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Math Help - Derive a power reduction formula

  1. #1
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    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...
    Last edited by Malaclypse; February 6th 2011 at 03:40 PM.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Malaclypse View Post
    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
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  3. #3
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    Of course! As soon as I saw that derivative I saw exactly where I went wrong. Thank you so much!
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