Evaluate: $$I=\iint_{[0,1]^2}\frac{dxdy}{(1+x^2+y^2)^{3/2}}.$$
Attempt. Working on substitution $x=\sqrt{1+y^2}\,\sinh u$ we get
$$I=\int\limits_{0}^{1}\frac{x}{(1+y^2)\sqrt{x^2+y^2+1}}\Bigg|_{0}^{1}\,dy=\int\limits_{0}^{1}\frac{dy}{(1+y^2)\sqrt{2+y^2}}$$ and then working on $y=\sqrt{2}\,\tan u$ we get:
$$I=\arctan\left(\frac{y}{\sqrt{y^2+2}}\right)\bigg|_{0}^{1}=\arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}.$$
On the other hand, polar coordinates give:
$$I=\int\limits_{0}^{\pi/4}\int\limits_{0}^{\frac{1}{\cos \phi}}\frac{r}{(1+r^2)^{3/2}}dr d\phi+\int\limits_{\pi/4}^{\pi/2}\int\limits_{0}^{\frac{1}{\sin \phi}}\frac{r}{(1+r^2)^{3/2}}dr d\phi$$
which gets us to:
$$I=\int\limits_{0}^{\pi/4}\left(1-\frac{\cos\phi}{\sqrt{1+\cos^2\phi}}\right)d\phi+\int\limits_{\pi/4}^{\pi/2}\left(1-\frac{\sin\phi}{\sqrt{1+\sin^2\phi}}\right)d\phi$$
and the computations seem difficult to handle.
Is there an easier way to approach the calculation of this integral?
Thank you in advance.
Best Answer
First, note that the two integrals in the last line are equal. Just substitute $\phi=\frac{\pi}{2}-\varphi$ and see. Therefore
$$I=2\int_{0}^{\frac{\pi}{4}}\left(1-\frac{\cos\phi}{\sqrt{2-\sin^{2}\phi}}\right){\rm d\phi}=\left[\begin{matrix}\sin\phi=\sqrt{2}\sin\theta\\\cos\phi{\rm d}\phi=\sqrt{2}\cos\theta{\rm d}\theta\end{matrix}\right]=$$
$$=\frac{\pi}{2}-2\int_{0}^{\frac{\pi}{6}}\frac{\cos\theta}{\sqrt{1-\sin^{2}\theta}}{\rm d}\theta=\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6}$$