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Definite Integral limit problems

  1. Nov 29, 2013 #1
    [tex] \int_0^{2\pi} \frac{1}{25cos^2(t) + 9sin^2(t)}dt [/tex]

    Substituted the variables twice and got the upper and lower boundaries to both be 0 (think i might have gone wrong there) [tex] \frac{1}{15} tan^{-1} \frac{3tan(t)}{5} [/tex] with upper and lower boundaries both being 0. I know the answer is [itex] 2\pi/15 [/itex] but I am not getting that answer, would it be my limits that are incorrect here?
     
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  3. Nov 29, 2013 #2

    Office_Shredder

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    We have no way of knowing if your boundaries were correct or not if you don't tell us what substitution you did. Write out all the steps you took and it will be a lot easier to spot where you went wrong.
     
  4. Nov 29, 2013 #3
  5. Nov 29, 2013 #4

    Student100

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    Your work is confusing; you should try to write it better for your own sake.

    It looks like you calculated the indefinite integral right, but why are you changing the limits of integration then back subbing at the end?

    The whole purpose of changing limits of integration is so that you don't have to make back substitutes at the end.

    [tex] \frac{1}{15} tan^{-1} \frac{3tan(t)}{5} [/tex]

    Evaluate this at [tex] 0, 2\pi [/tex]

    You still get 0, but what should you do? What is 0 to 2pi?
     
    Last edited: Nov 29, 2013
  6. Nov 30, 2013 #5
    But from the substitution, wouldn't the limit of [itex] 2\pi [/itex] become 0? Both points give us a value of 0 as [itex] tan(0) = tan(2\pi) [/itex]
     
  7. Nov 30, 2013 #6

    vanhees71

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    First calculate the indefinite integral. Note that
    [tex]\frac{1}{25 \cos^2 x+9 \sin^2 x}=\frac{1}{9 + 16 \cos^2 x}=\frac{1}{9} \cdot \frac{1}{1+\left(\frac{4}{3} \cos x \right)^2}.[/tex]
     
  8. Nov 30, 2013 #7
    A much better way, IMO, is to factor out the cos^2(x) i.e

    $$\frac{\sec^2x}{25+9\tan^2x}dx$$
    A very obvious substitution can be used now.
     
  9. Nov 30, 2013 #8

    Student100

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    They already have the indefinite integral, if you look at their work they have what you wrote. What they need to understand is what a net signed area is, and why the integral from 0 to 2pi is obviously zero.

    If they want the total area, then they need to find that.

    OP, while subbing if you change the limits of integration, then you don't need to make a back sub at the end. Does that make sense?
     
  10. Nov 30, 2013 #9

    vela

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    The integrand is positive over the entire interval of integration. There's no cancellation occurring leading to a vanishing result. This paradox has nothing to do with net signed area.
     
  11. Nov 30, 2013 #10

    Student100

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    [tex]\frac{1}{15} tan^{-1} \frac{3tan(t)}{5}[/tex] is positive from 0 to 2pi??
     
  12. Nov 30, 2013 #11

    vela

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    Is that the integrand?
     
  13. Nov 30, 2013 #12

    Student100

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    Haha, you're a clever one. I got so caught up in it. :/
     
  14. Nov 30, 2013 #13

    lurflurf

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    The function is positive so the integral must be increasing and continuous. There is a problem with the indefinite integral, it only holds for -pi/2<=t<=pi/2.
    To fix it we can either break the integral up or make sure the integral is increasing and continuous.

    Break it up
    $$\int_0^{2\pi}\frac{\mathrm{d}t}{25\cos^2(t)+9\sin^2(t)}=4\int_0^{\pi/2}\frac{\mathrm{d}t}{25\cos^2(t)+9\sin^2(t)}=4\left[\frac{1}{15}\tan^{-1}\left(\frac{3}{5}\tan(t)\right)\right]_{t=0}^{t=(\pi/2)^-}$$
    It turns out that
    $$\int_0^{x}\frac{\mathrm{d}t}{25\cos^2(t)+9\sin^2(t)}=\frac{1}{15}\tan^{-1}\left(\frac{3}{5}\tan(x)\right)\\\text{is valid when}\\-\frac{\pi}{2}<x<\frac{\pi}{2}$$
     
    Last edited: Dec 1, 2013
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