Connect the curve 4

Let the Green shaded Region's Area be A
$A \text{ = Area(1) - Area(2) + Area(3) - Area(4)}$ (There is a -ve sign before Area(4) as it is negative but we need to add it.)
We can rewrite this as
$A = \displaystyle \int_{0}^{\pi} {cos(cos{x})dx} - \int_{0}^{\pi} {sin(sin(x))} + \int_{\pi}^{2\pi}{cos(cos(x))dx} \\ \displaystyle - \int_{\pi}^{2\pi}{sin(sin{x})dx}$

Changing limits of the integrals we get
$A = \displaystyle \int_{0}^{\pi} {cos(cos{x})dx} - \int_{0}^{\pi} {sin(sin(x))} + \int_{0}^{\pi}{cos(cos(x))dx} \\ \displaystyle + \int_{0}^{\pi}{sin(sin{x})dx}$

Simplifying we get
$A = \displaystyle 2\int_{0}^{\pi}{cos(cos{x})dx}$

Hence
$A = \displaystyle 2\pi J_{\alpha} (n)$

I used the Residue theorem.

The area is $\displaystyle A = \int_{0}^{2\pi} (\cos(\cos(x)) - \sin(\sin(x))) d x$ . . . (i)

Using the property $\displaystyle \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ ,we get

$\displaystyle A=\int_{0}^{2\pi} (\cos(\cos(x))+\sin(\sin(x)))dx$ . . . (ii).

Adding (i) and (ii), we get $\displaystyle A = \int_{0}^{2\pi} \cos(\cos(x)) dx$

Now take $\displaystyle z=e^{ix}$ and $\displaystyle \frac{dz}{iz} = dx$

$\displaystyle A=\oint \cos(\frac{z+(1/z)}{2}) \frac{dz}{iz} = \frac{1}{i} \oint \frac{1}{z} \cos(\frac{z^2+1}{2z}) dz$

The contour is the unit circle centred at the origin and is traversed in the counterclockwise (anticlockwise) direction.

There is only one pole of the integrand inside this contour, namely 0.

Hence we find the residue of the integrand $\displaystyle\frac{1}{z} \cos(\frac{z^2+1}{2z})$ at $\displaystyle z=0$ .

We know that the residue at a point $\displaystyle c$ is the $\displaystyle (z-c)^{-1}$ coefficient in the laurent series about that point

Writing down the laurent series of the integrand about 0, we find the $\displaystyle z^{-1}$ coefficient (i.e. the residue) to be

$\displaystyle \sum_{m=0}^{\infty} \frac{(-1)^m {2m \choose m}}{2^{2m} (2m)!}$

$\displaystyle=\sum_{m=0}^{\infty} \frac{(-1)^m}{2^{2m} (m!)(m!)}$

$\displaystyle=\sum_{m=0}^{\infty} \frac{(-1)^m}{m! \Gamma(m+0+1)} (\frac{1}{2})^{2m+0}$

$\displaystyle=J_{0} (1)$

Hence by the residue theorem,

$\displaystyle A = \frac{1}{i} 2\pi i(\sum_{z_0 \in P} \mathrm{Res}_{z=z_0}(\frac{1}{z} \cos(\frac{z^2+1}{2z})))$ where $\displaystyle P$ is the set of all poles inside the contour

$\displaystyle A = 2\pi \mathrm{Res}_{z=0} (\frac{1}{z} \cos(\frac{z^2+1}{2z})) = \boxed{2 \pi J_{0}(1)}$

So finally we have $\displaystyle A=2, B=1, \alpha = 0, n=1$

$\displaystyle A+B+\alpha+n=\boxed{4}$