Question

How would you evaluate the integral of dx/(a^2+x^2) by first changing it into a more familiar integral using the substitution x=au?

Answer 1

Let's do what the question suggests :)
\[\Large \int \frac{dx}{a^2+x^2}\]

Answer 2

It says let x = au right?

Answer 3

Following that, must be dx?

Answer 4

When I tried this myself I ended up with that substitution and du = (1/a)dx but when I plugged things in it wasn't turning out for me. I know it's supposed to end up as (1/a) arctan(x/a) but I have no idea where the arctan comes from.

Answer 5

Well, to change it to 'something more familiar', it would seem you actually have to be familiar with it first :D

Having said that, you know what the derivative of arctan is?
\[\Large \frac{d}{dx}\tan^{-1}(x) = \color{red}?\]

Answer 6

Yes, it's \[1/(1 + x^2) \]

Answer 7

Oh... okay, great :D
Now, after we let x = au
it follows that dx = a du right?

Answer 8

yes

Answer 9

Right, let's make that substitution then...
\[\large x = \color{red}{au}\]\[\large dx = \color{blue}{a \ du}\]
\[\Large \int \frac{\color{blue}{a \ du}}{a^2 + (\color{red}{au})^2}\]

Answer 10

Catch me so far?

Answer 11

yep I got that far

Answer 12

Okay, at the denominator, we can 'expand', and then factor out:
\[\Large \int \frac{a \ du }{a^2 + a^2u^t}\]

Answer 13

\[\Large \int \frac{a \ du}{a^2(1+u^2)}\]

Answer 14

Can you take it from here?

Answer 15

I think so, let me look a bit.

Answer 16

Remember that a is just a constant...

Answer 17

Yeah I got it now, thank you so much!

Answer 18

Okay, awesome ^_^