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Homework Help: Calculus Problem Help Integral of (secx)^3 dx

  1. Oct 4, 2004 #1
    Hi all, I am stuck on this trig integral problem. The answer is provided in the book, but I do not know how to get it. The problem is this:


    Integral of (secX)^3 dx


    it says first use integration by parts:

    u = sec x du = sec x tan x dx

    dv = (secx)^2 dx v = tan x

    uv - integral (v du) = sex x tan x - integral sec x((secx)^2-1) dx


    Then it says secx tanx - integral (sec x)^3 dx + integral sec x dx


    Now there is another (sec x)^3 like in the original problem, after this step they just provide the answer, but how ?




    The answer is 1/2(secxtanx + ln |secx+tanx|) + C
     
  2. jcsd
  3. Oct 4, 2004 #2

    arildno

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    Let
    [tex]I=\int\frac{1}{\cos^{3}x}dx[/tex]
    Hence, you have shown:
    [tex]I=sec(x)tan(x)-I+\int\frac{1}{\cos{x}}dx[/tex]
    Can you take it from there?
     
  4. Oct 4, 2004 #3
    hmm, let me try, thats diff. method from the book but it looks easier, brb. thx.
     
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