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Homework Help: Finding the general integral

  1. Sep 6, 2006 #1

    I have a problem that I am having difficulties with. I'm told to find the general integral.

    So here is the problem:
    [itex]\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} \frac {sinx} {1 - sin^2 x} dx[/itex]

    Here is my partial solution:

    [itex]\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} \frac {sinx} {cos^2 x} dx[/itex]

    [itex]\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} \frac {1} {cosx} * \frac {sinx} {cosx} dx[/itex]

    [itex]\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} secxtanx dx[/itex]

    [itex]= [tanx][/itex]

    But here is where I am stuck, because when you evaluate [itex]tanx[/itex] from [itex]\frac {\pi} {2}[/itex] to [itex]\frac {-\pi} {2}[/itex] you get an error because no value exists for [itex]tan \frac {\pi} {2}[/itex]

    Any help would be greatly appreciated. :smile:
  2. jcsd
  3. Sep 6, 2006 #2
    wouldn't it be zero? you're integrating from pi/2 to -pi/2 & tan is an odd function.
  4. Sep 6, 2006 #3
    Well that is what I was unsure of. I just double checked the math on my calculator, and I realized I made an error inputting the values. So that would explain the error.

    Thanks for the input. :smile:
  5. Sep 6, 2006 #4
    The problem might be that you have the wrong antiderivative of sec(x)tan(x). What is the derivative of sec(x)?
  6. Sep 7, 2006 #5
    OOPS i guess it has been a while since i last did calculus... sorry about that :eek: :redface:
  7. Sep 7, 2006 #6
    But surely sec (pi/2)=1/0 as well??:yuck:
  8. Sep 7, 2006 #7


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    I'm a little puzzled by your saying you are asked to find the "general integral" and then asking for a specific value for a definite integral.

    Of course, since the denominator, 1- sin2x is 0 at both [itex]\frac{pi}{2}[/itex] and [itex]-\frac{\pi}{2}[/itex], this is an "improper integral". It's value is defined as
    [tex]\lim_{\alpha\rightarrow -\frac{\pi}{2}} \lim_{\beta\rightarrow \frac{\pi}{2}}\int_\beta^\alpha \frac{sin x}{1- sin^2 x}dx[/tex]
    So what you want to do is find the "general anti-derivative" and then take limits. Of course, that is precisely what you did- you can't evaluate tan(x) at the endpoints precisely because those limits do not exist.

    The "Cauchy Principal Value" for this integral is defined as
    [tex]\lim_{\alpha\rightarrow \frac{pi}{2}}\int_\alpha^{-\alpha}\frac{sin x}{1- sin^2x} dx[/tex]
    and since the integrand is odd, that is 0.
    Last edited by a moderator: Sep 7, 2006
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