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Math Help - Integral question

  1. #1
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    Integral question

    I'm sorry in advance that I don't know how to make this look nice like the other posts do.

    I have this integral $ xdy/(x^2 + y^2)^(3/2) where x is a constant.

    The answer is given to me as y/x(x^2 + y^2)^(1/2)

    I am able to work backwards by taking the derivative of the answer to get the original integral but I can't figure out how to work the integral.

    Any help would be appreciated,

    Thank you
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Quasar View Post
    I'm sorry in advance that I don't know how to make this look nice like the other posts do.

    I have this integral $ xdy/(x^2 + y^2)^(3/2) where x is a constant.

    The answer is given to me as y/x(x^2 + y^2)^(1/2)

    I am able to work backwards by taking the derivative of the answer to get the original integral but I can't figure out how to work the integral.

    Any help would be appreciated,

    Thank you
    they used a trigonometric substitution, y = x \tan \theta

    (remember, we treat x as a constant throughout the entire problem)

    can you do the integral now that you know that?

    to make the math symbols "look nice," we use LaTex
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  3. #3
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    I still can't do the integral but atleast I know where to start looking. I have my math book open to the Trigonometric Substitutions chapter now. I'm taking Physics II and Calc II this semester and this is actually part of a physics problem I have involving electric fields. It looks like I'm 2 to 3 weeks away from learning the Trig Substitution method in my Calc II class. I worked through this physics problem and I got the integral set up but then I couldn't figure out what to do with the part of it that I just posted.

    I've been going around and around with this problem now for hours. I thought that I must know how to work that integral but I just couldn't figure it out. I have a test in the morning so I need to get this figured out tonight...if I don't pass out first. I'm exhausted.

    Thanks for your help!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Quasar View Post
    I still can't do the integral but atleast I know where to start looking. I have my math book open to the Trigonometric Substitutions chapter now. I'm taking Physics II and Calc II this semester and this is actually part of a physics problem I have involving electric fields. It looks like I'm 2 to 3 weeks away from learning the Trig Substitution method in my Calc II class. I worked through this physics problem and I got the integral set up but then I couldn't figure out what to do with the part of it that I just posted.

    I've been going around and around with this problem now for hours. I thought that I must know how to work that integral but I just couldn't figure it out. I have a test in the morning so I need to get this figured out tonight...if I don't pass out first. I'm exhausted.

    Thanks for your help!
    if you haven't done it in your class, why are you expected to know it for the test?

    \int \frac {x~dy}{\left( x^2 + y^2 \right)^{3/2}}

    Let y = x \tan \theta \implies \tan \theta = \frac yx ....remember this, we will get back to it

    \Rightarrow dy = x \sec^2 \theta ~d \theta

    So our integral becomes:

    \int \frac {x^2 \sec^2 \theta}{\left[ x^2 \left( 1 + \tan^2 \theta \right) \right]^{3/2}}~d \theta = \frac 1x \int \frac {\sec^2 \theta}{\sec^3 \theta}~d \theta

    = \frac 1x \int \frac 1{\sec \theta}~d \theta

    = \frac 1x \int \cos \theta ~d \theta

    = \frac 1x \sin \theta + C

    now we have an answer in terms of \theta, but we want an answer in terms of y

    now recall that we said \tan \theta = \frac yx

    so if we draw a right triangle with an acute angle \theta, the side opposite to the angle will be y and the side adjacent the angle will be x. By Pythagoras, the hypotenuse will be \sqrt{x^2 + y^2}.

    Now \sin \theta = \frac {\mbox {Opposite}}{\mbox {Hypotenuse}} = \frac y{\sqrt {x^2 + y^2}}

    plugging this in for \sin \theta in our answer, we find that:

    \int \frac {x~dy}{\left( x^2 + y^2 \right)^{3/2}} = \frac y{x \sqrt {x^2 + y^2}} + C

    as desired
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  5. #5
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    This integral's getting interesting, look at here.
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