If $\int\limits_{-\pi / 2}^{\pi / 2} \frac{8 \sqrt{2} \cos x \mathrm{~d} x}{\left(1+\mathrm{e}^{\sin x}\right)\left(1+\sin ^4 x\right)}=\alpha \pi+\beta \log _{\mathrm{e}}(3+2 \sqrt{2})$, where $\alpha, \beta$ are integers, then $\alpha^2+\beta^2$ equals :

$$ I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{\left(1+e^{\sin x}\right)\left(1+\sin ^4 x\right)} d x $$ ............(1) <br/><br/>Apply king's rule <br/><br/>$$ I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x\left(e^{\sin x}\right)}{\left(1+e^{\sin x}\right)\left(1+\sin ^4 x\right)} d x $$ ..........(2) <br/><br/>Adding (1) and (2), we get <br/><br/>$$ \begin{aligned} & 2 I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} d x \\\\ & I=\int\limits_0^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} d x, \end{aligned} $$ <br/><br/>Let $\sin x=t$ <br/><br/>$$ \begin{aligned} & I=8 \sqrt{2} \int\limits_0^1 \frac{d t}{1+t^4} \\\\ & =4 \sqrt{2} \int\limits_0^1 \frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}} d t \end{aligned} $$ <br/><br/>$\begin{aligned} & =4 \sqrt{2} \int\limits_0^1 \frac{\left(1+\frac{1}{t^2}\right) d t}{\left(t-\frac{1}{t^2}\right)^2+2}-4 \sqrt{2} \int\limits_0^1 \frac{\left(1-\frac{1}{t^2}\right) d t}{\left(t+\frac{1}{t^2}\right)^2-2} \\\\ & =4 \sqrt{2} \cdot \frac{1}{\sqrt{2}}\left(\left.\tan ^{-1} \frac{t-\frac{1}{t}}{\sqrt{2}}\right|_0 ^1-4 \sqrt{2} \cdot \frac{1}{2 \sqrt{2}}\left[\log \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|\right]_0^1\right. \\\\ & =2 \pi-2 \log \left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right|\end{aligned}$ <br/><br/>$\begin{aligned} & =2 \pi+2 \log (3+2 \sqrt{2})=\alpha \pi+\beta \log _e(3+2 \sqrt{2}) \\\\ & \Rightarrow \alpha=2, \beta=2 \\\\ & \Rightarrow \alpha^2+\beta^2=8\end{aligned}$