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Difficult integral

  1. Mar 7, 2008 #1
    [SOLVED] Difficult integral

    1. The problem statement, all variables and given/known data
    [tex]\int x^3 tan^6(x^4)sec(x^4) dx[/tex]


    3. The attempt at a solution
    I notice how x^3 is the derivative of x^4 (Considering that we can pull a 4/4 out, and move 4/1 outside, leaving 1/4 inside to make it true).

    Beyond that, I'm stuck as hell.

    I notice that the power of tan is even, while the power of sec is odd, therefore we need to use the rules (First repeatedly to reduce the exponent to 1, then the second to deal with a 1):
    [tex]\int sec^{2k+1}x dx = \frac{1}{2k}tanx sec^{2k-1}x - \frac{2k-1}{2k}\int sec^{2k-1}xdx[/tex]

    and

    [tex]\int secx dx = ln|tanx + secx| + c[/tex]

    Also, due to the chapter I've found this in, we're likely to have to make use of the Pythagorean Identity, [itex]1 + tan^2x = sec^2x[/itex].

    Edit: The correct answer should be...
    [tex]\frac{1}{24}tan(x^4)sec^5(x^4) - \frac{13}{96}tan(x^4)sec^3(x^4) + \frac{11}{64}tan(x^4)sec(x^4) - \frac{5}{64}ln|sec(x^4) + tan(x^4)| + C[/tex]

    Looking at the final answer, it looks like they've used the two rules above. I believe convert the tan^6 into (sec^2 - 1), leaving us will all x^3 and sec, then use those rules until finished.

    Assuming I'm right on that, the x^3 and x^4 being there still throw me off.
     
    Last edited: Mar 7, 2008
  2. jcsd
  3. Mar 7, 2008 #2
    [tex]\int x^3 tan^6(x^4)sec(x^4) dx[/tex]
    [tex]= 4\int \frac{1}{4}x^3(tan^2(x^4))^3sec(x^4) dx[/tex]
    [tex]= 4\int \frac{x^3}{4}(sec^2(x^4) - 1)^3 sec(x^4) dx[/tex]
    [tex]= 4\int \frac{x^3}{4}(sec^2(x^4) - 1)^2 (sec^2(x^4) - 1) sec(x^4) dx[/tex]
    [tex]= 4\int \frac{x^3}{4}(sec^2(x^4) - 1)^2 (sec^3(x^4) - sec(x^4)) dx[/tex]
    [tex]= 4\int \frac{x^3}{4}(sec^2(x^4) - 1)(sec^2(x^4) - 1) (sec^3(x^4) - sec(x^4)) dx[/tex]
    [tex]= 4\int \frac{x^3}{4}(sec^2(x^4) - 1)(sec^5(x^4) + sec(x^4) - sec^3(x^4) - sec^3(x^4)) dx[/tex]
    [tex]= 4\int \frac{x^3}{4}(sec^2(x^4) - 1)(sec^5(x^4) + sec(x^4) - 2sec^3(x^4)) dx[/tex]
    [tex]= 4\int \frac{x^3}{4}(sec^7(x^4) + sec^3(x^4) - 2sec^5(x^4) - sec^5(x^4) - sec(x^4) + 2sec^3(x^4)) dx[/tex]
    [tex]= 4\int \frac{x^3}{4}(sec^7(x^4) - 3sec^5(x^4) + 3sec^3(x^4) - sec(x^4)) dx[/tex]

    How do I deal with this x^3 / 4? I can turn it into D(x^4) if I like, but I'm unclear how that'd help me.
     
  4. Mar 7, 2008 #3

    Dick

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    Homework Helper

    First do the substitution u=x^4. The x^3 will then be absorbed into the du. (or d(x^4) as you said).
     
  5. Mar 7, 2008 #4
    CORRECTION: Should've pulled 1/4 out, before, not 4, such that we have 4x^3, not (1/4)x^3. Thanks to Dick for pointing this out. Correcting it from here on in.

    [tex]= \frac{1}{4}\int 4x^3}(sec^7(x^4) - 3sec^5(x^4) + 3sec^3(x^4) - sec(x^4)) dx[/tex]
    [tex]= \frac{1}{4}\int (sec^7u - 3sec^5u + 3sec^3u - secu) du[/tex]
    [tex]= \frac{1}{4}\int sec^7u du - \frac{3]{4}\int sec^5u du + \frac{3]{4}\int sec^3u du - \frac{1]{4}\int secu du[/tex]
    [tex]= \frac{1}{4}A - \frac{3]{4}B + \frac{3]{4}C - \frac{1]{4}D[/tex]


    [tex]D = \int secu du[/tex]
    [tex]D = ln|secu + tanu|[/tex]

    [tex]C = \int sec^3u du[/tex]
    [tex]C = \frac{1}{2}tanu*secu - \frac{1}{2}\int secu du[/tex]
    [tex]C = \frac{1}{2}tanu*secu - \frac{1}{2}D[/tex]

    [tex]B = \int sec^5u du[/tex]
    [tex]B = \frac{1}{4}tanu*sec^3u - \frac{3}{4}\int sec^3u du[/tex]
    [tex]B = \frac{1}{4}tanu*sec^3u - \frac{3}{4}C[/tex]

    [tex]A = \int sec^7u du[/tex]
    [tex]A = \frac{1}{6}tanu*sec^5u - \frac{5}{6}\int sec^5u du[/tex]
    [tex]A = \frac{1}{6}tanu*sec^5u - \frac{5}{6}B[/tex]


    [tex]= \frac{1}{4}A - \frac{3]{4}B + \frac{3]{4}C - \frac{1]{4}D[/tex]
    [tex]= \frac{1}{4}(\frac{1}{6}tanu*sec^5u - \frac{5}{6}B) - \frac{3]{4}B + \frac{3]{4}C - \frac{1]{4}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{5}{24}B - \frac{3}{4}B + \frac{3]{4}C - \frac{1]{4}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{5}{24}B - \frac{18}{24}B + \frac{3]{4}C - \frac{1]{4}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{23}{24}B + \frac{3}{4}C - \frac{1]{4}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{23}{24}(\frac{1}{4}tanu*sec^3u - \frac{3}{4}C) + \frac{3]{4}C - \frac{1]{4}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{23}{96}tanu*sec^3u - \frac{69}{96}C + \frac{3]{4}C - \frac{1]{4}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{23}{96}tanu*sec^3u - \frac{69}{96}C + \frac{72]{96}C - \frac{1]{4}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{23}{96}tanu*sec^3u + \frac{3}{96}C - \frac{1]{4}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{23}{96}tanu*sec^3u + \frac{3}{96}(\frac{1}{2}tanu*secu - \frac{1}{2}D) - \frac{1]{4}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{23}{96}tanu*sec^3u + \frac{3}{192}tanu*secu - \frac{3}{192}D - \frac{1]{4}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{23}{96}tanu*sec^3u + \frac{3}{192}tanu*secu - \frac{3}{192}D - \frac{48]{192}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{23}{96}tanu*sec^3u + \frac{3}{192}tanu*secu - \frac{51}{192}D[/tex]
    [tex]= \frac{1}{24}tanu*sec^5u - \frac{23}{96}tanu*sec^3u + \frac{3}{192}tanu*secu - \frac{51}{192}ln|secu + tanu|[/tex]


    My answer:
    [tex]= \frac{1}{24}tan(x^4)sec^5(x^4) - \frac{23}{96}tan(x^4)sec^3(x^4) + \frac{1}{64}tan(x^4)sec(x^4) - \frac{17}{64}ln|sec(x^4) + tan(x^4)| + C[/tex]





    Actual answer:
    [tex]\frac{1}{24}tan(x^4)sec^5(x^4) - \frac{13}{96}tan(x^4)sec^3(x^4) + \frac{11}{64}tan(x^4)sec(x^4) - \frac{5}{64}ln|sec(x^4) + tan(x^4)| + C[/tex]

    It looks similar, but the fractions don't match. Where is my mistake?
    I believe it's in my calculation of A, using the sec^2k+1 rule from the original post.
     
    Last edited: Mar 7, 2008
  6. Mar 7, 2008 #5
    argh PF makes me sooo angry! had it all typed up, hehe ... been crashing lately.

    [tex]\int x^3 tan^6(x^4)sec(x^4) dx[/tex]

    [tex]t = x^4[/tex]
    [tex]\frac 1 4 dt = x^3dx[/tex]

    [tex]I =\frac 1 4\int\tan^{6}t\sec tdt=\frac 1 4\int\tan^{5}t\sec t\tan tdt[/tex]

    [tex]u = \tan^{5}t[/tex]
    [tex]du = 5\tan^{4}t\sec^{2}tdt[/tex]

    [tex]dV = \sec t\tan tdt[/tex]
    [tex]V = \sec t[/tex]

    [tex]I = \frac 1 4\sec t\tan^{5}t - \frac 5 4\int\tan^{4}t\sec^{3}tdt[/tex]

    From here you can either use a trig identity or do parts again ... not sure which is the best route ...

    1.
    [tex]I = \frac 1 4\sec t\tan^{5}t - \frac 5 4\int\tan^{4}t\sec tdt-\frac 5 4\int\tan^{6}t\sec t[/tex]

    or

    2.
    [tex]u=\sec t[/tex]
    [tex]du=\sec t\tan tdt[/tex]

    [tex]dV=\tan^{4}t\sec^{2}tdt[/tex]
    [tex]V=\frac 1 5 \tan^{5}t[/tex]

    [tex]I = \frac 1 4\sec t\tan^{5}t - \frac 14(\tan^{5}t\sec t-\int\tan^{6}t\sec tdt)[/tex]
     
    Last edited: Mar 7, 2008
  7. Mar 7, 2008 #6

    Dick

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    Homework Helper

    Don't know yet. But x^3*dx=du/4, not du*4. Beyond that you must just be adding some fractions wrong.
     
  8. Mar 7, 2008 #7
    Ok, done typing. I'm not even doing this on paper, sorry.
     
  9. Mar 7, 2008 #8
    Don't be sorry. I appreciate the effort you and others have put into this.
     
  10. Mar 7, 2008 #9
    Ok another update ... uh I wish had a tablet pc :(
     
  11. Mar 7, 2008 #10
    Ok let me do this on the board ... brb
     
  12. Mar 7, 2008 #11
    Ok I did it on the board and that's what I got ... weird.
     
  13. Mar 7, 2008 #12
    While some of my TeX images aren't updating (Maybe just on my computer), the 1/23 came out correct. Due to the way the formula is structured, if the 23/96 is incorrect, I must've made my mistake somewhere after calculating the 1/23.

    The simple math of it, is -5/96 - 18/96 = -23/96
    I believe whoever computed the answers for these must've been mistaken, seeing 18 and 5, some negative signs, then thinking "18-5=13..... / 96" or something.

    I think we can call the book wrong on this one :)

    I spoke to my professor, showing him what sort of practice I've been doing in preparation for our upcoming exam, and asked if things like the crazy-difficult sample problems could be on the exam. He said that these crazy-difficult questions were "fair play", so I choose one (This one) out of the book.

    Besides needing help on substitution (Plus mixing up 1/4 for 4/1), and some simple fractional algebra, I'm quite impressed to see how well this question went.

    Despite not matching the book's answer, I believe this to be the correct answer, and thank the both of you for the effort you've contributed to this :)

    This sucker's /solved. ;)
     
  14. Mar 7, 2008 #13

    Ben Niehoff

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    A little note on LaTeX: all of the trig functions are actually implemented with \cos, \sin, etc. Put the backslash before them and they will turn out pretty, with the correct spacing:

    [tex]\int x^3 \tan^4 x \sec x \; dx[/tex]
     
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