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Particular Integral of arctan example

  1. Aug 17, 2004 #1
    I know for a fact that

    [tex]
    \int \frac{(du)}{(a^2 + u^2)} = \frac{1}{a} \cdot arctan \frac{u}{a} + C
    [/tex]

    I was given the problem of solving the indefinite integral of

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

    First, I multiplied the integral by (1/2) / (1/2) to eliminate the coefficient of the x^2 in the denominator, so now I am left with

    [tex]
    \frac{1}{2} \cdot \int \frac{(dx)}{(x^2 + x + \frac{5}{2})}
    [/tex]

    Now, in completing the square of the denominator, I added [tex] \frac{1}{4} - \frac{1}{4} [/tex] (which is zero) so that the equation would look like this:

    [tex]
    = \frac{1}{2} \int \frac{dx}{x^2 + x + \frac{5}{2} + \frac{1}{4} - \frac{1}{4}}
    [/tex]

    Simplifying, I got:

    [tex]
    = \frac{1}{2} \int \frac{dx}{x^2 + x + \frac{1}{4} + (\frac{5}{2} - \frac{1}{4})}
    [/tex]

    [tex]
    = \frac{1}{2} \int \frac{dx}{(x + \frac{1}{2})^2 + \frac{9}{4}}
    [/tex]

    [tex]
    = \frac{1}{2} \int \frac{dx}{(x + \frac{1}{2})^2 + ( \frac{3}{2})^2}
    [/tex]

    If we let [tex](x + \frac{1}{2})^2 = u^2[/tex] and [tex]( \frac{3}{2})^2 = a^2[/tex] we now have the integrable form stated above, so

    [tex]
    = \frac{1}{2} \int \frac{du}{u^2 + a^2}
    [/tex]

    [tex]
    = \frac{1}{2} \cdot \frac{2}{3} \cdot arctan \frac{u}{a} + C
    [/tex]

    [tex]
    = \frac{1}{3} \cdot arctan \frac{x^2 + \frac{1}{2}}{\frac{3}{2}} + C
    [/tex]

    [tex]
    = \frac{1}{3} \cdot arctan \frac{2(x^2 + \frac{1}{2})}{3} + C
    [/tex]

    My questions are:
    (1) Is this the correct solution?
    (2) If my solution is correct, how do you get the necessary constant in order to make the denominator of this problem a complete square? the (1/4 - 1/4) I added to the denominator just popped out of my mind. Is there any way to get it without resorting to trial and error?

    thank you
     
  2. jcsd
  3. Aug 17, 2004 #2
    1) No, there seems to be some confusion regarding the subsitution of variables, you've chosen (x + 1/2)^2 = u^2, this means that u = x + 1/2, not u = (x + 1/2)^2.

    2) Suppose you wanted to write x^2 + bx + c in the form (x + p)^2 + q, i.e. you want to choose p, q such that x^2 + bx + c = (x + p)^2 + q = x^2 + 2px + p^2 + q. Comparing coefficients on both sides of the equation gives 2p = b and p^2 + q = c. So you can take p = b/2 and q = c - p^2 = c - (b/2)^2.
     
    Last edited: Aug 17, 2004
  4. Aug 17, 2004 #3

    Zurtex

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    Homework Helper

    When I was doing this I found trying to complete all the quadratic equations I had by completing the square rather than any other method. Either actually solving them or just putting them in the form [itex]a(x+b)^2 + c[/itex], remember good algebra techniques greatly help calculus.
     
  5. Aug 17, 2004 #4
    um... since [tex] u^2 = ( x + \frac{1}{2} )^2 [/tex] you can directly place this in the integrable form because the form needs a [tex] u^2 [/tex],

    but I understand the process of solving the quadratic eq'n or completing the square. thanx.

    I was just wondering if the answer I got was correct.
     
  6. Aug 17, 2004 #5

    Zurtex

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    Science Advisor
    Homework Helper

    In the final two steps you have ended up replaces u with [itex]x^2 + 1/2[/itex], that was the only mistake I think.
     
  7. Aug 18, 2004 #6
    You also picked the wrong u originally. But then when you went to substitute you put in the right one. But later when you replaced u with x you went back to the wrong u.
     
  8. Aug 18, 2004 #7
    ahh.. ok, I see the mistake. Sorry. And the title should be Particular Integral that yields arctan. My mistake as well...

    the u should be

    [tex] u = x + \frac{1}{2} [/tex]

    and the real answer should be

    [tex]= \frac{1}{3} \cdot arctan \frac{2(x + \frac{1}{2})}{3} + C[/tex]


    thanx again for the clarifications and replies. ^_^
     
    Last edited: Aug 18, 2004
  9. Aug 18, 2004 #8
    I like your approach though; I probably would have used a ugly trig sub. I should probably memorize a few of the basic integral in tables.
     
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