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Homework Help: Integral of sec(x)^3

  1. Feb 13, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm just trying to solve this

    [tex] \int {\sec^3{x} dx [/tex]

    2. Relevant equations
    [tex] sec^2{x} = 1 + tan^2{x} [/tex]

    3. The attempt at a solution

    well i was able to simplify it to this:

    [tex] \int {\sec{x}*\tan^2{x}} dx + \ln{|\sec{x} + \tan{x}|} [/tex]

    but I still was not able to find that new integral
    Last edited: Feb 13, 2007
  2. jcsd
  3. Feb 13, 2007 #2
    = INT[secx.sec^2x]dx
    By parts (differentiating secx and integrating sec^2x):
    = INT[secx.sec^2x]dx
    = secx.tanx - INT[secx.tan^2x]dx
    = secx.tanx - INT[secx(sec^2x - 1)]dx
    = secx.tanx - INT[sec^3x - secx]dx
    So we have:

    INT[sec^3x]dx = secx.tanx - INT[sec^3x - secx]dx
    INT[sec^3x]dx = secx.tanx - INT[sec^3x]dx + INT[secx]dx
    2 INT[sec^3x]dx = secx.tanx + INT[secx]dx
    2 INT[sec^3x]dx = secx.tanx + Ln|secx + tanx|
    INT[sec^3x]dx = 1/2 secx.tanx + 1/2 Ln|secx + tanx|
  4. Feb 14, 2007 #3


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    [tex]\int \frac{\cos x}{\left( 1-\sin ^{2}x\right) ^{2}}dx=\allowbreak \int \frac{1}{\left( 1-t^{2}\right) ^{2}}\,dt [/tex]

    for the last integral, use simple fractions. [itex] \sin x=t [/itex] has been used.
  5. Feb 14, 2007 #4
    Ah, alright thanks. It was simpler than i thought it would be
  6. Feb 14, 2007 #5


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    Another way: Since sec(x)= 1/cos(x),
    [tex]\int sec^3(x)dx= \int\frac{dx}{cos^3(x)}[/tex]
    which is an odd power of cos(x). Multiply numerator and denominator by cos(x):
    [tex]\int \frac{cos(x)dx}{cos^4(x)}= \int\frac{cos(x)dx}{(1- sin^2(x))^2}[/tex]
    Let u= sin(x) so du= cos(x)dx
    [tex]\int \frac{du}{(1-u^2)^2}= \int\frac{du}{(1-u)^2(1+u)^2}[/tex]
    and, again, use partial fractions.
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