Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{1}{2+\cos x}dx$ equals

(a) $\frac{1}{3}{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(b) $\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(c) $\sqrt{3}{\tan }^{-1}\left(\sqrt{3}\right)$

(d) $2\sqrt{3}{\tan }^{-1}\sqrt{3}$

Answer 1

$\left(\mathrm{b}\right)\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{2+\cos x}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{2+{\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+{\tan }^2\frac{x}{2}}{2+2{\tan }^2{\displaystyle \frac{x}{2}}+1-{\tan }^2{\displaystyle \frac{x}{2}}}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sec }^2{\displaystyle \frac{x}{2}}}{3+{\tan }^2{\displaystyle \frac{x}{2}}}dx \end{gathered}$$

$$\begin{gathered} \mathrm{Putting}\tan \frac{x}{2}=t \\ \Rightarrow \frac{1}{2}{\sec }^2\frac{\mathrm{x}}{2}dx=dt \\ \Rightarrow {\sec }^2\frac{\mathrm{x}}{2}dx=2dt \\ \mathrm{When},x\to 0;t\to 0 \\ \mathrm{and}x\to \frac{\mathrm{\pi }}{2};t\to 1 \\ \therefore I={\int }_0^1\frac{2}{3+t^2}dt \\ =2{\int }_0^1\frac{1}{{\left(\sqrt{3}\right)}^2+t^2}dt \\ =\frac{2}{\sqrt{3}}{\left[{\tan }^{-1}\frac{t}{\sqrt{3}}\right]}_0^1 \\ =\frac{2}{\sqrt{3}}\left[{\tan }^{-1}\frac{1}{\sqrt{3}}-{\tan }^{-1}\frac{0}{\sqrt{3}}\right] \\ =\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{gathered}$$
3√tan−1(13√)