

Here, let me help,
$$\int \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\;dv \rightarrow -mc^2\left(1-\frac{v^2}{c^2}\right)$$... you are trying to work the maths from the LHS to the RHS.
... yes :)

try $u=v/c$ to start with.

3. Aug 1, 2014

### DiracPool

Ok, now I see.. If I take U=1-v^2/c^2, then I get dU/dx=-2v/c^2, and then just follow the protocol, and sure enough, there's the right answer.

Thanks Simon!

4. Aug 1, 2014

### Simon Bridge

Well done - you usually want to look for some sort of substitution for the bit that gives you problems.
Don't be frightened to try several different ones.

Since you will be doing a lot of this, it is best practice to learn LaTeX :)

5. Aug 2, 2014

### vanhees71

To be fair one should say that the original statement is not correct. There's a square root missing in the given result of the integral!

6. Aug 2, 2014

### HallsofIvy

Staff Emeritus
Also, since the integral on the left is an indefinite integral, there should be an added "constant of integration". What you are trying to prove simply isn't true!