The value of the integral ∫ _ - (π)/(4) ^ (π)/(4) (x + (π)/(4))/(2...

Let,

$I = {∫}_{\dfrac{- \pi}{4}}^{\dfrac{\pi}{4}}\dfrac{x + \dfrac{\pi}{4}}{2 - \cos 2x}dx \ldots \left(1\right)$

Now replacing $x \to - x$ we get,

$⇒ I = {∫}_{\dfrac{- \pi}{4}}^{\dfrac{\pi}{4}}\dfrac{- x + \dfrac{\pi}{4}}{2 - \cos 2x}dx \ldots \left(2\right)$

Now both the equation we get,

$⇒ 2I = {∫}_{\dfrac{- \pi}{4}}^{\dfrac{\pi}{4}}\dfrac{\dfrac{\pi}{2}}{2 - \cos 2x}dx$

$⇒ I = \dfrac{\pi}{4} \cdot 2{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{dx}{2 - \cos 2x}dx\left\{\text{as}\cos 2x\text{is even function}\right\}$

$⇒ I = \dfrac{\pi}{4} \cdot 2{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{\left(1 + \tan^{2}x\right)dx}{2\left(1 + \tan^{2}x\right) - \left(1 - \tan^{2}x\right)}$

$⇒ I = \dfrac{\pi}{4} \cdot 2{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{\sec^{2}xdx}{3\tan^{2}x + 1}$

Now let $\tan x = t ⇒ \sec^{2}xdx = dt$

$⇒ I = \dfrac{\pi}{2}{∫}_{0}^{1}\dfrac{dt}{3{t}^{2} + 1}$

$⇒ I = \dfrac{\pi}{2\sqrt{3}}\tan^{- 1}\sqrt{3}$

$⇒ I = \dfrac{{\pi}^{2}}{6\sqrt{3}}$