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Math Help - integral(tan^(2m)(x)*sec^(2k+1)(x),x)

  1. #1
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    integral(tan^(2m)(x)*sec^(2k+1)(x),x)

    Hello,

    How would I find the integral of an even powered tangent function multiplied by an odd powered secant function?

    integral(tan^(2m)(x)*sec^(2k+1)(x),x)

    For instance, how would we find
    integral(tan^2(x)*sec(x),x)?

    I was informed the method would be integration by parts.

    Using int(udv) = uv - int(vdu)

    if i choose dv to be sec(x), v is not fun to have for the next integral.
    if i choose dv to be tan(x)sec(x), i have to integrate sec^3(x).
    if i choose dv to be tan(x), the next integral is not fun.

    So it seems that i could solve this if would learn how to integrate sec^n(x).

    Are there any better methods then what I have outlined above?

    Thanks,
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  2. #2
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    Quote Originally Posted by n0083 View Post
    For instance, how would we find
    integral(tan^2(x)*sec(x),x)?
    If u=\sec(x) then du=\sec(x)\tan(x)dx and this becomes simple.
    Last edited by Jameson; October 3rd 2008 at 12:27 AM.
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  3. #3
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    i don't see how u=sec(x) helps.

    we only have one secant, the du requires one too.

    Please provide the next 'simple' step and educate me please.
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  4. #4
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    I made a mistake. Misread the problem. Sorry.

    Quote Originally Posted by n0083 View Post
    For instance, how would we find
    integral(tan^2(x)*sec(x),x)?
    This is really hard to see without knowing the "trick".

    \sec(x) \times \frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}=\frac{\sec  ^2(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}

    This is now \frac{1}{u}du

    I think you can put it together now.
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  5. #5
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    I'm not sure what did you choose for u and v' but if it was like this:
    u = tan^2(x)
    v' = sec(x) dx

    You'll get a more complicated integral. Here's one way to solve \int \tan^2{x} \sec{x} dx:

    \int (\sec^2{x} - 1)\sec{x} dx

    \int \sec^2{x}\sec{x}~dx - \int \sec{x}~dx

    The second integral can be solved the way Jameson told you to, so let's put all our focus to the first one, which is an obvious integration by parts.

    \int \sec^2{x} \sec{x}~dx = \int (\tan{x})'\sec{x}~dx = \sec{x}\tan{x} - \int \tan^2{x}\sec{x}~dx<br />

    = \sec{x}\tan{x} - \int (\sec^3{x} - \sec{x})~dx

    Now you should notice that you can treat sec^3(x) as the unknown and then solve for it algebraically. Problem solved.

    You can also use a reduction formula for sec^3(x)
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