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I don't understand this integral

  1. Feb 3, 2008 #1
    I was just starting to get comfortable with integrals in the easy situations, and then with substitution in some of the more difficult cases. I thought it might be kind of fun to try and calculate pi with the area under the semi circle:


    I figured the proper integral for this would be:

    [tex]\begin{math}4\int_0^3 \sqrt{9-x^2}\end{math}[/tex]
    but then I realized I'm not fully equipped to take that anti derivative. Substitution never fully gets rid of the x, and I'm unaware of a standard anti derivative that could be used to calculate such an integral. How would you go about this?
  2. jcsd
  3. Feb 3, 2008 #2
    Is this Calculus 1? In Calculus 2, you learn about trig substitution.

    Which one so we can approach it by the method you're suppose to be learning.

    Anyways, I want to sleep so I'm just going to assume Calc 1.

    Well just draw it, then you will see from x = 0 to 3, that it is a quarter of a circle.
    Last edited: Feb 3, 2008
  4. Feb 3, 2008 #3
    I'm taking calculus 2, but this isn't for school. We haven't covered trig substitution.

    A key here is that I'm not interested in the area of the circle. I'm only interested in the area of the circle, and the radius of the circle because they are needed for this:

    So yes the integral calculates the area of a quarter circle, but you'll see I've multiplied that area by 4, and now you can see why I feel like doing it the hard way.
  5. Feb 3, 2008 #4
    kentm, you forgot the dx in the integral. Try to avoid this, you will loose points on an exam.

    Anyway, you have written the correct integral for calculating the area:

    [tex]A=4\int_0^3 \sqrt{9-x^2}dx[/tex]

    The way to proceed is indeed a trigonometric substitution. It isn't as difficult as it sounds, instead of setting p.e. x=t^2 you set:

    [tex]x=3\cdot sin(t)[/tex]

    Using the same principle as for "simpler" substitutions you need to find the dt and new limits of the integral. These are thus:

    [tex]dx=3\cdot cos(t)dt[/tex]
    [tex]x=0 \rightarrow t=0[/tex]
    [tex]x=3 \rightarrow t=\frac{\pi}{2}[/tex]

    Using this in the integral you get:

    [tex]A=4\int_0^{\frac{\pi}{2}} \sqrt{9-9\cdot sin^2(t)}\cdot 3\cdot cos(t) dt[/tex]
    [tex]A=4\cdot 3 \cdot 3 \int_0^{\frac{\pi}{2}} \sqrt{1-sin^2(t)}\cdot cos(t) dt[/tex]
    [tex]A=36 \int_0^{\frac{\pi}{2}} cos^2(t)} dt[/tex]

    Why did I use this substitution? It is a matter of practice to "see" that something like I did is turning the integral into a simpler one. Very often you will encounter integrals with square roots like this and it is best to try these first, it might be the solution. This is certainly not true in general, but an attempt should be made. To solve the remaining integral think of the following:

    [tex]cos^2(t)=\frac{1}{2}\left( 1+cos(2t) \right)[/tex]

    Put this in the integral and you will end up with two basic integrals you should be able to tackle.
  6. Feb 3, 2008 #5
    Cool, that actually makes a bit of sense. If you told me what the substitution was I could have done it, I think. Where that substitution came from, you're right, is beyond me. It must take a little practice. Now, after having learned something I see that:

    [tex]\begin{math}A=36 \int_0^{\frac{\pi}{2}} cos^2(t)} dt=36\left[tan(t)\right]_0^{\frac{\pi}{2}}=36[\frac{1}{0}-0]\end{math}[/tex]

    ...and to check:


    Why would the area of an existing circle be undefined?
    Last edited: Feb 3, 2008
  7. Feb 3, 2008 #6
    The integral of cos^2 (x) is not tan(x). You even differentiated it to check: [itex]sec^2(x) \neq cos^2(x)[/itex]

    Use what coomast said last to integrate cos^2.

    As for knowing which trig sub to use, simply start with this identity:

    [tex]sin^2(x) + cos^2(x) = 1[/tex]

    Divide both sides by cos^2(x).

    [tex]tan^2(x) + 1 = sec^2(x)[/tex]

    Rearrange either of these two equations the following three ways:

    [tex]1 - sin^2(x) = cos^2(x)[/tex]
    [tex]tan^2(x) + 1 = sec^2(x)[/tex]
    [tex]sec^2(x) - 1 = tan^2(x)[/tex]

    Note the three forms of the left hand side: constant - variable, constant + variable, and variable - constant. Your integral, [itex]\sqrt{9 - x^2}[/itex], is in form constant - variable. So, you make the substitution [itex]x = 3 sin(\theta)[/itex] and the integral will become simple to do, because the square root turns into [itex]3cos(\theta)[/itex]
    Last edited: Feb 3, 2008
  8. Feb 3, 2008 #7
    Haha, I don't know where that came from. So:

    [tex]\frac{1}{2}\int_0^{\frac{\pi}{2}} 1+cos2x\,dx=\frac{1}{2}\left[x+\frac{1}{2}sin2x\right]_0^{\frac{\pi}{2}} [/tex]

    It's good to know this stuff, but the way this ended up working out doesn't seem to work. I'm assuming that I don't know what pi is, hence the purpose of all of this. It ends up working out to this:

    [tex]\begin{math}\frac{1}{2}\left[(\frac{\pi}{2}+\frac{1}{2}sin2\frac{\pi}{2}) - (0+\frac{1}{2}sin2(0))\right] \end{math}[/tex]

    [tex]sin\frac{2\pi}{2}[/tex] works out to be 0, along with sin(0), but there is no way to figure out [tex]\frac{1}{2}*\frac{\pi}{2}[/tex] because we are assuming we don't know the value of pi. There is probably something wrong with the math as well. I'm nearly sure that [tex]3^2\pi>\frac{\pi}{4}[/tex], and realistically [tex]9\pi [/tex] is the correct answer.
    Last edited: Feb 3, 2008
  9. Feb 3, 2008 #8
    I forgot to bring down the 36, so the answer is correct:

    plugging in the numbers to the equation: [tex]Area=\pi r^2[/tex], we end up with a true statement:


    Oh well, it was fun anyways.
  10. Feb 3, 2008 #9

    Gib Z

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    I'm sure you were hoping you would find a different expression for pi from the integral, instead of exactly just pi again? I used to hope that too lol. We never know until we try =]
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