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Homework Help: Trig Integral Problem

  1. Sep 3, 2007 #1
    1. The problem statement, all variables and given/known data

    If [tex]\int_{0}^{pi/4}\tan^{6}(x)sec(x)dx = I[/tex] then express [tex]\int_{0}^{pi/4}\tan^{8}(x)sec(x)dx[/tex] in terms of I.

    2. Relevant equations



    3. The attempt at a solution
    I figured you would have to get the second equation to look like the first one, so I pulled out a tan^2(x) to make it
    [tex]\int_{0}^{pi/4}\((sec^{2}(x)-1)tan^{6}(x)sec(x)dx[/tex] then
    [tex]\int_{0}^{pi/4}\sec^{2}{x}tan^{6}(x)sec(x)dx - I[/tex]

    And here's where I get stuck. My thought is to use integration by parts, setting sec^2 = u, and tan^6xsecxdx = dv. However this just leads me to a dead end when I try to figure out what v could be. Using the opposite substitution (u = tan^6... etc) I end up with

    [tex]tan^{7}(x)sec(x) - 6\int_{0}^{pi/4}\tan^{6}(x)sec^{3}(x)dx = 2\int_{0}^{pi/4}\tan^{8}(x)sec(x)dx[/tex]

    which gives me the same problem I had to begin with (an even power to tangent and an odd power of secant).

    Is there some substitution I could make somewhere?
     
  2. jcsd
  3. Sep 4, 2007 #2
    Have you tried integrating by parts using dv = sec x tan x?
     
  4. Sep 4, 2007 #3
    Yeah... that gets me:

    [tex]\int_{0}^{pi/4}\sec^{2}{x}tan^{6}(x)sec(x)dx - I[/tex]

    u = sec^2xtan^5x dv = secxtanx
    du = 2sec^2xtan^6x + 5tan^4xsec^4x

    [tex]sec^{3}(x)tan^{5}(x) - \int_{0}^{pi/4}\sec(x)[2sec^{2}(x)tan^{6}(x) + 5tan^{4}(x)sec^{4}(x)]dx[/tex]

    simplified...

    [tex]sec^{3}(x)tan^{5}(x) - 2\int_{0}^{pi/4}\sec^{3}(x)tan^{6}(x) - 5\int_{0}^{pi/4}\tan^{4}(x)sec^{5}(x)dx[/tex]

    Which still leaves me with odd powers of secant and even powers of tangent...

    EDIT: Woops, dropped the "- I" after all of those, but it's pretty irrelevant.
     
    Last edited: Sep 4, 2007
  5. Sep 4, 2007 #4
    Hmm.. I just tried a different approach, would like to know if this is a correct approach...

    [tex]\int_{0}^{pi/4}\sec^{2}{x}tan^{6}(x)sec(x)dx - I[/tex]

    using u = secx, dv = tan^6xsec^2x
    du = secxtanx, v = (1/7)tan^7x

    [tex]\frac{sec(x)tan^{7}(x)}{7} - \frac{1}{7}\int_{0}^{pi/4}\sec(x)tan^{8}(x)dx - I = \int_{0}^{pi/4}tan^{8}(x)sec(x)dx[/tex]

    So...

    [tex]\frac{\sqrt{2}}{7} - I = \frac{8}{7}\int_{0}^{pi/4}tan^{8}(x)sec(x)dx[/tex]

    [tex]= \frac{\sqrt{2}-7I}{8}[/tex]


    this look right?
     
    Last edited: Sep 4, 2007
  6. Sep 4, 2007 #5
    Looks good to me. (And for the record, you can get the same result by using using dv = sec x tan x in the *original* equation. You'll get an integral with sec^3 x, which you can rewrite as sec x(1 + tan^2 x).)
     
  7. Sep 4, 2007 #6
    Ah... yeah I see. I did that at one point but didn't think to write sec^2 as (1+tan^2). Anyway, thanks a ton.
     
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