What's wrong with my integration for the work-energy theorem?

[DiracPool]

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

Homework Statement

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

Homework Equations

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?

 

[Simon Bridge]

I'm having trouble with an integral involved in deriving the work-energy theorem
...
I'm trying to get from ∫mv/√(1-v^2/c^2)dv to -mc^2(1-v^2/c^2).

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.

Should I be doing a U-substitution or chain rule?

... yes :)

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

 

[DiracPool]

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.

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!

 

[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 :)

 

[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!

 

[HallsofIvy]

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!