If ∫ − π / 2 π / 2 8 2 cos x d x 1 + e sin x 1 + sin 4 x = α π + β log e 3 + 2 2 , where α , β are integers,…

Given: $I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{8\sqrt{2}\cos x}{\left(1+e^{\sin x}\right)\left(1+{\sin }^{4}x\right)}dx ...\left(i\right)$

Applying ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$ rule we get,

$\Rightarrow I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{8\sqrt{2}\cos x\left(e^{\sin x}\right)}{\left(1+e^{\sin x}\right)\left(1+{\sin }^{4}x\right)}dx ...\left(ii\right)$

Adding $\left(i\right) \text{and} \left(ii\right)$

$\Rightarrow 2I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{8\sqrt{2}\cos x}{1+{\sin }^{4}x}dx$

$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{8\sqrt{2}\cos x}{1+{\sin }^{4}x}dx$

Putting, $\sin x=t$

$\Rightarrow I=\underset{0}{\overset{1}{\int }}\frac{8\sqrt{2}}{1+t^4}dx$

$\Rightarrow I=4\sqrt{2}\underset{0}{\overset{1}{\int }}\left(\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}-\frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}\right)dt$

$\Rightarrow I=4\sqrt{2}\underset{0}{\overset{1}{\int }}\left(\frac{1+\frac{1}{t^2}}{{\left(t-\frac{1}{t}\right)}^2+2}-\frac{\left(1-\frac{1}{t^2}\right)}{{\left(t+\frac{1}{t}\right)}^2-2}\right)dt$

Let $t-\frac{1}{t}=z$ and $t+\frac{1}{t}=k$

$\Rightarrow I=4\sqrt{2}\left[\underset{-\infty }{\overset{0}{\int }}\frac{dz}{z^2+2}-\underset{\infty }{\overset{2}{\int }}\frac{dk}{k^2-2}\right]$

$\Rightarrow I=4\sqrt{2}{\left[\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{z}{\sqrt{2}}\right]}_{-\infty }^0-{\left[\frac{1}{2\sqrt{2}}\ln \left(\frac{k-\sqrt{2}}{k+\sqrt{2}}\right)\right]}_{\infty }^2$

$\Rightarrow I=4\sqrt{2}\left[\frac{\pi }{2\sqrt{2}}-\frac{1}{2\sqrt{2}}\left[\ln \frac{2-\sqrt{2}}{2+\sqrt{2}}\right]\right]$

$\Rightarrow I=2\pi +2\log \left(3+2\sqrt{2}\right)$

Comparing this value with $\left[\alpha \pi +\beta {\log }_e\left(3+2\sqrt{2}\right)\right]$

$\Rightarrow \alpha =2$ and $\beta =2$

$\Rightarrow {\alpha }^2+{\beta }^2=8$