Jump to content

# Formula Derivation Proof

## Recommended Posts

In class today, we had a problem in which we needed to integrate a function following the form

$\int{\sqrt{u^2 \pm a^2}\,du}$

and it said to refer to the appendices which gave that integral was equal to

$\frac{1}{2}(u\sqrt{u^2+a^2}\pm a^2\ln{|u+\sqrt{u^2+a^2}}|)$.

I tried to do this (using $+$ for the $\pm$) and I first did some rationalizing

$\int{\sqrt{u^2 \pm a^2}} \frac{\sqrt{u^2 \pm a^2}}{\sqrt{u^2 \pm a^2}}\,du$

as to split it into

$\int{\frac{u^2+a^2}{\sqrt{u^2+a^2}}du} = \int{\frac{u^2}{\sqrt{u^2+a^2}}du} + \int{\frac{a^2}{\sqrt{u^2+a^2}}du}$

and the second fraction is, if I understand correctly,

$\int{\frac{a^2}{\sqrt{u^2+a^2}}du} = a^2 \sinh^{-1}\frac{u}{a} = a^2\ln{|u+\sqrt{u^2+a^2}|}$

which would reason that

$\int{\frac{u^2}{\sqrt{u^2+a^2}}du} = \frac{1}{2}(u\sqrt{u^2+a^2}-a^2\ln{|u+\sqrt{u^2+a^2}|})$

but I can't find how to do this. Am I overlooking a simple technique? I also can't look up the proof because I don't know the name or whatever. Can anyone explain this last step, or did I start wrong?

(Either the whole derivation or just how to integrate $\int{\frac{u^2}{\sqrt{u^2+a^2}}du}$ )

##### Share on other sites

I don't think this is the approach you should be taking. The very last integral you state is quite non-obvious to integrate at first sight. However, you can solve your original equation by simply letting $u = a\sinh t$. Then it transforms into:

$a^2 \int \cosh^2 t \, dt$

by using $\cosh^2 t - \sinh^2 t = 1$. Evaluating this integral is easy by using the double angle formulae for cosh2 in terms of cosh(2x). The result follows from taking the inverse of your substitution.

##### Share on other sites

I don't think this is the approach you should be taking. The very last integral you state is quite non-obvious to integrate at first sight. However, you can solve your original equation by simply letting $u = a\sinh t$. Then it transforms into:

$a^2 \int \cosh^2 t \, dt$

by using $\cosh^2 t - \sinh^2 t = 1$. Evaluating this integral is easy by using the double angle formulae for cosh2 in terms of cosh(2x). The result follows from taking the inverse of your substitution.

Oh thanks! I'm new to the hyperbolic functions so I didn't think that at first. It looks much simpler now.

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account

## Sign in

Already have an account? Sign in here.

Sign In Now
×

• #### Activity

• Leaderboard
×
• Create New...

## Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.