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Calculus II - Trigonometric Integrals - Evaluate Integral tan(x)^5*sec(x)^4 dx

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi,
    I'm trying to solve this problem and guess I'm doing something wrong.

    Evaluate Integral tan(x)^5*sec(x)^4 dx

    2. Relevant equations

    integral tan(x) dx = ln(|sec(x)|)

    integral tan(x)^n dx = tan(x)^(n-1)/(n-1) - integral tan(x)^(n-1) dx

    tan(x)^2+1=sec(x)^2

    3. The attempt at a solution

    My Answer
    Integral tan(x)^5*sec(x)^4 dx = 1/8*tan(x)^8 + 1/6*tan(x)^6 + 1/2*tan(x)^2 + ln(|sec(x)|)+c

    I don't see what I'm doing wrong...
    You can see my work attached. Thanks in advance for any assistance you can provide! Don't forget you can click on the window that pops up after clicking attachments to open the image in a new tab to view it at a larger scale.
     

    Attached Files:

    Last edited: Aug 9, 2011
  2. jcsd
  3. Aug 9, 2011 #2
    ∫ tan5(x) sec4(x) dx
    = ∫ tan5(x) (tan2(x)+1) sec2(x) dx
    = ∫ [tan7(x) + tan5(x)] sec2(x) dx
     
  4. Aug 10, 2011 #3
    ∫ tan5(x) sec4(x) dx
    = ∫ tan5(x) (tan2(x)+1) sec2(x) dx
    = ∫ [tan7(x) + tan5(x)] sec2(x) dx
    = ∫ [tan7(x) + tan5(x)] (tan2(x)+1) dx
    = ∫ [tan9(x) + tan7(x)+tan7(x)+tan5(x)] dx
    = ∫ [tan9(x) + 2tan7(x)+tan5(x)] dx
    is this not correct?
     
  5. Aug 10, 2011 #4
    Why would you do that when d/dx tan(x) = sec2(x)?
     
  6. Aug 10, 2011 #5
    I didn't think of making the substitution at the time <_< it would of made the problem much easier @_@ I should still be able to obtain the correct answer though, yes? I find it odd that I don't get the right answer

    as you can see in my work
    ∫ tan5(x) sec4(x) dx
    = ∫ tan5(x) (tan2(x)+1) sec2(x) dx
    = ∫ [tan7(x) + tan5(x)] sec2(x) dx
    = ∫ [tan7(x) + tan5(x)] (tan2(x)+1) dx
    = ∫ [tan9(x) + tan7(x)+tan7(x)+tan5(x)] dx
    = ∫ [tan9(x) + 2tan7(x)+tan5(x)] dx

    one i get here i evaluated each integral separately using
    integral tan(x)^n dx = tan(x)^(n-1)/(n-1) - integral tan(x)^(n-1) dx
    I should be able to get the correct answer and i don't think my answer
    1/8*tan(x)^8 + 1/6*tan(x)^6 + 1/2*tan(x)^2 + ln(|sec(x)|)+c
    is correct for whatever strange reason when i can't find a error in my work at all
     
  7. Aug 10, 2011 #6
    It's not correct. The bolded terms are extraneous. Here's your work:

    ∫ tan5(x) sec4(x) dx
    = ∫ tan5(x) (tan2(x)+1)2 dx
    = ∫ tan5(x) (tan2(x)+1)2 dx
    = ∫ tan5(x) (tan4(x)+2 tan2(x) +1) dx
    = ∫ [tan9(x) + 2 tan7(x)+tan5(x)] dx

    Let's evaluate each individually, using your formula. I'm not even sure whether or not your formula is true, but let's assume that it is.

    ∫ tann(x) dx = tann-1(x)/(n-1) - ∫ tann-1(x) dx
    ∫ [tan9(x) dx = tan8(x)/8 - ∫ tan8(x) dx
    ∫ [tan8(x) dx = tan7(x)/7 - ∫ tan7(x) dx
    ∫ [tan7(x) dx = tan6(x)/6 - ∫ tan6(x) dx
    ∫ [tan6(x) dx = tan5(x)/5 - ∫ tan5(x) dx

    The problem here is that you used your own formula incorrectly. Using the formula correctly:

    ∫ [tan9(x) + 2 tan7(x)+tan5(x)] dx
    = tan8(x)/8 - ∫ tan8(x) dx + ∫ [2 tan7(x)+tan5(x)] dx
    = tan8(x)/8 - tan7(x)/7 + ∫ tan7(x) dx + ∫ [2 tan7(x)+tan5(x)] dx
    = tan8(x)/8 - tan7(x)/7 + ∫ [3 tan7(x)+tan5(x)] dx
    = tan8(x)/8 - tan7(x)/7 + 3 [tan6(x)/6 - ∫ tan6(x) dx] + ∫ tan5(x) dx
    = tan8(x)/8 - tan7(x)/7 + 3 [tan6(x)/6 - tan5(x)/5 + ∫ tan5(x) dx] + ∫ tan5(x) dx
    = tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 ∫ tan5(x) dx
    = tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 tan4(x)/4 - 4 ∫ tan4(x) dx
    = tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 tan4(x)/4 - 4 tan3(x)/3 + 4 ∫ tan3(x) dx
    = tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 tan4(x)/4 - 4 tan3(x)/3 + 4 tan2(x)/2 - 4 ∫ tan2(x) dx
    = tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 tan4(x)/4 - 4 tan3(x)/3 + 4 tan2(x)/2 - 4 tan(x) - 4 ∫ tan(x) dx
    = tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 tan4(x)/4 - 4 tan3(x)/3 + 4 tan2(x)/2 - 4 tan(x) + 4 log|cos(x)| +C

    It seems like something is seriously wrong with your formula.
     
    Last edited: Aug 10, 2011
  8. Aug 10, 2011 #7
    I thought the formula was correct, see attachment
    ah yes the correct formula is
    ∫ tann(x) dx = tann-1(x)/(n-1) - ∫ tann-2(x) dx
    which although i wrote the wrong formula in my posts i still used this one in my work and the correct

    I don't see how you get rid of terms... isn't this correct

    ∫ tan9(x) dx = tan8(x)/8 - ∫ tan7(x) dx
    = tan8(x)/8 - tan6(x)/6 +∫ tan5(x) dx
    = tan8(x)/8 - tan6(x)/6 + tan4(x)/4 - ∫ tan3(x) dx
    = tan8(x)/8 - tan6(x)/6 + tan4(x)/4 - tan2(x)/2 + ∫ tan(x) dx
    = tan8(x)/8 - tan6(x)/6 + tan4(x)/4 - tan2(x)/2 + ln(|sec(x)|)

    2∫ tan7(x) = 2*tan6(x)/6 - 2 ∫ tan5(x) dx
    = tan6(x)/3 - 2*tan4(x)/4 + 2 ∫ tan3(x) dx
    = tan6(x)/3 - tan4(x)/2 + 2*tan2(x)/2 - 2 ∫ tan(x) dx
    = tan6(x)/3 - tan4(x)/2 + tan2(x) - 2 ln(|sec(x)|)

    ∫ tan5(x) dx = tan4(x)/4 - ∫ tan3(x) dx
    = tan4(x)/4 - tan2(x)/2 + ∫ tan(x) dx
    = tan4(x)/4 - tan2(x)/2 + ln(|sec(x)|)

    ∫ [tan9(x) + 2 tan7(x)+tan5(x)] dx

    = tan8(x)/8 - tan6(x)/6 + tan4(x)/4 - tan2(x)/2 + ln(|sec(x)|) + tan6(x)/3 - tan4(x)/2 + tan2(x) - 2 ln(|sec(x)|) + tan4(x)/4 - tan2(x)/2 + ln(|sec(x)|)

    = tan8(x)/8 - tan6(x)/6 + tan6(x)/3 + tan4(x)/4 - tan4(x)/2 + tan4(x)/4 - tan2(x)/2 + tan2(x) - tan2(x)/2 + ln(|sec(x)|) - 2 ln(|sec(x)|) + ln(|sec(x)|)


    = tan8(x)/8 + tan6(x)/6

    hmm interesting
     

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