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Integration by parts headache

  1. Sep 30, 2005 #1
    Integration by parts....

    I just started Calc. II and though I struggle a bit, it's fascinating. I have been fooling with a problem lately...one of those standard problems that professors like to assign, and it usually appears in calculus texts:

    Have ya'll ever done integration by parts with secx? A friend of mine worked it out for me, but I have had trouble reaching the solution on my own. Just thought I'd throw it out there in case ya'll hadn't run into it! It is a fun kind of headache.

    -Gin
     
  2. jcsd
  3. Sep 30, 2005 #2

    Tom Mattson

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    Hi Ginny,

    I'm scooting this over to our Homework section.

    Now, on to your question: Are you talking about integrands that contain powers of [itex]\sec(x)[/itex], or just the [itex]\sec(x)[/itex] itself? The reason I'm asking is that it is normal to integrate odd powers greater than 1 of the secant function by parts, but not so normal to integrate the secant function itself by parts.
     
  4. Sep 30, 2005 #3
    That integral doesn't really require integration by parts... You have to mulitply by a form of one, which can be hard to see.

    [tex]\int \sec{x}dx= \int \sec(x)*\frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}dx[/tex]

    You can do a u-substitution from there.
     
  5. Sep 30, 2005 #4

    TD

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    Or use the substitution [itex]t = \tan \left( {x/2} \right)[/itex]
     
  6. Sep 30, 2005 #5
    Or [tex]\sec x = \frac{1}{{\cos x}}\left( {\frac{{\cos x}}{{\cos x}}} \right) = \frac{{\cos x}}{{\cos ^2 x}} = \frac{{\cos x}}{{1 - \sin ^2 x}}[/tex] and use substitution + partial fractions. It's one of the longer ways of doing it but it is an alternative. :biggrin:
     
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