How do you find the integral of $\int_a^bdx\int_a^b \frac{y^2}{(x^2+y^2)^\frac{3}{2}}dy$ $$\int_a^bdx\int_a^b \frac{y^2}{(x^2+y^2)^\frac{3}{2}}dy$$
I've tried using the trigonometric substitution $x=\cos t, y=\sin t$ but it didn't work out
 A: Interchanging the order of integration, we have
\begin{align}
I&=\int_a^b dyy^2\int_a^b\frac{dx}{(x^2+y^2)^{\frac{3}{2}}}
\end{align}
Now let $x=y\tan\phi$, then $dx=y\sec^2\phi d\phi$. Plugging this in, we have
\begin{align}
I &= \int_a^b dyy^2\int_{\arctan\left(\frac{a}{y}\right)}^{\arctan\left(\frac{b}{y}\right)}d\phi\frac{y\sec^2\phi}{(y^2\tan^2\phi + y^2)^{\frac{3}{2}}}\\
&=\int_a^b dyy^2\int_{\arctan\left(\frac{a}{y}\right)}^{\arctan\left(\frac{b}{y}\right)}d\phi\frac{y\sec^2\phi}{y^3(\tan^2\phi + 1)^{\frac{3}{2}}}\\
&=\int_a^b dy\int_{\arctan\left(\frac{a}{y}\right)}^{\arctan\left(\frac{b}{y}\right)}d\phi\frac{\sec^2\phi}{\sec^3\phi}\\
&=\int_a^b dy\int_{\arctan\left(\frac{a}{y}\right)}^{\arctan\left(\frac{b}{y}\right)}d\phi\cos\phi\\
&=\int_a^b dy\left[\sin\left(\arctan\left(\frac{b}{y}\right)\right)-\sin\left(\arctan\left(\frac{a}{y}\right)\right)\right]\\
&=\int_a^b dy\left[\frac{b}{\sqrt{b^2+y^2}}-\frac{a}{\sqrt{a^2+y^2}}\right]\\
&=\left[b\log(\sqrt{a^2+y^2}+y)-a\log(\sqrt{b^2+y^2}+y)\right]_{y=a}^{y=b}\\
&=b\log(\sqrt{a^2+b^2}+b) - a\log((\sqrt{2}+1)b)-b\log((\sqrt{2}+1)a) + a\log(\sqrt{a^2+b^2}+a)\\
&=a\log\left(\frac{\sqrt{a^2+b^2}+a}{(\sqrt{2}+1)b}\right) + b\log\left(\frac{\sqrt{a^2+b^2}+b^2}{(\sqrt{2}+1)a}\right)
\end{align}
This presumes that the substitution $x=y\tan\phi$ is invertible over the interval $(a,b)$ - otherwise, you'll have to split up the integral, which changes things a little. The gist of the calculation should stay the same, though.
A: By symmetry,
$$ I=\iint_{(a,b)^2}\frac{y^2}{(x^2+y^2)^{3/2}}\,d\mu = \iint_{(a,b)^2}\frac{x^2}{(x^2+y^2)^{3/2}}\,d\mu = \frac{1}{2}\iint_{(a,b)^2}\frac{d\mu}{\sqrt{x^2+y^2}} \tag{1} $$
hence, assuming $a,b>0$:
$$ I = \frac{1}{2}\int_{a}^{b}\log\left(\frac{b+\sqrt{b^2+y^2}}{a+\sqrt{a^2+y^2}}\right)\,dy \tag{2}$$
and:
$$\boxed{ I = (a+b)\log(1+\sqrt{2})+a\log\left(\frac{a}{b+\sqrt{a^2+b^2}}\right)+b\log\left(\frac{b}{a+\sqrt{a^2+b^2}}\right).}\tag{3} $$
A: Let $y=x \tan(u)$ and $dy=x \sec^2(u)du$:
$$\int\frac{y^2}{(x^2+y^2)^\frac{3}{2}}dy=\int\frac{x^3tg^2(u)\sec^2(u)du}{x^3\sec^3(u)}$$
$$\int \sin(u)\tan(u)du = \int \sec(u)du - \int \cos(u)du$$
