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Math Help - Definate integral calculation difficulty

  1. #1
    Member Furyan's Avatar
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    Definate integral calculation difficulty

    Hello,

    The question is to evaluate the following integral.

    \int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\ 7 \cot^2(7x) dx

    by writing it as

    7\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\ \csc^2(7x) - 1 dx

    I got

     -\cot(7x) -7x

    which I think is right, my difficulty seems to be with evaluating. The answer is given as 1+\dfrac{7\pi}{4}, but I keep getting  -1 - \dfrac{7\pi}{4}.

    I'd be very grateful if someone would check this.

    Thank you.
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Definate integral calculation difficulty

    Hi Furyan!

    You have a couple of singularities in there.
    Perhaps you could split the integral in such a way you can take care of these singularities?
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  3. #3
    Member Furyan's Avatar
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    Re: Definate integral calculation difficulty

    Quote Originally Posted by ILikeSerena View Post
    Hi Furyan!

    You have a couple of singularities in there.
    Perhaps you could split the integral in such a way you can take care of these singularities?
    Hi ILikeSerena, I certainly feel very close to a black hole, but I'm afraid I can't resist the field strength. I figure since I've got the integral it should just be plain sailing. What do you mean by singularities?

    Thank you
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Definate integral calculation difficulty

    Quote Originally Posted by Furyan View Post
    Hi ILikeSerena, I certainly feel very close to a black hole, but I'm afraid I can't resist the field strength. I figure since I've got the integral it should just be plain sailing. What do you mean by singularities?

    Thank you
    Actually, now that I'm taking a better look, it's even worse. Your integral does not exist.
    Your given answer is wrong.

    The problem is that \cot 7x is not defined for all the points in the range.
    It has so called singularities in x=0, \frac \pi 7, \frac {2\pi} 7, \frac {3\pi} 7.

    It means that you have to split your integral up at these singularities into so called improper integrals (see included wiki link).
    Your integral expression 7 \cot^2(7x) has infinitely high mountains in these singularities.
    Now it could still be possible that these mountains have a finite area, but in this case they don't.


    If we split up your integral at the singularities, we get:

    \int_{\pi / 4}^{\pi / 2} 7 \cot^2(7x) dx = \int_{\pi / 4}^{2\pi / 7} 7 \cot^2(7x) dx + \int_{2\pi / 7}^{3\pi / 7} 7 \cot^2(7x) dx + \int_{3\pi / 7}^{\pi / 2} 7 \cot^2(7x) dx


    Let's pick the first term:

    \int_{\pi / 4}^{2\pi / 7} 7 \cot^2(7x) dx
    = \lim_{b~ \uparrow~ {2\pi / 7}} \int_{\pi / 4}^{b} 7 \cot^2(7x) dx

    = \lim_{b~ \uparrow~ {2\pi / 7}} (-\cot(7x)-7x)|_{\pi / 4}^b

    =\text{does not exist}


    This improper integral does not exist, because \cot(7\cdot {2\pi \over 7}) = \cot(2 \pi) is not defined.

    None of these improper integrals exist, so your integral does not exist either \Box.
    Last edited by ILikeSerena; January 26th 2013 at 01:29 AM.
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  5. #5
    Member Furyan's Avatar
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    Re: Definate integral calculation difficulty

    Quote Originally Posted by ILikeSerena View Post
    Actually, now that I'm taking a better look, it's even worse. Your integral does not exist.
    Your given answer is wrong.
    Hi IlikeSerena,

    Thank you for taking the time to post such a comprehensive reply. I now understand that the integral does not exist because you cannot integrate across a discontinuity. I also now see that it's important to sketch a graph when finding a definite integral and doing so helped me to see why I was getting a couple of other questions wrong. One thing that was confusing me was that when evaluating:

    \dfrac{1}{\tan\tfrac{7\pi}{2}} I got an error

    but when evaluating

    \dfrac{\cos\tfrac{7\pi}{2}}{\sin\tfrac{7\pi}{2}} I got an answer of 0

    I now have to evaluate:

    \int_0^2\dfrac{5 - 2x}{(x - 1)(2x + 1)}

    Integrating I get \ln(x - 1) - 2\ln(2x + 1)

    The given answer is \ln\dfrac{1}{25}

    Am I right in thinking this integral does not exist either since y = \dfrac{1}{x - 1} is not defined when x = 1?

    Thank you
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  6. #6
    Super Member ILikeSerena's Avatar
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    Re: Definate integral calculation difficulty

    Quote Originally Posted by Furyan View Post
    Hi IlikeSerena,

    Thank you for taking the time to post such a comprehensive reply. I now understand that the integral does not exist because you cannot integrate across a discontinuity. I also now see that it's important to sketch a graph when finding a definite integral and doing so helped me to see why I was getting a couple of other questions wrong.
    Good!


    One thing that was confusing me was that when evaluating:

    \dfrac{1}{\tan\tfrac{7\pi}{2}} I got an error

    but when evaluating

    \dfrac{\cos\tfrac{7\pi}{2}}{\sin\tfrac{7\pi}{2}} I got an answer of 0
    The point is that you cannot divide by zero (or something bad happens to the universe, like black holes and singularities and stuff ).

    The tan function is actually a division of the sine by the cosine.
    It does not exist if the cosine is zero, which is the case at 7pi/2, or more generally at pi/2 + k pi, where k is an integer.

    When the tan does not exist, you can't take its inverse any more, even if that would have existed as in your case.

    So 1 \over \displaystyle ({\sin 7\pi/2 \over \cos 7\pi/2}) does not exist, because it contains a division by zero.
    But {\cos 7\pi/2 \over \sin 7\pi/2} does exist, since there is no division by zero.




    I now have to evaluate:

    \int_0^2\dfrac{5 - 2x}{(x - 1)(2x + 1)}

    Integrating I get \ln(x - 1) - 2\ln(2x + 1)

    The given answer is \ln\dfrac{1}{25}

    Am I right in thinking this integral does not exist either since y = \dfrac{1}{x - 1} is not defined when x = 1?

    Thank you
    Yes, you are correct. The integral does not converge due to the singularity at x=1.


    Btw, your anti-derivative is not entirely correct, since it should also be defined for x < 1.
    You lost that somewhere in your integration.
    The anti-derivative you have only works for the part to the right of the singularity.
    You would probably need to find another anti-derivative for the left part.
    Last edited by ILikeSerena; January 26th 2013 at 11:34 AM.
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  7. #7
    Member Furyan's Avatar
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    Re: Definate integral calculation difficulty



    Yes, you are correct. The integral does not converge due to the singularity at x=1.


    Btw, your anti-derivative is not entirely correct, since it should also be defined for x < 1.
    You lost that somewhere in your integration.
    The anti-derivative you have only works for the part to the right of the singularity.
    You would probably need to find another anti-derivative for the left part.
    Thank you very much. Yes I can see that the anti-derivative only works for the part to the right of the singularity. For now I have no idea how I would go about finding another one for the left part, but I will think about it.

    Thanks again for all your help.
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