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Integral concept question

  1. Jul 30, 2015 #1
    int_0^(pi) sin(t)/[3 + cos(t)] dt

    u = cos(t); du = -sin(t) dt; a = 1; b = -1

    int_1^(-1) -1/(u + 3) du

    int_(-1)^1 1/(u + 3) du = ln(u + 3) from F(1) - F(-1)

    ln(4) - ln(2) = ln(2)

    I had help on this. And there are a few aspects I don't understand. What is the purpose of a=1; b=-1 and why does the integral separate into int_1^(-1) and then int_(-1)^1 why does it switch and what happened to pi and 0? and finally F(1)-F(-1). I know this is asking a lot but I am trying to grasp these concepts for the first time.

    Thank you
  2. jcsd
  3. Jul 31, 2015 #2
    This is really hard to read. Is your integral?:

    [tex][\int_{0}^{\pi }\frac{sin(t)}{3+cos(t)} dt[/tex]

    edit: and what are a and b?
    Last edited: Jul 31, 2015
  4. Jul 31, 2015 #3
    [tex]\int_{0}^{\pi }\frac{sin(t)}{3+cos(t)} dt[/tex]
    let [tex]u=cos(t)[/tex]
    so [tex]du=-sin(t)[/tex]
    which means: [tex]-du=sin(t)[/tex]
    Now make the substitution:
    [tex]\int_{1}^{-1 }\frac{-du}{3+u} [/tex]
    Now evaluate:
    [tex]\int_{1}^{-1}\frac{-du}{3+u} = -ln(3-1)+ln(3+1)=ln(2)[/tex]
    Last edited: Jul 31, 2015
  5. Jul 31, 2015 #4


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    You can't evaluate the integral in u with limits for t. The a and b come from transforming the limits in t to equivalents in u:

    Since u = cos t, and t runs from t = 0 to t = π, then the lower limit of the integral after substitution is found by u = cos (0) = 1 = a, and the upper limit is u = cos (π) = -1 = b.

    These limits in u can be switched by taking the negative of the original integral in u, using the lower limit of -1 and the upper limit of 1:

    [tex]\int_{1}^{-1}\frac{-du}{3+u} = -\int_{-1}^{1}\frac{-du}{3+u} = ln(3+1)-ln(3-1)=ln(2)[/tex]
  6. Jul 31, 2015 #5
    Gotcha. I am aware of changing bounds on an integral....I suppose Im just used to doing it informally(or just in my head) with elementary integrals. For some reason, it sounded as if he was doing partial fractions or something..I suppose I needed clarification as well!
  7. Jul 31, 2015 #6
    Thanks for the replies I am understanding. But I need a few things addressed, I see that the bounds or extremes run from 0 to pi so that is the limit of the integral. But why is 0 and pi plugged into cos(t) not sin(t). I guess I need more of a conceptual understanding of why this works this way. Also, when u=cos(0)=1 becomes "a" and u=cos(pi)=-1 becomes "b" does "a" represent the upper bound and "b" the lower bound automatically? or could you have plugged pi in first and 0 seconded? And finally why does the int first have -1 as the upper and 1 as the lower then swapped but then a negative sign is placed outside. Sorry I know these are basic questions just really want to understand them.
  8. Jul 31, 2015 #7
    a and b are arbitrary - the reason "a" becomes the upper bound is becuase in order to find "a" we plugged in the upper bound for t (the same goes for "b")

    if you plugged in pi first and 0 second you are reversing the order on your bounds - your integral would come out negative

    this sort of goes back to your last question - if we DID plug in the bounds backwards, the integral comes out negative. Isn't this like plugging in the bounds in the correct order, then multiplying by a negative and reversing the order of the bounds? - This is done simply to make your work look a little nicer. You dont have to multiply by (-1): notice you would just have -ln(3-1)+ln(3+1) instead of ln(3+1)-ln(3-1).
    Last edited: Jul 31, 2015
  9. Jul 31, 2015 #8


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    Because you are making the substitution u = cos t and then integrating with respect to u. The variable t disappears in the new integral expression, and therefore, the limits for t from the original integral must be transformed into limits in u for the new integral.

    I thought the steps shown in Post #4 explained this clearly.

    You find out what the upper limit in t is when the u-substitution is made; same for the lower limit. Often, the upper and lower limits in the original integral transform so that the upper limit in u > the lower limit in u. This integral switched limits, but these u-limits can be switched back by multiplying the u-integral by -1. This is just the basic application of the properties of definite integrals:

  10. Aug 1, 2015 #9
    Ok, so essentially the -1 reverses the integral which you must do because the lower bound must be lower, or start lower for the limit to make logic. What I still don't understand is the last part, which may be very simple. How does -int_(-1)^(1) -du/(3+u) equal a natural log of (3+1) where does the natural log come from? Also, when doing a U sub how do I know which to pick or which one is the most effective?

    Thank you
  11. Aug 1, 2015 #10


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    From the posts you are making, it appears that you have only just started to learn calculus, at least the integral variety.

    After some study, it should be apparent that one of the more familiar indefinite integrals is:

    ##\int \frac {du} {u} = ln |u| + C ##

    It doesn't matter that the denominator of the original u-sub integral was (u + 3), since the derivative of (u+3), indeed (u + C), is still du.
    This is where the natural log comes from.
  12. Aug 1, 2015 #11
    What I don't entirely understand is why the integral goes from -du/3+u to equal ln(3+1)-ln(3-1)=ln(2) I understand the log rules that when you divide and expand it then it becomes subtraction. But for the fraction is it ln because 1/x=ln(x) something like that? what happens to the -du, wouldn't that be sin/4 if the values were plugged in. Thanks
  13. Aug 1, 2015 #12


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    Apparently, you did not understand my Post #10.

    The integral of [d(mess) / mess] = ln | mess | + C

    In this case, mess = u + 3, and d(mess) = du

    The derivative of (u + C) = du, for any constant value of C (including C = 0).
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