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Trigonometric integration question

  1. Aug 8, 2012 #1
    The question asks to find ∫secxtan2x

    I rewrote tan2x as (sec2x-1). Then I expanded the equation having sec3x-secx and I know the integral of secx which is 0.5ln|tanx+secx|,

    but my question is, is integrating sec3x by parts the correct path? or not?

    Thanks
     
  2. jcsd
  3. Aug 9, 2012 #2

    AGNuke

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    Gold Member

    [tex]\int secx{dx} = ln(tanx+secx)[/tex]

    And as for ∫sec3x dx; You can try parts, but you thought how you can break it?
    OR you can look for some formula of finding integrals of powers of trigonometric functions (Reduction formulas)
     
  4. Aug 9, 2012 #3

    ehild

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    sec(x)tan2(x)=sec3(x)sin2(x)=[sec3(x)sin(x)]sin(x), which you can integrate by parts.

    ehild
     
  5. Aug 9, 2012 #4
    The integral of secant cubed can be evaluated as follows (it is a common integral) with using integration by parts, applying [itex]u=\sec(x)[/itex] and [itex]dv=\sec^2(x)dx[/itex]:
    [tex]\begin{align}
    \int \sec^3(x)dx=\sec(x)\tan(x)-\int \sec(x)\tan^2(x)dx \\
    = \sec(x)\tan(x)-\int \sec^3(x)dx + \int \sec(x)dx \\
    = \sec(x)\tan(x)-\int \sec^3(x)dx + \log(\sec(x)+\tan(x))
    \end{align}[/tex]
    Now solve that equation for the integral.
     
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