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Homework Help: Antiderivative of -tanx

  1. Jul 4, 2008 #1


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    Hi, I'm just starting to learn integration (I went to the first class in my summer course yesterday), and I'm already a bit confused. I was doing some practice problems in the text, and I found this:

    Find the most general antiderivative of f on the indicated interval:

    [tex]f(x) = -\tan x[/tex] on [tex](-\frac{\pi}{2}, \frac{\pi}{2})[/tex]

    I'm...really unsure as to how to do this problem. All this time I'm thinking, "What do I differentiate to get tan x in the first place?" ...heh, I'm still not sure. Could someone please give me some pointers? Thanks!
  2. jcsd
  3. Jul 4, 2008 #2
    Try multiplying by (secx + tanx)/(secx + tanx).
  4. Jul 4, 2008 #3
    [tex]-\int\frac{\sin x}{\cos x}dx[/tex]

    [tex]u=\cos x \rightarrow -du=\sin xdx[/tex]
    Last edited: Jul 4, 2008
  5. Jul 4, 2008 #4


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    I took your suggestion and worked it out on paper, but...I don't seem to be able to see how this would help.

    f(x) = (-sec x tan x - tan^2 x)/(sec x + tan x)...

    I'm afraid I've run into something that's a bit too advanced (this problem was listed under the "challenger" questions..and I was just curious). I will come back to this problem once I'm more...knowledgeable? Nevertheless, thank you for your help. I shall ask my teacher to explain it to me in detail later.
  6. Jul 4, 2008 #5
    The method used to solved that is called integration by substitution. Basically, you make a part of the equation equal to u (I'd choose cos x) and find du (which is equal to -sinx*dx). Substitute in and integrate in terms of u.
  7. Jul 4, 2008 #6


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    Homework Helper

    I believe you're thinking for the trick for integrating sec x ...

    You may be having trouble with this one because it's not generally a function students are asked to find the antiderivative of before they have seen the basic antiderivatives. And tan x is not the derivative of any of the elementary functions. Were you assigned this problem for practice or did you pick it out yourself? I ask because you will quickly discover that, while after having learned the rules of differentiation, you can find the derivative of any messy function you can cook up out of the elementary functions, it is rather easy to pick out or make up a function that is extremely difficult to anti-differentiate (and there are some no one has found a method for: these are integrated numerically).

    As some of the posters have mentioned already, the trick here is to exploit the fact that the derivative of ln(x) is 1/x , so we can say d[ ln(x) ] = 1/x dx = dx / x . This gives us a method for taking anything that can be write as

    ( derivative of u ) / u ,

    and expressing its general antiderivative as ln |u| + C . You will see this used a lot in your course before too much longer...
    Last edited: Jul 4, 2008
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