# Derive a power reduction formula

• Feb 6th 2011, 03:28 PM
Malaclypse
Derive a power reduction formula
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):

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

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

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

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 integral like this:

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

That'll give me the $${{{\sec }^2}x}$$ 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...
• Feb 6th 2011, 03:50 PM
TheEmptySet
Quote:

Originally Posted by 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):

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

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

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

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:

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

That'll give me the $${{{\sec }^2}x}$$ 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...

Let $\displaystyle I_n=\int \sec^n(x)dx$

as per the hint rewrite as

$\displaystyle I_n=\int \sec^n(x)dx=\int \sec^{n-2}(x)\sec^2(x)dx=$

Now use integration by parts with
$u=\sec^{n-2}(x) \implies du=(n-2)\sec^{n-2}\tan(x)dx$ and
$dv=\sec^2(x)dx \implies v=\tan(x)$

This gives

$I_n=\sec^{n-2}(x)\tan(x)-(n-2)\int \sec^{n-2}(x)\tan^2(x)dx = \sec^{n-2}(x)\tan(x)-(n-2)\int (\sec^n(x)-\sec^{n-2}(x))dx$

This gives

$I_n= \sec^{n-2}(x)\tan(x)-(n-2)I_n +(n-2)\int \sec^{n-2}(x))dx$

Solving for $I_n$ gives

$(n-1)I_n= \sec^{n-2}(x)\tan(x)+(n-2)\int \sec^{n-2}(x))dx$

$\displaystyle I_n= \frac{\sec^{n-2}(x)\tan(x)}{n-1}+\frac{n-2}{n-1}\int \sec^{n-2}(x))dx$
• Feb 6th 2011, 04:02 PM
Malaclypse
Of course! As soon as I saw that derivative I saw exactly where I went wrong. Thank you so much!