$$\int_0^{\frac{\pi}{4}}(\cos{x} - 2\sin{x}\sin(2x))dx=\int_0^{\frac{\pi}{4}}(\cos{x} - 4(1-\cos^2{x})\cos{x})dx$$

Let $u=\cos x$, then the above integral equals $$\int_0^{\frac{\sqrt{2}}{2}}(u -4u + 4u^3)du=\boxed{\frac{\sqrt{2}}{6}}$$