it's Trigonometric Integral

The given integrand is as follows. The first steps are to simplify the denominator using trigonometric identities.

$I = \frac{8 \cot\left(\frac{\pi}{4} + x\right)}{\cos{4x} + 4\sin^2{x}\sin{2x} - 8\sin^2{x} - 4\sin{2x} + 7}$ $\implies I = \frac{8 \cot\left(\frac{\pi}{4} + x\right)}{1 - 2\sin^2{2x} + 4\sin^2{x}\sin{2x} - 8\sin^2{x} - 4\sin{2x} + 7}$ $\implies I = \frac{8 \cot\left(\frac{\pi}{4} + x\right)}{8 - 8\sin^2{x}- 2\sin^2{2x} -4\sin{2x}( 1-\sin^2{x}) }$ $\implies I = \frac{8 \cot\left(\frac{\pi}{4} + x\right)}{8\cos^2{x}- 2\sin^2{2x} -4\sin{2x}\cos^2{x} }$ $\implies I = \frac{8 \cot\left(\frac{\pi}{4} + x\right)}{8\cos^2{x}- 8\sin^2{x}\cos^2{x} -8\sin{x}\cos^3{x} }$ $\implies I = \frac{8 \cot\left(\frac{\pi}{4} + x\right)}{\cos^2{x}(8- 8\sin^2{x}) -8\sin{x}\cos^3{x} }$ $\implies I = \frac{8 \cot\left(\frac{\pi}{4} + x\right)}{8\cos^4{x} -8\sin{x}\cos^3{x} }$ $\implies I = \frac{8 \cot\left(\frac{\pi}{4} + x\right)}{8\cos^3{x} (\cos{x} -\sin{x})}$ $\implies I = \frac{8 \cot\left(\frac{\pi}{4} + x\right)}{8\sqrt{2}\cos^3{x} \left(\frac{\cos{x}}{\sqrt{2}} -\frac{\sin{x}}{\sqrt{2}}\right)}$ $\implies I = \frac{8 \cot\left(\frac{\pi}{4} + x\right)}{8\sqrt{2}\cos^3{x}\cos\left(\frac{\pi}{4} + x\right)}$ $\implies I = \frac{1}{\sqrt{2}\cos^3{x}\sin\left(\frac{\pi}{4} + x\right)}$ $\implies I = \frac{\sec^3{x}}{\cos{x}+\sin{x}}$ $\implies I = \frac{\sec^4{x}}{1+\tan{x}}$ $\implies I = \frac{(1+\tan^2{x})\sec^2{x}}{1+\tan{x}}$

Now:

$A=\int_{0}^{\pi/4} I \ dx = \int_{0}^{\pi/4}\frac{(1+\tan^2{x})\sec^2{x}}{1+\tan{x}} \ dx$

Taking $z = \tan{x}$ transforms the integral to:

$A=\int_{0}^{1} \frac{1+z^2}{1+z} \ dz = \int_{0}^{1} \frac{1}{1+z} \ dz + \int_{0}^{1} \frac{z^2}{1+z} \ dz$ $A=\int_{0}^{1} \frac{1}{1+z} \ dz + \int_{0}^{1} \frac{z^2}{1+z} \ dz = \int_{0}^{1} \frac{1}{1+z} \ dz + \int_{0}^{1} \frac{z^2-1 + 1}{1+z} \ dz$ $A=2\int_{0}^{1} \frac{1}{1+z} \ dz + \int_{0}^{1} (z-1) \ dz$

From this point, it is easy to show that the integral evaluates to:

$A=\int_{0}^{\pi/4} \frac{8 \cot\left(\frac{\pi}{4} + x\right)}{\cos{4x} + 4\sin^2{x}\sin{2x} - 8\sin^2{x} - 4\sin{2x} + 7} \ dx =\boxed{2\ln{2} - \frac{1}{2}}$