Solve Double Integration: \int\frac{x^2-y^2}{(x^2+y^2)^2}dx

Now, you'll see that the first term is the complex conjugate of the second term. So, \frac{1}{2}\left[\frac{1}{x^2+2ixy-y^2} + \frac{1}{x^2-2ixy-y^2}\right] = \frac
  • #1
sadris
2
0
[tex]\int \frac{x^2 - y^2}{(x^2 + y^2)^2} dx[/tex]

Above is an integration involved in a double integration, I know the answers via TI-89, but I am trying to find out how to get them :frown: I have tried trig substitution, u sub, integration by parts, etc. etc. And I am out of ideas. Can anyone please help?

Thanks!
 
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  • #2
Do you know partial fractions?
 
  • #3
First of all, label your original integral as, say, [itex]I[/itex]

[tex]I=\int \frac{x^2 - y^2}{(x^2 + y^2)^2} dx[/tex]

Then split the integral into two separate integrals, one where [itex]x^2[/itex] is in the numerator, and other where [itex]y^2[/itex] is in the numerator.

With the first integral (the numerator [itex]x^2[/itex] one), integrate this by parts with the aim of maxing the denominator of your integral [itex](x^2 + y^2)[/itex].

After you have done this, look at the complete expression for [itex]I[/itex] (substituting into this the integration-by-parts you just carried out).The integrals which remain (the one involving [itex]y^2[/itex] in the numerator, and the integral which is left from integration by parts), combine them both back into one single integral, with the integrand expressed as a single fraction. Compare this integral with the the expression for [itex]I[/itex] on the first line (i.e. the expression above). You should then have an algebraic equation in [itex]I[/itex], which you can solve.
 
  • #4
Do a trig substitution with x = y Tan[theta], dx = y (Sec[theta])^2. This will reduce your function to something resembling a trig identity that can easily be integrated.
 
  • #5
Partial fractioning is probably the easiest way as StatusX suggested.
 
  • #6
StatusX said:
Do you know partial fractions?

Yea I tried that but you end up with

[tex]x^2 - y^2 = A(x^2 + Y^2) + B(x^2 + Y^2)[/tex]

And when you set x^2 = -y^2 you would end up with -2y^2 =0 which really isn't a helpful expression :(
 
  • #7
Why would you set x^2=-y^2? You need to write:

[tex]\frac{x^2-y^2}{(x^2+y^2)^2} = \frac{Ax+B}{x^2+y^2}+\frac{Cx+D}{(x^2+y^2)^2}[/tex]

Then solve for A,B,C,D.
 
Last edited:
  • #8
There's a quicker way to manipulate the integrand into a form which ia easier to integrate if you say that

[tex]
(x^2 + y^2) = (x+iy)(x-iy)
[/tex]

and

[tex]
x^2 - y^2 = \frac{1}{2}\left[(x+iy)^2 + (x-iy)^2\right]
[/tex]

So,

[tex]
\frac{x^2 - y^2}{(x^2 + y^2)^2} = \frac{1}{2}\frac{(x+iy)^2 + (x-iy)^2}{(x+iy)^2(x-iy)^2}
[/tex]
 

What is the meaning of double integration?

Double integration is a mathematical process where the integral of a function is taken with respect to two variables. It involves finding the area under a surface in a three-dimensional space.

How do I solve double integration?

To solve double integration, you need to first identify the limits of integration for both variables. Then, use the appropriate integration techniques, such as u-substitution or integration by parts, to evaluate the integral. Make sure to check for any symmetry or special properties of the function to simplify the integration process.

What is the purpose of solving double integration?

Double integration is used in many areas of science and engineering to find the volume, surface area, or average value of a function over a given region. It is also used to solve problems involving motion, such as finding the position or velocity of an object over time.

What are the common challenges in solving double integration?

Some common challenges in solving double integration include identifying the correct limits of integration, choosing the appropriate integration technique, and handling complex functions or regions. It is important to practice and understand the fundamental principles of integration to overcome these challenges.

How can I improve my skills in solving double integration?

Practice is key to improving your skills in solving double integration. Start with simple problems and gradually move on to more complex ones. It is also helpful to review the fundamental principles of integration and seek help from a tutor or online resources if needed.

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