Let $\int_0^\pi \frac{1}{4+3 \cos x} d x$
Put $\frac{x}{2}=t \Rightarrow \cos x=\frac{1-t^2}{1+t^2}$ and $\sec ^2 \frac{x}{2}\left(\frac{1}{2}\right) d x=d t \Rightarrow d t=\frac{2}{1+t^2} d t$
When $\mathrm{x}=0, \mathrm{t}=0$ and when $\mathrm{x}=\pi, \mathrm{t}=\infty$
$$
\begin{aligned}
& \therefore \quad \mathrm{I}=\int_0^{\infty} \frac{1}{4+3\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)} \times \frac{2}{1+\mathrm{t}^2} \\
& =\int_0^{\infty} \frac{\left(1+\mathrm{t}^2\right)}{4\left(1+\mathrm{t}^2\right)+3\left(1-\mathrm{t}^2\right)} \times \frac{2}{1+\mathrm{t}^2} \mathrm{dt}=\int_0^{\infty} \frac{2}{7+\mathrm{t}^2} \mathrm{dt}=\frac{2}{7} \int_0^{\infty} \frac{\mathrm{dt}}{1+\left(\frac{1}{\sqrt{7}}\right)^2} \\
& =\frac{2}{7}\left[\tan ^{-1}\left(\frac{\mathrm{t}}{7}\right)\right]_0^{\infty} \frac{1}{\left(\frac{1}{\sqrt{7}}\right)}=\frac{2}{\sqrt{7}}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right]=\frac{2}{\sqrt{7}} \times \frac{\pi}{2}=\frac{\pi}{\sqrt{7}}
\end{aligned}
$$