\(\int_0^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x=\)

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 \(\mathrm{B}=\frac{1}{10}, \mathrm{~A}=\frac{3}{10}\)
\(
\begin{aligned}
& =\frac{3}{10} \int_0^{\frac{\pi}{2}} d x+\frac{1}{10} \int_0^{\frac{\pi}{2}} \frac{d}{\frac{d x}{d x}(3 \cos x+\sin x)} d x \\
& =\frac{3}{10}[x]_0^{\frac{\pi}{2}}+\frac{1}{10}[\log |3 \cos x+\sin x|]_0^{\frac{\pi}{2}} \\
& =\frac{3}{10}\left(\frac{\pi}{2}\right)+\frac{1}{10}[\log |1|-\log |3|] \\
& =\frac{3 \pi}{20}-\frac{1}{10} \log 3\end{aligned}
\)