If ∫ _ - π / 2 ^ π / 2 8 2 cosxdx (1 + e ^ x )(1 + ^ 4 x) = απ + β...

Given: $I = \mathop{∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\dfrac{8\sqrt{2}\cos x}{\left(1 + {e}^{\sin x}\right)\left(1 + \sin^{4}x\right)}dx . . . \left(i\right)$

Applying ${∫}_{a}^{b}f\left(x\right) d x = {∫}_{a}^{b}f\left(a + b - x\right) d x$ rule we get,

$⇒ I = \mathop{∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\dfrac{8\sqrt{2}\cos x\left({e}^{\sin x}\right)}{\left(1 + {e}^{\sin x}\right)\left(1 + \sin^{4}x\right)}dx . . . \left(ii\right)$

Adding \left(i\right)\text{and}\left(ii\right)$</p><p>$⇒ 2I = \mathop{∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\dfrac{8\sqrt{2}\cos x}{1 + \sin^{4}x}dx$</p><p>$⇒ I = \mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{8\sqrt{2}\cos x}{1 + \sin^{4}x}dx$</p><p>Putting,$\sin x = t$</p><p>$⇒ I = \mathop{∫}_{0}^{1}\dfrac{8\sqrt{2}}{1 + {t}^{4}}dx$</p><p>$⇒ I = 4\sqrt{2}\mathop{∫}_{0}^{1}\left(\dfrac{1 + \dfrac{1}{{t}^{2}}}{{t}^{2} + \dfrac{1}{{t}^{2}}} - \dfrac{1 - \dfrac{1}{{t}^{2}}}{{t}^{2} + \dfrac{1}{{t}^{2}}}\right)dt$</p><p>$⇒ I = 4\sqrt{2}\mathop{∫}_{0}^{1}\left(\dfrac{1 + \dfrac{1}{{t}^{2}}}{{\left(t - \dfrac{1}{t}\right)}^{2} + 2} - \dfrac{\left(1 - \dfrac{1}{{t}^{2}}\right)}{{\left(t + \dfrac{1}{t}\right)}^{2} - 2}\right)dt$</p><p>Let$t - \dfrac{1}{t} = z$and$t + \dfrac{1}{t} = k$</p><p>$⇒ I = 4\sqrt{2}\left[\mathop{∫}_{- \infty}^{0}\dfrac{dz}{{z}^{2} + 2} - \mathop{∫}_{\infty}^{2}\dfrac{dk}{{k}^{2} - 2}\right]$</p><p>$⇒ I = 4\sqrt{2}{\left[\dfrac{1}{\sqrt{2}}\tan^{- 1}\dfrac{z}{\sqrt{2}}\right]}_{- \infty}^{0} - {\left[\dfrac{1}{2\sqrt{2}}\ln \left(\dfrac{k - \sqrt{2}}{k + \sqrt{2}}\right)\right]}_{\infty}^{2}$</p><p>$⇒ I = 4\sqrt{2}\left[\dfrac{\pi}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}\left[ln\dfrac{2 - \sqrt{2}}{2 + \sqrt{2}}\right]\right]$</p><p>$⇒ I = 2\pi + 2\log \left(3 + 2\sqrt{2}\right)$</p><p>Comparing this value with \left[\alpha\pi + \beta{\log}_{e}\left(3 + 2\sqrt{2}\right)\right]$

$⇒ \alpha = 2$ and $\beta = 2$

$⇒ {\alpha}^{2} + {\beta}^{2} = 8$