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Thread: 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 $\displaystyle u=\sec(x)$ then $\displaystyle du=\sec(x)\tan(x)dx$ and this becomes simple.
    Last edited by Jameson; Oct 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".

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

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

    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 $\displaystyle \int \tan^2{x} \sec{x} dx$:

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

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

    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.

    $\displaystyle \int \sec^2{x} \sec{x}~dx = \int (\tan{x})'\sec{x}~dx = \sec{x}\tan{x} - \int \tan^2{x}\sec{x}~dx
    $

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

    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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