If the solution of the equation $$\log_{\cos x} \cot x + 4\log_{\sin x} \tan x = 1$$, $$x \in (0, \frac{\pi}{2})$$ is $$\sin^{-1}\frac{\alpha + \sqrt{\beta}}{2}$$, where $$\alpha, \beta$$ are integers, then $$\alpha + \beta$$ is equal to:

Given,

$$\log_{\cos x}(\cot x) + 4\log_{\sin x}(\tan x) = 1$$

Using,

$$\cot x = \frac{\cos x}{\sin x}$$

and

$$\tan x = \frac{\sin x}{\cos x}$$

we get

$$\log_{\cos x}\left(\frac{\cos x}{\sin x}\right) + 4\log_{\sin x}\left(\frac{\sin x}{\cos x}\right) = 1$$

Using the logarithmic property

$$\log_a\left(\frac{m}{n}\right)=\log_a m-\log_a n$$

$$\log_{\cos x}\cos x - \log_{\cos x}\sin x + 4\left(\log_{\sin x}\sin x - \log_{\sin x}\cos x\right)=1$$

Since,

$$\log_{\cos x}\cos x = 1$$

and

$$\log_{\sin x}\sin x = 1$$

we obtain

$$1-\log_{\cos x}\sin x + 4\left(1-\log_{\sin x}\cos x\right)=1$$

Let

$$t=\log_{\cos x}\sin x$$

Using the reciprocal property of logarithms,

$$\log_{\sin x}\cos x = \frac{1}{t}$$

Substituting,

$$1-t+4\left(1-\frac{1}{t}\right)=1$$

$$1-t+4-\frac{4}{t}=1$$

$$5-t-\frac{4}{t}=1$$

$$4-t-\frac{4}{t}=0$$

Multiplying throughout by $$t$$,

$$4t-t^2-4=0$$

$$t^2-4t+4=0$$

$$(t-2)^2=0$$

$$t=2$$

Therefore,

$$\log_{\cos x}\sin x = 2$$

Converting to exponential form,

$$\sin x = \cos^2 x$$

Using,

$$\cos^2 x = 1-\sin^2 x$$

we get

$$\sin x = 1-\sin^2 x$$

$$\sin^2 x + \sin x -1 =0$$

Using the quadratic formula,

$$\sin x = \frac{-1\pm\sqrt{1+4}}{2}$$

$$\sin x = \frac{-1\pm\sqrt{5}}{2}$$

Therefore,

$$x=\sin^{-1}\left(\frac{-1+\sqrt5}{2}\right)$$

Comparing with

$$x=\sin^{-1}\left(\frac{\alpha+\sqrt\beta}{2}\right)$$

we get

$$\alpha=-1,\qquad \beta=5$$

Hence,

$$\alpha+\beta = -1+5 = 4$$

Therefore, the required answer is

$$\boxed{4}$$