dx=sec^2(p)*dp

1+tan^2(p)=sec^2(p)

dx/(1+tan^2(p))^2=1/(sec^2(p))=cos^2(p)

cos(2p)=2cos^2(p)-1]]>

You would let x = 1/2 u and so dx = 1/2du. So the integral becomes:

We do the same thing for your equation, expect you use x = tan u. To get I, you have:

Because:

If you are still unsure, can you name specifically what part you don't get?

]]>Although it's smack bang in front of me, I still can't understand it, so could anyone work it through for me? Cheers, Tredici.

]]>It's probably very simple, as I've said, and I'd like to add, I'd really appreciate it if it were worked through thoroughly, as I'm not looking for answers; I'm looking to understand how you've answered it, which can then be applied to similar questions. Thanks chaps, you've always been of great help to me.

]]>