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Some simple integration (trig)

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    [1] integral of 2x / (4+x^2)
    [2] [tex]\int \frac{3}{\abs{x}\sqrt{x^{6}-1}}[/tex]
    (the x above is absolute |x|, can't find the latex code....

    2. Relevant equations

    3. The attempt at a solution

    [1] can be done easily using natural log method.
    ln|4+x^2| +c

    The reason I asked this was because this question belong to "The Caclulus of The Inverse Trig Functions", where it introduced how to take d/dx and integral of arcsin, arccos, arctan, etc.

    I was wondering if [1] can be done by arctan... arctan = integral of 1/1+x^2

    [2] It was very similar to integral of sec, integral of 1/ (|x| sqrt(x^2-1))
    I tried to factor out those x and reduce it to x^3, but still no luck, because of |x|

    Any help is appreciated.
  2. jcsd
  3. Apr 7, 2010 #2

    Gib Z

    User Avatar
    Homework Helper

    Sorry no, theres no "nice" way to express it in terms of the arctangent function. The book was maybe trying to show you that just because you see some (x^2 + a^2) term doesn't mean you should try [itex]x= a \tan t[/tex] straight away, but still check that the methods you used before don't work first. If they do, then use that!

    Your hunch is good. So if you want to make the integral you have now look like the one for arcsec x, what substitution could you make that does that? With experience you will learn that when you want to make two algebraic expressions "similar" enough for a u-substitution, the best things to make the same is whats under the root sign, and hopefully the rest will fall into place.
  4. Apr 7, 2010 #3


    Staff: Mentor

    Don't omit the dx in your integrals. If you do, you can run into problems, especially in trig substitutions.

    Gib Z,
    Regarding [tex]\int \frac{2x dx}{4 + x^2}[/tex]
    this can be integrated using trig substitution. This is not the most efficient way to do things, because an ordinary substitution is quicker for this integral, but a trig substitution works and isn't much more involved.

    Using an ordinary substitution, u = 4 + x2, and du = 2xdx, you arrive at an antiderivative of ln(4 + x2) + C

    Using trig substitution, with 2tan w = x, and 2sec2 w = dx, you arrive eventually at an antiderivative of ln( (4 + x2)/4 ) + C'.

    These antiderivatives appear to be different, and they are, but they differ only by a constant, namely ln(4).
  5. Apr 7, 2010 #4
    Thank you both of your helps. I understood them now.
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