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Math Help - [SOLVED] Trig Integral

  1. #1
    Senior Member mollymcf2009's Avatar
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    [SOLVED] Trig Integral

    \int 3x sec (x) tan(x) dx

    Not sure what I need to use for my substitution.
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  2. #2
    Member Abu-Khalil's Avatar
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    dv=\sec x\tan x dx \Rightarrow v=\sec x, u=3x\Rightarrow du=3dx and to calculate \int \sec x dx rewrite as \int \sec x \frac{\sec x + \tan x}{\sec x + \tan x}dx
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  3. #3
    Senior Member mollymcf2009's Avatar
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    So I need to integration by parts?
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    Quote Originally Posted by mollymcf2009 View Post
    \int 3x sec (x) tan(x) dx

    Not sure what I need to use for my substitution.
    \int 3x sec (x) tan(x) dx

    u =secx == > arcsec(u) = x<br />

     du = secxtanx



    \int 3x sec (x) tan(x) dx = 3\int  arcsec (u)du

    Integration by parts

    3\int  arcsec (u)du = \frac{3u}{arcsec (u)} - 3 \int \frac{1}{sqrt(u^2-1)}du

    Use trig substitution and i think you should get an answer.
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  5. #5
    Member Abu-Khalil's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    So I need to integration by parts?
    Yes.
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  6. #6
    Senior Member mollymcf2009's Avatar
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    Here is what I am getting before I do any u substitutions or integration by parts:

    \int 3x \frac{sec^2xtanx +secxtan^2x}{secx + tanx} dx

    Is that what I need to do and then do my integraton by parts?


    Edit 1: Wait,
    Ok, so from my original problem, I need to so my integration by parts first, THEN when secx is in my integral from my integ. by parts, that is when I need to multiply top and bottom by secx + tanx
    Right?

    So my answer is:

    3xsecx - 3 \ln|secx+tanx| + C

    Edit 2: GOT IT! Thanks y'all!
    Last edited by mr fantastic; February 8th 2009 at 07:52 PM.
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  7. #7
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    I did garbage before sorry about that i don't know what i was thinking.


    \int3xsecxtanxdx = 3xsecx - 3\intsecx

    secx = 1/cosx = cosx/ cos^2x = cosx/ (1 -sin^2x)

    u = sinx
    du =cosx

    \int3xsecxtanxdx = 3xsecx - 3\int1/(1-u^2)


    using partial fractions

    1/(1-u^2) = \frac{1/2}{1-u} + \frac{1/2}{1+u}


    \int3xsecxtanxdx = 3xsecx - 3/2[   ln | 1+ sinx| +ln|1-sinx|   ] +C

    Both answers are right although they appear different.
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