TITLE: Proving $\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx=\frac{π}{2}-1$
QUESTION [5 upvotes]: Proving $$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx= \frac{π}{2}-1$$
My attempt is: 
I assumed the $1-x=u$
$du =-dx$
$$\int_0^1 \sqrt{\frac{u}{2+u}}\,(-du)$$

REPLY [3 votes]: $$
\begin{align}
\int_0^1\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=2\int_{1/2}^1\sqrt{\frac{1-x}{x}}\,\mathrm{d}x\tag{1}\\
&=4\int_{1/\sqrt2}^1\sqrt{1-x^2}\,\mathrm{d}x\tag{2}\\
&=4\int_0^{\pi/4}\sin^2(x)\,\mathrm{d}x\tag{3}\\
&=2\int_0^{\pi/4}1-\cos(2x)\,\mathrm{d}x\tag{4}\\
&=\left[2x-\sin(2x)\vphantom{\int}\right]_0^{\pi/4}\tag{5}\\[6pt]
&=\frac\pi2-1\tag{6}
\end{align}
$$
Explanation:
$(1)$: $x\mapsto2x-1$
$(2)$: $x\mapsto x^2$
$(3)$: $x\mapsto\cos(x)$
$(4)$: $\cos(2x)=1-2\sin^2(x)$
$(5)$: integrate
$(6)$: evaluate