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**Malaclypse** Hi all... I've been working on a problem for the past several hours and haven't had much luck solving it...

I'm trying to derive this formula (for n>2):

$\displaystyle \[\int {{{\sec }^n}x{\rm{d}}x = \frac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{{\sec }^{n - 2}}x{\rm{d}}x} } \]$

And the hint that came with the challenge pointed out this trig identity:

$\displaystyle \[1 + {\tan ^2}x = {\sec ^2}x\]$

I've played around with it a bit, but where I'm running into trouble is my u and dv values for integration by parts. I am fairly confident that I need to set my original integrand like this:

$\displaystyle \[\int {{{\sec }^{n - 2}}x{{\sec }^2}x{\rm{d}}x} \]$

That'll give me the $\displaystyle \[{{{\sec }^2}x}\]$ I know I need, but no matter how I try, I'm not getting to the end result. I've played around with the identity, but it doesn't seem to be doing me a lot of good.

Any thoughts/solutions (in a hidden spoiler would be preferred, but at this point I'll take what I can get!)?

Thanks all...