Trig Substitution: int [1 / sqrt(x^2 - 1)] dx

[flakine]

∫ 1/√(x^2-1)dx

I don't really understand these types of problems.

√(x^2-1)= √(sec^2x - 1) = √(tan^2x) = tan x, x=sec a, dx= sec a tan a da

∫1/tan x dx
∫(1/tan a) sec a tan a da
∫sec a da
∫1/cos a da u=cos a, du=-sin a da
????
I'm lost, can someone point me in the right direction?

 

[galactus]

You're on the right track.

Let \(\displaystyle \L\\x=sec{\theta}\) and \(\displaystyle \L\\dx=sec{\theta}tan{\theta}d{\theta}\)

\(\displaystyle \L\\\int\frac{dx}{\sqrt{x^{2}-1}\)

\(\displaystyle \L\\\int\frac{sec{\theta}tan{\theta}}{\sqrt{sec^{2}{\theta}-1}}d{\theta}\)

\(\displaystyle \L\\\int\frac{sec{\theta}tan{\theta}}{tan{\theta}}d{\theta}\)

\(\displaystyle \L\\\int\sec{\theta}d{\theta}\)

Now, you want to derive the integral of sec?.

There's a trick you can use if you must.

\(\displaystyle \L\\\int\sec{\theta}d{\theta}=\int\sec{\theta}\left(\frac{sec{\theta}+tan{\theta}}{sec{\theta}+tan{\theta}}\right)d{\theta}\)

=\(\displaystyle \L\\\int\frac{sec^{2}{\theta}+sec{\theta}tan{\theta}}{sec{\theta}+tan{\theta}}d{\theta}\)

Now, let \(\displaystyle \L\\u=sec{\theta}+tan{\theta}\) and \(\displaystyle \L\\du=(sec^{2}{\theta}+sec{\theta}tan{\theta})d{\theta}\)

And you get:

\(\displaystyle \L\\ln|sec{\theta}+tan{\theta}|\)

Now, put it n terms of x if you have to.

 

[flakine]

Thanks, that was great!

How do I get my fonts and math symbols to display like yours?

 

[galactus]

That's LaTex. Click on the quote icon at the upper right of my post to see the code. There is a tutorial on this site as well as numerous sites on the web.

It may take a little practice, but it's well worth it if you're going to post to any extent.