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[itex]\int tan^{5}x dx[/itex] , Two Methods, are both correct?

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data


    Are both methods and final results correct?


    [itex]\int tan^{5}xdx[/itex]


    Method 1:

    [itex]\int tan^{3}xtan^{2}xdx[/itex]

    [itex]\int tanx^{3} (sec^{2}x - 1) dx[/itex]

    [itex]\int tan^{3}xsec^{2}x dx - \int tanxtan^{2}x dx[/itex]

    [itex]\int tan^{3}xsec^{2}x dx - \int tanxsec^{2}x dx + \int tanx[/itex]

    [itex]\int u^{3}du - \int u du - \int \frac{1}{u} du[/itex]

    [itex]\frac{1}{4}tan^{4}x - \frac{1}{2}tanx - ln|cosx| + C[/itex]



    Method 2:

    [itex]\int tan^{5}x dx[/itex]

    [itex]\int tan^{4}tanx[/itex]

    [itex]\int (sec^{2}x - 1)^{2}tanxdx [/itex]

    [itex]\int (sec^{4}x - 2sec^{2}x + 1)tanx dx [/itex]

    [itex]\int (u^{2} - 2u +1) du[/itex]

    [itex]\frac{1}{3}tan^{3}x - tan^{2}x + tanx + C[/itex]
     
  2. jcsd
  3. Oct 13, 2013 #2
    Although Method 2 is much easier and a lot more straightforward, both methods look correct, despite having different answers.
     
  4. Oct 13, 2013 #3
    Thanks Yosty.
     
  5. Oct 13, 2013 #4

    arildno

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    I don't understand what you are doing.
    Tip:
    Make absolutely clear WHAT substitution you are using!
    Here is how I would do it:
    [tex]u=\tan(x)\to{du}=\sec^{2}(x)dx=(u^{2}+1)dx \to{dx}=\frac{du}{u^{2}+1}[/tex]
    Thus, you are to evaluate the integral:
    [tex]\int\frac{u^{5}}{u^{2}+1}du= \int\frac{u^{3}(u^{2}+1-1)}{u^{2}+1}du =\int{u^{3}}du-\int\frac{u^{3}}{u^{2}+1}du[/tex]
    The last integral can be written as:
    [tex]-\int\frac{u^{3}}{u^{2}+1}du=-\int{u}du+\int\frac{u}{u^{2}+1}du[/tex]
    Thus, collecting together, you'll have:
    [tex]\frac{1}{4}\tan^{4}(x)-\frac{1}{2}\tan^{2}(x)+\frac{1}{2}\ln(\tan^{2}(x)+1)[/tex]
    where the latter term can be simplified to [itex]-\ln|\cos(x)|[/itex]
     
  6. Oct 13, 2013 #5

    arildno

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    Thus, NEITHER of your expressions are correct, although the first comes closer.
     
  7. Oct 13, 2013 #6

    D H

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    Method 2 is also incorrect. Your u-substitution, ##u=\sec x## lost something in the translation. With this, you should get ##du = \sec x \tan x\, dx##, but you apparently used ##du = \tan x \, dx##.
     
  8. Oct 13, 2013 #7
    Thank you Arildno,


    Method 1:

    [itex]\int tan^{3}xtan^{2}xdx[/itex]

    [itex]\int tanx^{3} (sec^{2}x - 1) dx[/itex]

    [itex]\int tan^{3}xsec^{2}x dx - \int tanxtan^{2}x dx[/itex]

    [itex]\int tan^{3}xsec^{2}x dx - \int tanxsec^{2}x dx + \int tanx dx[/itex]


    I did three separate substitutions.

    For [itex]\int tan^{3}xsec^{2}x dx[/itex]

    I let u = tanx, and [itex] du = sec^{2}x dx [/itex]

    For [itex]\int tanxsec^{2}x dx [/itex]

    I let u = tanx, and [itex] du = sec^{2}x dx [/itex]

    For [itex] \int tanx dx[/itex],

    I solved by writing as [itex] \int \frac{sinx}{cosx} dx [/itex]

    and then letting u = cosx, so [itex] -du = sinx dx [/itex]

    This produces:

    [itex]\int u^{3} du - \int u du - \int \frac{1}{u} du[/itex]
     
  9. Oct 14, 2013 #8

    arildno

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    Well, int udu goes to 1/2u^2, doesn't it?, rather than your 1/2u?
     
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