Let $I=\int_0^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x$
Put $\cos x=A(3 \cos x+\sin x)+B \frac{d}{d x}(3 \cos x+\sin x)$
$$
=\mathrm{A}(3 \cos \mathrm{x}+\sin \mathrm{x})+\mathrm{B}(-3 \sin \mathrm{x}+\cos \mathrm{x})
$$
Thus $3 \mathrm{~A}+\mathrm{B}=1$ and $\mathrm{A}-3 \mathrm{~B}=0$
Solving, we get $\mathrm{B}=\frac{1}{10}, \mathrm{~A}=\frac{3}{10}$
$$
\begin{aligned}
& =\frac{3}{10} \int_0^{\frac{\pi}{2}} d x+\frac{1}{10} \int_0^{\frac{\pi}{2}} \frac{d}{\frac{d x}{d x}(3 \cos x+\sin x)} d x \\
& =\frac{3}{10}[x]_0^{\frac{\pi}{2}}+\frac{1}{10}[\log |3 \cos x+\sin x|]_0^{\frac{\pi}{2}} \\
& =\frac{3}{10}\left(\frac{\pi}{2}\right)+\frac{1}{10}[\log |1|-\log |3|] \\
& =\frac{3 \pi}{20}-\frac{1}{10} \log 3 \\
&
\end{aligned}
$$