My third integration problem

Because $\cos 4x=2\cos^2 2x-1$ we can get $3-4\cos 2x+ \cos 4x=2-4\cos 2x+\cos^2 2x=2(\cos 2x -1)^2$ $3+4\cos 2x+ \cos 4x=2+4\cos 2x+\cos^2 2x=2(\cos 2x +1)^2$ because $\cos 2x=2\cos^2 x-1=1-2\sin^2x$ so $3-4\cos 2x+ \cos 4x=2(\cos 2x -1)^2=2(-2\sin^2 x)^2=4\sin^4 x$ $3+4\cos 2x+ \cos 4x=2(\cos 2x +1)^2=2(2\cos^2 x)^2=4\cos^4 x$ so the integral change to $\int^{\frac{\pi}{4}}_0 \frac{4\sin^4 x}{4\cos^4 x} dx = \int^{\frac{\pi}{4}}_0 (\tan^2 x) (\tan^2 x) dx =\int^{\frac{\pi}{4}}_0 (\tan^2 x)\left(1-\frac{1}{\cos^2 x}\right) dx$ $=\int^{\frac{\pi}{4}}_0 (\tan^2 x) dx-\int^{\frac{\pi}{4}}_0 \frac{(\tan^2 x)}{\cos^2 x} dx =\int^{\frac{\pi}{4}}_01-\frac{1}{\cos^2 x} dx-\int^{\frac{\pi}{4}}_0(\tan^2 x) d(\tan x)$ $= \left. x-\tan x \right|^{\frac{\pi}{4}}_0+\left.\frac{\tan^3x}{3} \right|^{\frac{\pi}{4}}_0=\frac{\pi}{4}-1+\frac{1}{3}=\frac{\pi}{4}-\frac{2}{3}$ so we can get $a=4, b=2,c=3$ and $a+b+c=9$