If $$\int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2}\cos x\,dx}{(1+e^{\sin x})(1+\sin^4 x)} = \alpha\pi + \beta\log_e(3+2\sqrt{2})$$, where $$\alpha, \beta$$ are integers, then $$\alpha^2 + \beta^2$$ equals:

Use the property $$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$.

Here $$a+b = 0$$, so replace $$x$$ with $$-x$$:

$$I = \int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2} \cos(-x)}{(1+e^{\sin(-x)})(1+\sin^4(-x))} dx = \int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2} \cos x}{(1+e^{-\sin x})(1+\sin^4 x)} dx$$

$$2I = \int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2} \cos x}{1+\sin^4 x} \left( \frac{1}{1+e^{\sin x}} + \frac{e^{\sin x}}{1+e^{\sin x}} \right) dx = \int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2} \cos x}{1+\sin^4 x} dx$$

Since the integrand is even: $$I = \int_{0}^{\pi/2} \frac{8\sqrt{2} \cos x}{1+\sin^4 x} dx$$.

Let $$\sin x = t \implies \cos x dx = dt$$. Limits: $$0$$ to $$1$$.

$$I = 8\sqrt{2} \int_{0}^{1} \frac{dt}{1+t^4}$$

Using the standard integral form for $$1/(1+t^4)$$:

$$I = 4\sqrt{2} \int_{0}^{1} \frac{(t^2+1)-(t^2-1)}{t^4+1} dt = 4\sqrt{2} \left[ \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right) + \frac{1}{2\sqrt{2}}\log \left|\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right| \right]_0^1$$

After applying limits: $$I = 2\pi + 2 \log_e(3+2\sqrt{2})$$.

Comparing with $$\alpha\pi + \beta \log_e(3+2\sqrt{2})$$, we get $$\alpha=2, \beta=2$$.

$$\alpha^2 + \beta^2 = 4 + 4 = 8$$.