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Why Is My Answer Wrong (trig integral)

  1. Feb 23, 2009 #1
    agh, trig integrals are driving me nuts.

    1. The problem statement, all variables and given/known data

    the integral of

    tanx sec^4x dx


    2. Relevant equations



    3. The attempt at a solution

    tanx sec^4x dx
    = sec^3 tanx secx dx

    u = secx
    du = tanx secx dx

    so,
    = u^3 du
    = (u/4)^4
    = (1/4)sec^4x
     
  2. jcsd
  3. Feb 23, 2009 #2

    D H

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    The final result looks good to me. You did make a mistake in an intermediate step, however; you should have had u^4/4, rather than (u/4)^4.

    What makes you think your answer is wrong?
     
  4. Feb 23, 2009 #3
    Yes, your final answer should be sec^4x/4, but you forgot one thing: +C. This is important.
     
  5. Feb 23, 2009 #4
    O, ok. The book substitutes u = tanx to get (1/4)tan^4x + (1/2)tan^2x + C, and that got me confused.

    Thanks
     
  6. Feb 24, 2009 #5
    Just the same, since sec^2x=1+tan^2x
     
  7. Feb 24, 2009 #6

    HallsofIvy

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    The way I would have done this would have been to convert immediately to sine and cosine (I get so confused with those other trig functions!): tan(x) sec^4(x)= sin(x)/cos^5(x)[/itex] and then the substitution u= cos(x), du = -sin(x) dx gives the integral
    [tex]-\int du/u^5= -\int u^{-5}du= (1/4)u^{-4}+ C= (1/4)cos^{-4}(x)+ C= (1/4)sec^4(x)+ C[/itex]
    again, "just the same".
     
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