<s>[INST] Computing $I=\iint_D\sqrt{\frac{1-x^2-y^2}{1+x^2+y^2}}dxdy$ 
Compute $$I=\iint_D\sqrt{\frac{1-x^2-y^2}{1+x^2+y^2}}dxdy,$$ where $D=\{(x,y)\in\Bbb R^2\mid x^2+y^2\le 1, x\ge 0, y\ge 0\}.$
Source: Berman, task 3537

My attempt:
I switched to polar coordinates $$\begin{cases}x=r\cos\varphi\\y=r\sin\varphi\end{cases}, 0\le\varphi\le \pi/2,0\le r\le 1$$
$$\begin{aligned}I&=\int_0^{\pi/2}\int_0^1\sqrt{\frac{1-r^2}{1+r^2}}rdrd\varphi\\&=\frac\pi2\int_0^1\sqrt{\frac{1-r^2}{1+r^2}}rdr\\&=\begin{bmatrix}t&=r^2\\dt&=2rdr\end{bmatrix}\\&=\frac\pi4\int_0^1\sqrt{\frac{1-t}{1+t}}dt\\&=\begin{bmatrix}s=\sqrt{\frac{1-t}{1+t}}\implies(1+t)s^2=1-t\implies s^2+ts^2=1-t\implies(1+s^2)t=1-s^2\implies t=\frac{1-s^2}{1+s^2}\\\implies dt=-\frac{4sds}{(1+s^2)^2}\end{bmatrix}\\&=-\frac\pi4\int_1^0\frac{4s}{(1+s^2)^2}ds\\&=\frac\pi2\int_0^1\frac{2sds}{{(1+s^2)}^2}\\&=\begin{bmatrix}v=1+s^2\\dv=2sds\end{bmatrix}\\&=\frac\pi2\int_1^2\frac{dv}{v^2}\\&=-\frac\pi2\frac1v\Big|_1^2\\&=\frac\pi4\end{aligned}$$
The solution in the book is $\frac{\pi(\pi-2)}8.$ What did I do wrong?
 [/INST] Under the change of variables
$$t \mapsto \frac{1 - s^2}{1 + s^2} \quad \left(\textrm{that is, } s = \sqrt{\frac{1 - t}{1 + t}}\right), \qquad dt = -\frac{4 s\,ds}{(1 + s^2)^2},$$
the integral transforms as
$$\int_0^1 \sqrt{\frac{1 - t}{1 + t}} \,dt = \int_1^0 s \cdot -\frac{4 s\,ds}{(1 + s^2)^2} = 4 \int_0^1 \frac{s^2 \,ds}{(1 + s^2)^2}.$$
In your solution, the numerator in the integral in $s$ only contains one factor of $s$. The correctly transformed integral does lead to the text's solution, $\frac{(\pi - 2) \pi}{8}$.
Remark The correctly transformed integral in $s$ can be evaluated efficiently via a standard trigonometric substitution.
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