The integral $\int_\limits0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} \mathrm{~d} x$ is equal to :

<p>$\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$</p> <p>$$\begin{aligned} & \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \\ & \begin{array}{l} 3 A-5 B=1 \\ 5 A+3 B=0 \end{array}>A=\frac{3}{34} \quad B=\frac{-5}{34} \end{aligned}$$</p> <p>$$\begin{aligned} & \int_0^{\pi / 4} \frac{136\left[\frac{3}{34}(3 \sin x+5 \cos x)-\frac{5}{34}(3 \cos x-5 \sin x)\right]}{3 \sin x+5 \cos x} d x \\ & \int_0^{\pi / 4} 12 d x-20 \int_0^{\pi / 4} \frac{3 \cos x-5 \sin x}{3 \sin x+5 \cos x} d x \\ & 12 \times \frac{\pi}{4}-20\left[\ln \left|\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right|-\ln 5\right] \\ & 3 \pi-20 \ln 2^{5 / 2}+20 \ln 5 \\ & \Rightarrow 3 \pi-50 \ln 2+20 \ln 5 \end{aligned}$$</p>