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Help trig integral question

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral

    integral of (tanx)^5*(secx)^4

    2. Relevant equations

    3. The attempt at a solution

    what I did was broke the sec^4x into 2 sec^2's multiplied by each other
    then one of these terms became tan^2x +1
    so I had tan^5x*(tan^2x + 1)*sec^2x

    then u = tanx
    du = sec^2x

    eventual answer tan^8x/8 + tan^6x/6 + c

    however the book shows this problem by taking a (secxtanx) out of eveyrthing then doing tan^4x = (sec^2x -1)^2*sec^3x
    u= secx
    du = tanxsecx

    eventual answer sec^8x/8 - sec^6x/3 + sec^4x/4 + C

    ...is there something wrong with the way I did it; I don't see anything wrong with the way I'm doing it but the book insists on doing all problems like this the otherway. I wouldn't mind but I don't like memorizing things and the way I always seem to think of doing these problems is the first method I wrote out please respond I have an exam on this stuff tomarrow night
  2. jcsd
  3. Feb 26, 2008 #2

    [tex]\int\sec^{3}x\tan^{4}x(\sec x\tan x)dx[/tex]

    [tex]\int\sec^{3}x(\sec^{2}x-1)^2(\sec x\tan x)dx[/tex]
  4. Feb 26, 2008 #3
    yes that is how the book does it but why is the way that I did it wrong I don't get it?
  5. Feb 26, 2008 #4
    Your way ...



    I guess just have confidence in yourself. You did it all the steps correctly, but through a different route.
    Last edited: Feb 26, 2008
  6. Feb 26, 2008 #5
    ok thanks alot; and wow thats a good way to proove it I didn't think of that
  7. Feb 26, 2008 #6
    I take that back. I tried it and it didn't work, lol. But your solution is still right, but the limits of Integration is a different story.
  8. Feb 27, 2008 #7
    ah yea that does make sense that you would need different limits
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