anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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terenzreignz
  • terenzreignz
Let's do what the question suggests :) \[\Large \int \frac{dx}{a^2+x^2}\]
terenzreignz
  • terenzreignz
It says let x = au right?
terenzreignz
  • terenzreignz
Following that, must be dx?

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anonymous
  • anonymous
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.
terenzreignz
  • terenzreignz
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? \[\Large \frac{d}{dx}\tan^{-1}(x) = \color{red}?\]
anonymous
  • anonymous
Yes, it's \[1/(1 + x^2) \]
terenzreignz
  • terenzreignz
Oh... okay, great :D Now, after we let x = au it follows that dx = a du right?
anonymous
  • anonymous
yes
terenzreignz
  • terenzreignz
Right, let's make that substitution then... \[\large x = \color{red}{au}\]\[\large dx = \color{blue}{a \ du}\] \[\Large \int \frac{\color{blue}{a \ du}}{a^2 + (\color{red}{au})^2}\]
terenzreignz
  • terenzreignz
Catch me so far?
anonymous
  • anonymous
yep I got that far
terenzreignz
  • terenzreignz
Okay, at the denominator, we can 'expand', and then factor out: \[\Large \int \frac{a \ du }{a^2 + a^2u^t}\]
terenzreignz
  • terenzreignz
\[\Large \int \frac{a \ du}{a^2(1+u^2)}\]
terenzreignz
  • terenzreignz
Can you take it from here?
anonymous
  • anonymous
I think so, let me look a bit.
terenzreignz
  • terenzreignz
Remember that a is just a constant...
anonymous
  • anonymous
Yeah I got it now, thank you so much!
terenzreignz
  • terenzreignz
Okay, awesome ^_^

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