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

Question

Aicharya · Accepted Answer

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 $\sec ^2 \frac{x}{2}\left(\frac{1}{2}ight) d x=d t \Rightarrow d t=\frac{2}{1+t^2} d t$ When $\mathrm{x}=0, \mathrm{t}=0$ and when $\mathrm{x}=\pi, \mathrm{t}=\infty$ $\begin{aligned} & 	herefore \quad \mathrm{I}=\int_0^{\infty} \frac{1}{4+3\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}ight)} 	imes \frac{2}{1+\mathrm{t}^2} \ & =\int_0^{\infty} \frac{\left(1+\mathrm{t}^2ight)}{4\left(1+\mathrm{t}^2ight)+3\left(1-\mathrm{t}^2ight)} 	imes \frac{2}{1+\mathrm{t}^2} \mathrm{dt}=\int_0^{\infty} \frac{2}{7+\mathrm{t}^2} \mathrm{dt}=\frac{2}{7} \int_0^{\infty} \frac{\mathrm{dt}}{1+\left(\frac{1}{\sqrt{7}}ight)^2} \ & =\frac{2}{7}\left[	an ^{-1}\left(\frac{\mathrm{t}}{7}ight)ight]_0^{\infty} \frac{1}{\left(\frac{1}{\sqrt{7}}ight)}=\frac{2}{\sqrt{7}}\left[	an ^{-1} \infty-	an ^{-1} 0ight]=\frac{2}{\sqrt{7}} 	imes \frac{\pi}{2}=\frac{\pi}{\sqrt{7}} \end{aligned}$

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

Solution