This form pleases me

Using basic Definite integral rules we get the answer $\boxed{\frac{\pi ^{2}}{6\sqrt{3}}}$ =0.9497 And here lies the problem, the answer, going by the question should be 0. But the answer given here is $\left\lfloor 6\sqrt { 3 } \right\rfloor$ =10. @SandeepBhardwaj , you might want to change either the question or the answer.

$\begin{matrix} \int _{ \frac { -\pi }{ 4 } }^{ \frac { \pi }{ 4 } }{ \frac { x+\frac { \pi }{ 4 } }{ 2-\cos (2x) } dx } =\int _{ \frac { -\pi }{ 4 } }^{ \frac { \pi }{ 4 } }{ \frac { x }{ 2-\cos (2x) } dx } +\frac { \pi }{ 4 } \int _{ \frac { -\pi }{ 4 } }^{ \frac { \pi }{ 4 } }{ \frac { 1 }{ 2-\cos (2x) } dx } & & \color{#D61F06} // \int _{ \frac { -\pi }{ 4 } }^{ \frac { \pi }{ 4 } }{ \frac { x }{ 2-\cos (2x) } dx }=0 & & \color{#D61F06} \text{odd function} \end{matrix}$

$\text{let I}=\frac { \pi }{ 4 } \int _{ \frac { -\pi }{ 4 } }^{ \frac { \pi }{ 4 } }{ \frac { 1 }{ 2-\cos (2x) } dx }=\frac { \pi }{ 2 } \int _{0}^{ \frac { \pi }{ 4 } }{ \frac { 1 }{ 2-\cos (2x) } dx }=\frac { \pi }{ 4 } \int _{0}^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ 2-\cos (u) } du } \color{#D61F06}{\text{ // u=2x}}$

let t = $tan(\frac{x}{2})$ Weierstrass Substitution

$I=\frac { \pi }{ 4 } \int _{0}^{1}{ \frac { 1 }{ 2-\left(\frac{1-t^2}{1+t^2}\right ) } \left (\frac{2dt}{1+t^2}\right ) }=\frac {\pi}{2} \int _{0}^{1} {\frac { dt }{ 3t^2+1}}=\frac {\pi}{2\sqrt{3}} \int _{0}^{1} {\frac { \sqrt{3} dt }{ \left (\sqrt{3}t \right )^2+1}}=\frac {\pi}{2\sqrt{3}}\left . \tan^{-1}(\sqrt{3}t) \right |_{0}^1=\frac {\pi^2}{6\sqrt{3}}$

1st we use king. Then basic techniques of indefinite integration. Its comes out to be. .94