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

Question

Aicharya · Accepted Answer

Let $\mathrm{I}=\int_0^\pi \frac{\mathrm{dx}}{4+3 \cos x}$ Put $	an \frac{x}{2}=\mathrm{t}$ $	herefore \quad \mathrm{d} x=\frac{2 \mathrm{dt}}{1+\mathrm{t}^2}$ and $\cos x=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}$ $\begin{aligned} 	herefore \quad \mathrm{I}=\int_0^\pi \frac{\mathrm{d} x}{4+3 \cos x} & =\int_0^{\infty} \frac{2 \mathrm{dt}}{7+\mathrm{t}^2} \ & =\left[\frac{2}{\sqrt{7}} 	an ^{-1}\left(\frac{\mathrm{t}}{\sqrt{7}}ight)ight]_0^{\infty} \ & =\frac{2}{\sqrt{7}}\left[	an ^{-1} \infty-0ight] \ & =\frac{2}{\sqrt{7}} \cdot \frac{\pi}{2}=\frac{\pi}{\sqrt{7}}\end{aligned}$

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

Solution