Inspired by Carl Lian

We let the quantity be $\Re$ . We seek to make the expression homogeneous. Let $\theta = 2x$ . We have

$\begin{aligned} \Re & = 2 \int_0^\frac{\pi}{2} \frac{-3 \sin 2x + 4 \cos 2x + 4}{25 \cos 2x + 7} dx \\ & = 2 \int_0^\frac{\pi}{2} \frac{-6\sin x \cos x + 4(2\cos^2 x - 1) - 4}{25(\cos^2x - \sin^2 x) + 7(\cos^2 x + \sin^2 x)} dx\\ & = 2\int_0^\frac{\pi}{2}\frac{-6\sin x\cos x +8\cos^2 x}{32\cos^2x - 18\sin^2} dx = 2\int_0^\frac{\pi}{2}\frac{-3\sin x\cos x+4\cos^2 x}{16\cos^2x - 9\sin^2} dx \\ & = 2\int_0^\frac{\pi}{2} \frac{\cos x (4 \cos x - 3 \sin x)}{(4 \cos x + 3 \sin x)(4 \cos x - 3 \sin x)} dx = 2 \int_0^\frac{\pi}{2} \frac{\cos x}{4\cos x + 3\sin x} dx \\ & = 2\int_0^\frac{\pi}{2} \frac{ \frac{4}{25} (4 \cos x + 3 \sin x) + \frac{3}{25} (4 \cos x + 3 \sin x)'}{(4 \cos x + 3 \sin x)} dx \\ & = 2 \int_0^{\frac{\pi}{2}} \frac{4}{25} + \frac{3}{25} \cdot \frac{(4 \cos x + 3 \sin x)'}{(4 \cos x + 3 \sin x)} dx \\ & = 2\left ( \frac{4x}{25} + \frac{3}{25} \ln (4\cos x + 3 \sin x) \mid^{\frac{\pi}{2}}_0 \right ) \approx 4.34 \end{aligned}$