Correct answer: 8
\(\mathrm I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{\left(1+e^{\sin x}\right)\left(1+\sin ^{4} x\right)} d x \quad ....(1)\)
Apply king
\(\mathrm I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x\left(e^{\sin x}\right)}{\left(1+e^{\sin x}\right)\left(1+\sin ^{4} x\right)} d x\quad....(2)\)
Adding (1) & (2),
\(2 \mathrm I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^{4} x} d x\)
\(\mathrm I=\int\limits_{0}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^{4} x} d x\)
\(\sin x=t\)
\(\mathrm I=\int\limits_{0}^{1} \frac{8 \sqrt{2}}{1+t^{4}} d x\)
\(\mathrm I=4 \sqrt{2} \int\limits_{0}^{1}\left(\frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}}-\frac{1-\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}}\right) d t\)
\(\mathrm I=4 \sqrt{2} \int\limits_{0}^{1} \frac{\left(1+\frac{1}{t^{2}}\right)}{\left(t-\frac{1}{t}\right)^{2}+2}-\frac{\left(1-\frac{1}{t^{2}}\right)}{\left(t+\frac{1}{t}\right)^{2}-2} d t\)
Let \(t-\frac{1}{t}=z \ \&\ t+\frac{1}{t}=k\)
\(=4 \sqrt{2}\left[\int\limits_{-\infty}^0 \frac{d z}{z^2+2}-\int\limits_{\infty}^2 \frac{d k}{k^2-2}\right] \)
\(=4 \sqrt{2}\left[\frac{1}{\sqrt{2}} \tan ^{-1} \frac{z}{\sqrt{2}}\right]_{-\infty}^0-\left[\frac{1}{2 \sqrt{2}} \ln \left(\frac{k-\sqrt{2}}{k+\sqrt{2}}\right)\right]_{\infty}^2\)
\( =4 \sqrt{2}\left[\frac{\pi}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\left[\ln \frac{2-\sqrt{2}}{2+\sqrt{2}}\right]\right]\)
\(=2 \pi+2 \ln (3+2 \sqrt{2})\)
\(\alpha=2\)
\( \beta=2\)
\(\therefore \alpha^2 + \beta^2 = 8\)
