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Integration of trig function

  1. Jun 27, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the integral of tan x sec2x with respect to x?

    2. Relevant equations



    3. The attempt at a solution

    I have no idea as to how I should proceed!
     
  2. jcsd
  3. Jun 27, 2010 #2
    Hint: [tex]tanxsec^2xdx=\frac{sinxdx}{cos^3x}[/tex]
     
  4. Jun 27, 2010 #3
    How can that help? I have no idea!

    Also, if we let u = tan x, then we get the limit of sin x as x tends to infinity, which is nonsense.
     
  5. Jun 27, 2010 #4
    Another hint: sinxdx = d( ... )?
     
  6. Jun 27, 2010 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    As you were told in another thread, the "indefinite" integral is just the anti-derivative. It has nothing to do with a limit at infinity.
    To integrate
    [tex]\int \frac{sin x}{cos^3 x} dx[/tex]
    Let u= cos(x).
     
  7. Jun 27, 2010 #6

    Mark44

    Staff: Mentor

    Even more direct: If u = tanx, du = sec2x dx. The indefinite integral has the form [itex]\int u du[/itex].
     
  8. Jun 30, 2010 #7
    I don't understand what d(.....) actually means. I guess it's a clever way of using calculus that I'm not familiar with. But I have used the substitution u = cos x as follows.

    [tex]
    u = \cos x & \Rightarrow du = - sin x dx \\
    \int \frac{\sin x dx}{cos^{3} x} & = - \int\frac{1}{u^3} du \\
    & = \frac{1}{2} u^{-2} + c \\
    & = \frac{1}{2\cos^{2}x} + c
    [/tex]

    But if I use u = tanx I get the following.

    [tex]
    u = \tan x & \Rightarrow du = sec^{2} x dx \\
    \int tan x sec^{2} x dx & = - \intu du \\
    & = \frac{1}{2} u^{2} + c \\
    & = \frac{tan^{2} x}{2} + c

    [/tex]

    The two answers are contradictory. Where's the problem?
     
  9. Jun 30, 2010 #8
    Sorry I made a silly mistake in my Latex code. The correction is:

    [tex]
    \int tan x sec^{2} x dx & = \int u du \\
    [/tex]
     
  10. Jun 30, 2010 #9
    And I don't know how to break lines in my Latex code. Sorry for that!
     
  11. Jun 30, 2010 #10

    vela

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    They're not contradictory. Use the identity tan2 x + 1 = sec2 x.
     
  12. Jun 30, 2010 #11
    Ok, so

    [tex]
    \frac{1}{2}\sec^{2} x + c \\
    & = \frac{1}{2}\tan^{2} x + \frac{1}{2} + c \\
    [/tex]

    Therefore, the constant of integration resulting from my math is 1/2 + c, whereas the constant of integration in the other result is c. Should we not be worried abt that? Or is it simply an effect of the use of different substitutions at the start of the problem?
     
  13. Jun 30, 2010 #12

    Mark44

    Staff: Mentor

    If you get two different answers from an indefinite integral, they can differ by only a constant. (1/2)sec^2(x) and (1/2)tan^2(x) differ by a constant, which is what vela was saying.
     
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