Interesting Integral

During my time at university, I was asked to solve an integral as part of one of my assignments. The integral was as follows:

\(\text{Integrate the following:}\\
\int \int \frac{x^2 -y^2}{(x^2+y^2)^2} dx \; dy \;\;\; \text{for} \; x>0, y>0\)

If you’d like to try this out for yourself, then please do so now. I will provide the solution below.

The reason why this problem appears to be so difficult is because seeing the x² and the y² would incline someone to try and convert to cylindrical co-ordinates. I tried that myself and was not able to work it out.

The solution I have is as follows:

\int \int \frac{x^2 -y^2}{(x^2+y^2)^2} dx \; dy


\text{First, we generate a partial fraction expansion of the integrand:}\\
=\int \int \frac{1}{(x^2+y^2)} – \frac{2y^2}{(x^2+y^2)^2}dx \; dy\\
\text{Split the integral up and factor out the constants:}\\
=\int \int \frac{1}{(x^2+y^2)} dx \; dy + \int -2y^2 \int \frac{1}{(x^2+y^2)^2}dx \; dy\\
\text{Since y is constant, factor out } \frac{1}{y^2} \text{ from the first integral:}\\
=\int \frac{1}{y^2} \int \frac{1}{(\frac{x^2}{y^2}+1)} dx \; dy + \int -2y^2 \int \frac{1}{(x^2+y^2)^2}dx \; dy\\


\text{For } \frac{1}{\frac{x^2}{y^2} + 1} \text{ , sub } p=\frac{x}{y}\\
\text{Therefore, } dx =y\,dp\\


\text{For } \frac{1}{(x^2+y^2)^2} \text{ , sub } x=y \tan (u)\\
\text{Therefore, } dx =y \sec ^2(u) \; du\\
\text{Then, } \frac{1}{(x^2 + y^2)^2} = \frac{1}{(y^2 \tan ^2(u) + y^2)^2} = \frac{1}{(y^2 \sec ^2(u))^2} = \frac{1}{y^4 \sec ^4(u)}


\text{This gives us:}\\
=\int \frac{1}{y^2} \int \frac{y}{p^2+1} dp \; dy + \int -2y^2 \int \frac{y \sec ^2(u)}{y^4 \sec ^4 (u)}du \; dy\\
=\int \frac{1}{y} \int \frac{1}{p^2+1} dp \; dy + \int -\frac{2}{y} \int \frac{1}{\sec ^2 (u)}du \; dy\\
=\int \frac{1}{y} \int \frac{1}{p^2+1} dp \; dy + \int -\frac{2}{y} \int \cos ^2(u) du \; dy\\
\text{Now, we use the double angle formula } \cos (2u) \equiv 2 \cos ^2(u) -1\\
=\int \frac{1}{y} \int \frac{1}{p^2+1} dp \; dy + \int -\frac{2}{y} \int \frac{\cos (2u) + 1}{2} du \; dy\\
\text{We are now in a position to integrate:}\\
=\int \frac{\arctan (p)}{y} dy + \int -\frac{2}{y}  \frac{\frac{\sin (2u)}{2} + u}{2} dy\\
=\int \frac{\arctan (p)}{y} dy + \int -\frac{\sin (2u) + 2u}{2y} dy\\
\text{Now, we use the double angle formula } \sin (2u) \equiv 2 \sin (u) \cos (u)\\
=\int \frac{\arctan (p)}{y} dy + \int -\frac{2 \sin (u) \cos (u) + 2u}{2y} dy\\
=\int \frac{\arctan (p)}{y} dy + \int -\frac{\sin (u) \cos (u) + u}{y} dy\\
\text{Combine the integrals:}\\
=\int \frac{\arctan (p) – \sin (u) \cos (u) -u}{y} dy\\
\text{Sub back } p=\frac{x}{y} \text{ and } u= \arctan (\frac{x}{y})\\
=\int \frac{\arctan (\frac{x}{y}) – \sin (\arctan (\frac{x}{y})) \cos (\arctan (\frac{x}{y})) -\arctan (\frac{x}{y})}{y} dy\\
=\int \frac{- \sin (\arctan (\frac{x}{y})) \cos (\arctan (\frac{x}{y}))}{y} dy\\


\text{Now, use the inverse trigonometric identities:}\\
\sin (\arctan (x)) \equiv \frac{x}{\sqrt{1+x^2}}\\
\cos (\arctan (1)) \equiv \frac{1}{\sqrt{1+x^2}}\\


\text{This gives us:}\\
=\int \frac{- \frac{\frac{x}{y}}{\sqrt{1+(\frac{x}{y})^2}} \frac{1}{\sqrt{1+(\frac{x}{y})^2}} }{y} dy\\
=\int \frac{- \frac{\frac{x}{y}}{1+(\frac{x}{y})^2}}{y} dy\\
=\int \frac{- \frac{x}{y+ y(\frac{x}{y})^2}}{y} dy\\
=\int – \frac{x}{y^2+ y^2(\frac{x}{y})^2} dy\\
=\int – \frac{x}{y^2+ x^2} dy\\
\text{Since } x \text{ is a constant, we can factor it out:}\\
=-\frac{1}{x} \int \frac{1}{\frac{y^2}{x^2} +1} dy\\


\text{Sub } u=\frac{y}{x}\\
\text{Therefore, } dy =x \; du\\


\text{This gives us:}\\
=-\frac{1}{x} \int \frac{x}{u^2 +1} du\\
=-\int \frac{1}{u^2 +1} du\\
=- \arctan (u)\\
\text{Sub back } u = \frac{y}{x} \text{ :}\\
=- \arctan (\frac{y}{x})\\
\text{Add the constants of integration back in if needed:}\\
=- \arctan (\frac{y}{x}) + C_1x + C_2

And there we have it! In case you were curious as to how I was able to type out some very pretty mathematical symbols, I used LaTeX, which is an engine used to render mathematical symbols. If you are a mathematician and haven’t been exposed to it, you should check it out! It’s really cool!

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