Paneer Definite Masala 4

Let $\displaystyle I = \int_{-\pi/4}^{\pi/4} \dfrac { 2\pi - \sin{2\pi x}}{2 - \cos{2x}} dx$

Replacing $x$ by $-\pi/4 + \pi/4 -x = -x$ we get:

$\displaystyle I = \int_{-\pi/4}^{\pi/4} \dfrac { 2\pi + \sin{2\pi x}}{2 - \cos{2x}} dx$

Adding both expressions:

$\displaystyle 2I = \int_{-\pi/4}^{\pi/4} \dfrac { 4\pi}{2 - \cos{2x}} dx$

$\displaystyle \dfrac{I}{4\pi} = \int_{0}^{\pi/4} \dfrac {1}{2 - \cos{2x}} dx$

Replacing $2x$ by $x$

$\displaystyle \dfrac{I}{2\pi} = \int_{0}^{\pi/2} \dfrac {1}{2 - \cos{x}} dx$

Now as $\displaystyle \cos{x} = \dfrac{1 + \tan^2{\dfrac x2} }{ 1 + \tan^2{\dfrac x2} }$ Putting in above expression and using $\displaystyle 1 + \tan^2{\dfrac x2} = \sec^2{\dfrac x2}$ we get

$\displaystyle \dfrac{I}{2\pi} = \int_{0}^{\pi/2} \dfrac {\sec^2{\dfrac x2}}{1 + 3\tan^2{\dfrac x2}} dx$

Now putting $\tan{\dfrac x2} = y$

$\displaystyle \dfrac{I}{4\pi} = \int_{0}^{1} \dfrac {1}{1 + 3t^2} dt$

Integrating we get:

$\displaystyle I = \dfrac{4 \pi ^2 }{3 \sqrt{3}}$

$A =4 , B=2 , C=D =3 ,E=0$

$A+B+C+D+E = \boxed{12}$