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Integrating Trig functions (Namely tan^{2}x)

  1. Jan 14, 2007 #1
    I know that
    [tex]\int \tan^{2}x dx= \int \sec^{2}x-1\ dx = \tan x - x + C[/tex]
    but i don't completely understand how this is derived. Because of this lack of comprehension, I have no idea what to do with [tex]\int \sec^{4}5x\ dx[/tex]. I went from there to get:
    [tex]\int \sec^{4}5x\ dx=[/tex]
    [tex]\int [\sec^{2}5x]^{2}\ dx=[/tex]
    [tex]\int \tan^{2}5x\ dx = [/tex]
    but I have no idea where to go from here. Any help would be appreciated. Thanks in advance.
    Last edited: Jan 14, 2007
  2. jcsd
  3. Jan 14, 2007 #2


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    change from [text] to [tex]
  4. Jan 14, 2007 #3


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    ...and change from [\text] to [/tex].

    Edit: boy, are we helpful. :biggrin:
  5. Jan 14, 2007 #4
    Ha there we go! Thank you. How about help with the problem now?
  6. Jan 14, 2007 #5
    This is wrong, these two lines are not equivalent. What trigonometric identity do you know that relates secant and tangent?
  7. Jan 14, 2007 #6
    Doesn't [tex]\int \sec^{2}x\ dx=\int \tan x dx[/tex]?

    Edit: I answered the question as I was asking it. It equals [tex] \tan x[/tex] not [tex] \int \tan x dx[/tex]. So what should I do instead? I would also still like to know how to integrate [tex] \tan^2 x[/tex]
    Last edited: Jan 14, 2007
  8. Jan 14, 2007 #7
    No. What is the derivative of tan(x)?
  9. Jan 14, 2007 #8

    [tex]\sec^2 x[/tex]. roger. so what instead?
  10. Jan 14, 2007 #9
    Ok Starting with

    [tex] \int \sec^{4} 5x dx[/tex]

    split this up to

    [tex] \int (\sec^{2} 5x)(\sec^{2} 5x) dx[/tex]

    Then use an identity to simplify the first sec2(5x) and then it should be much simpler.
  11. Jan 14, 2007 #10


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    Use complex numbers via
    where [tex]i^2=-1[/tex]
    ... and make use of
    [tex]e^{ix}=\cos(x)+i\sin(x)[/tex] (to convert back)

    [Integrating the exponential function is much easier.]
  12. Jan 14, 2007 #11
    With sines and cosines this is easy, but with secants and tangents i think this would actually complicate things.
  13. Jan 14, 2007 #12


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    FYI, the only trick you have used here is the fact that
    [tex]\tan^2 x = \sec^2 x -1[/tex]
    which is a trig identity derived from
    [tex]\cos^2 x + \sin^2 x = 1[/tex]
    btw, knowing this is inconsequential to how you may solve
    [tex]\int \sec^4 (5x)\;dx[/tex]

    as previous posters mentioned, split them up and use the identity once...your aim is to rewrite the expression in terms of things you can readily integrate like
    [tex]\sec^2 x, \sec^2 x \tan^2 x [/tex]

    hope I haven't given away too much :smile:
  14. Jan 15, 2007 #13


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    [tex] =\frac{1}{5}\int \frac{du}{\cos^{4}u} [/tex]

    where [itex] u=5x [/itex].

    Now make the substitution

    [tex] \tan u =y [/tex]

  15. Jan 15, 2007 #14
    The easiest way to go about this is

    sec^4 (5x) dx
    let 5x = y (which you can differentialte and substitue accordingly).

    so lets integrate
    I= sec^4 x dx.
    = (sec^2x . sec^2x) dx
    = ((1+tan^2x) .sec^2x)dx
    = Sec^2 dx + sec^2x.tan^2x dx
    = tan x + c + J

    J = sec^2x.tan^2x dx

    Now let tan x =t
    therefore, sec^2xdx =dt
    J = t^2 dt
    J= t^3 / 3 + c

    I=tan x + c + (t^3 / 3) + c
    tan x +(tan^3x)/3 + constant of indefinate integration.
  16. Jan 15, 2007 #15
    that is good
  17. Jan 15, 2007 #16
    This is great guys, thank you all. I just need to memorize all those stupid trig identities and I think this will be a lot easier for me. Thanks again everyone!

  18. Jan 15, 2007 #17

    Please don't post complete solutions to homework questions.
  19. Jan 15, 2007 #18
    You dont have to memorize the identitites.
    I derive them when i need them

    the basic ones are

    Sin^2 x + cos^2 x= 1
    Cos2x = Cos^2 - sin^2x
    sin 2 x = 2. cosx. sinx

    from the first one we can divide throught by cos and get values for
    tan^2x +1=sec^x.
    dividing by sin gives
    1+cot^2x = csc^2x.

    from the second one
    you can substitute sin^2x or cos^2x as 1-cos^2x or 1-sin^2x
    this is how how you integrate
  20. Jan 15, 2007 #19
    thanks for the advice, i'll work on that.
  21. Jan 15, 2007 #20
    to add to what dickcruz said, the cos2x formulas (there are 3) are very useful in integration:

    take cos2x= 1 - 2sin^2x and solve for sin^2x
    (or take cos2x = 2cos^2x - 1 and solve for cos^2x
    This gives you the power reducing formulas which you'll use when you have even powers.
    i.e. to integrate (cos^2 x)
    you would first change it to the integral of (1 + cos2x)/2; a relatively simple integral to take care of.
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