$\int \frac{\mathrm{d} x}{2+\cos x}=$ (Where $C$ is a constant of integration.)
$\int \frac{\mathrm{d} x}{2+\cos x}=$ (Where $C$ is a constant of integration.)
- $\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{2 \sqrt{3}}\right)+C$
- $\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{2 \sqrt{3}}\right)+C$
- $\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{3}}\right)+C$
- $\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{3}}\right)+C$
Solution
$\begin{aligned} & \int \frac{\mathrm{d} x}{2+\cos x}=\int \frac{\mathrm{d} x}{2+\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}=\int \frac{\sec ^2 \frac{x}{2} \mathrm{~d} x}{3+\tan ^2 \frac{x}{2}} \\ & =2 \int \frac{\frac{1}{2} \sec ^2 \frac{x}{2} \mathrm{~d} x}{(\sqrt{3})^2+\left(\tan \frac{x}{2}\right)^2}=\frac{2}{\sqrt{3}} \cdot \tan ^{-1}\left(\frac{\tan \frac{x}{2}}{\sqrt{3}}\right)\end{aligned}$
Asked in: MHT CET 2022 (08 Aug Shift 1)
Practice more Indefinite Integration questions on Aicharya