\(\int_0^\pi \frac{1}{4+3 \cos x} d x=\)

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 \(\mathrm{x}=\pi, \mathrm{t}=\infty\)
\(\therefore \mathrm{I}=\int_0^{\infty} \frac{1}{4+3\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)} \times \frac{2}{1+\mathrm{t}^2} \)
\( =\int_0^{\infty} \frac{\left(1+\mathrm{t}^2\right)}{4\left(1+\mathrm{t}^2\right)+3\left(1-\mathrm{t}^2\right)} \times \frac{2}{1+\mathrm{t}^2} \mathrm{dt}=\int_0^{\infty}\) \(\frac{2}{7+\mathrm{t}^2} \mathrm{dt}=\frac{2}{7} \int_0^{\infty} \frac{\mathrm{dt}}{1+\left(\frac{1}{\sqrt{7}}\right)^2} \)
\( =\frac{2}{7}\left[\tan ^{-1}\left(\frac{\mathrm{t}}{7}\right)\right]_0^{\infty} \frac{1}{\left(\frac{1}{\sqrt{7}}\right)}=\frac{2}{\sqrt{7}}[\tan ^{-1} \infty-\) \(\tan ^{-1} 0]=\frac{2}{\sqrt{7}} \times \frac{\pi}{2}=\frac{\pi}{\sqrt{7}}\)