Using an Integral to Solve for a Variable a I am struggling to use the following equation:
$$
\int_0^a \sqrt{a^2-x^2}\,\,\text{sgn}(|x|-1)\, dx = 0
$$
where $a > 1$, to deduce that $a = \text{cosec}(\frac{\pi}{4} - \frac{\alpha}{2})$, where $\alpha$ satisfies $\alpha = \cos(\alpha)$.
I integrate the integrand, via
$$
\int_0^a \sqrt{a^2-x^2}\,\,\text{sgn}(|x|-1)\, dx = -\int_0^1 \sqrt{a^2-x^2}\, dx + \int_1^a \sqrt{a^2-x^2}\, dx 
$$
But once I calculate those integrals I cannot seem to get any closer to the answer.
Any help would be great.
 A: First, $a>1$ and $x\in[0,a]$, then $|x|=x\ge0$
$$
\renewcommand\sgn{\operatorname{sgn}}
\renewcommand\arcsec{\operatorname{arcsec}}
\begin{array}{ll}
0\!\!\!&=\int_0^a\sqrt{a^2-x^2}\sgn(x-1)\,\mathrm dx\\
&=\int_1^a\sqrt{a^2-x^2}\,\mathrm dx-\int_0^1\sqrt{a^2-x^2}\,\mathrm dx
\end{array}
$$
Then
$$\int_1^a\sqrt{a^2-x^2}\,\mathrm dx=\int_0^1\sqrt{a^2-x^2}\,\mathrm dx\tag{1}$$
But
$$\int_1^a\sqrt{a^2-x^2}\,\mathrm dx=\text{Area of }DAB\tag{2}$$
$$\int_0^1\sqrt{a^2-x^2}\,\mathrm dx=\text{Area of }ODBC\tag{3}$$

and area of $DAB+$ area of $ODBC=\dfrac14$area disc$(a)=\dfrac{\pi a^2}4$
Then $(1),(2),(3)$ yield
$$2\text{area of }DAB=\dfrac{\pi a^2}4\implies\text{ area of }DAB=\dfrac{\pi a^2}8=\int_1^a\sqrt{a^2-x^2}\,\mathrm dx\tag{★}$$
Moreover: Area of sector $OAB=\pi\theta= \pi\arccos\left(\frac1a\right)= \pi\arcsec a$
On the other hand, Area of sector $OAB=$ Area of triangle $ODB+$ Area of $DAB$.
Then,
$$\boxed{\displaystyle\pi\arcsec a=\dfrac12\sqrt{a^2-1}+\int_1^a\sqrt{a^2-x^2}\,\mathrm dx}\tag{★★}$$
From $($★$)$ and $($★★$)$ it follows that: $a$ is a solution of the equation
$$\boxed{\displaystyle\pi\arcsec a=\dfrac12\sqrt{a^2-1}+\frac{\pi a^2}{8}}$$
I transformed the initial problem into a simpler one (I hope!)
A: Your question is now: If $\alpha =2\sqrt{1-\frac{1}{a^{2}}}$ how can one
prove that $\cos \alpha =\alpha ?$
Your question in the title is :'' solve for a variable $a$ ''
 it means that: the problem is: solve for the variable $a$
the equation $\cos \alpha =\alpha ,$ where $a>1.$ So, lets go:
From $\frac{\alpha }{2}=\sqrt{1-\frac{1}{a^{2}}}$ then $\cos (\frac{\alpha }{%
2})=\frac{1}{a}$ (figure may help i will attach it below). But $\cos ^{2}(%
\frac{\alpha }{2})=\frac{1+\cos \alpha }{2}=\frac{1}{a^{2}}$ hence $\cos
\alpha =\frac{2}{a^{2}}-1.$ Therefore $\cos \alpha =\alpha $ if and only if $%
\frac{2}{a^{2}}-1=2\sqrt{1-\frac{1}{a^{2}}},$ or $\sqrt{1-\frac{1}{a^{2}}}=%
\frac{1}{a^{2}}-\frac{1}{2}.$ So solving this equation and taking only
solution $>1$ one obtains $a=\frac{\sqrt{2}}{\sqrt[4]{3}}\simeq
1.07>1.$

