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More trig substitution help

  1. Jan 31, 2008 #1
    [SOLVED] More trig substitution help...

    I've looked at this problem about 3 times and still can't figure it out...where identity did they use to substitute out the part in the red box? Thanks for the help

    [​IMG]
     
  2. jcsd
  3. Jan 31, 2008 #2
    What's the step you don't understand?
    [tex] \int \tan^{4}x \mathrm{d}x = \int \tan^{2}x \left(\sec^{2}x - 1\right) \mathrm{d}x [/tex]
    or
    [itex] \int \tan^{2}x \left(\sec^{2}x - 1\right) \mathrm{d}x = \int \tan^{2}x \sec^{2}x \mathrm{d}x - \int \tan^{2}x\mathrm{d}x[/itex].
     
  4. Feb 1, 2008 #3

    Gib Z

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    [tex]\sec^2 x - 1 = \tan^2 x[/tex] and [tex]\tan^2 x \cdot \tan^2 x = \tan^4 x[/tex].
     
  5. Feb 1, 2008 #4
    for me it's unclear how to integrate
    [tex]\int\tan^2 x\sec^2 x dx=\int (\sec^2 x-1)\sec^2 x dx=\int \frac{1}{\cos^4}dx-\tan x[/tex]
    So how to integrate
    [tex]\int \sec^4 dx[/tex]
    ?
     
  6. Feb 1, 2008 #5

    Gib Z

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    Integration by parts a few times does it, or write

    [tex]\int \sec^2 x (\tan^2 x + 1) dx [/tex] and let u= tan x.

    But rather than integrate sec^4, keep the original integral,
    [tex]\int \tan^2 x \sec^2 x dx = \int u^2 du[/tex] when u= tan x.
     
  7. Feb 1, 2008 #6
    I see
    [tex]\int\tan^2 x\sec^2 x dx=\int \tan^2 x d(\tan x)=\frac{1}{3}\tan^3 x+C[/tex]
     
    Last edited: Feb 1, 2008
  8. Feb 1, 2008 #7

    Gib Z

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    You forgot the x on the end of the tan, but other than that, its correct.
     
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