Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 9 maths textbook solution

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

Given:

$\int_{0}^{\frac{\pi}{2}} \frac{1}{2+\cos x} d x$

Hint:

Using $\int \frac{1}{1+x^{2}} d x$

Explanation:

Let

$I=\int_{0}^{\frac{\pi}{2}} \frac{1}{2+\cos x} d x$

Put

\begin{aligned} &\cos x=\frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2} \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{1}{2+\frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2}} d x \end{aligned}

$=\int_{0}^{\frac{\pi}{2}} \frac{1+\tan ^{2} x / 2}{2\left(1+\tan ^{2} x / 2\right)+1-\tan ^{2} x / 2} d x$

$=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2 } x/ 2}{3+\tan ^{2} x / 2} d x$

\begin{aligned} &\text { Put } \tan ^{x} /_{2}=t \\\\ &\sec ^{2} x / 2 \cdot 1 / 2 d x=d t \\\\ &\sec ^{2} x /{ }_{2} d x=2 d t \end{aligned}

When $x=0$ then $t= 0$

When $x=\frac{\pi }{2}$ then $t=1$

\begin{aligned} &=\int_{0}^{1} \frac{2 d t}{t^{2}+(\sqrt{3})^{2}} \\\\ &=2\left[\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)\right]_{0}^{1} \\\\ &=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\tan ^{-1}(0)\right] \end{aligned}

\begin{aligned} &=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)-0\right] \\\\ &=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{aligned}