#### Please solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.23 Question 1 maths textbook solution.

Answer : $\frac{2}{3} \tan ^{-1}\left(\frac{\tan \frac{x}{2}}{3}\right)+c$

Hint: To solve this equation we have to differentiate it and change cos in terms of tan

Given : $\int \frac{1}{5+4 \cos x} d x$

Solution : $\frac{1}{5+4 \cos x}$

Use the formula : $\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

$=\frac{1}{5+4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)}$

\begin{aligned} &=\frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)+4\left(1-\tan ^{2} \frac{x}{2}\right)} \\ &=\frac{1+\tan ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} \\ &=\frac{\sec ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} \\ &I=\int \frac{\sec ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} d x \end{aligned}

Let,

\begin{aligned} &\tan \frac{x}{2}=t \\ &\frac{1}{2} \sec ^{2} \frac{x}{2}=\frac{d t}{d x} \\ &d x=\frac{d t}{\frac{1}{2} \sec ^{2} \frac{x}{2}} \\ &I=\int \frac{\sec ^{2} \frac{x}{2}}{9+t^{2}} \cdot\frac{d t}{\frac{1}{2} \sec ^{2} \frac{x}{2}} \end{aligned}

\begin{aligned} &I=\int \frac{2 d t}{9+t^{2}} \\ &I=\frac{2}{3} \tan ^{-1}\left(\frac{t}{3}\right)+c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right] \\ &I=\frac{2}{3} \tan ^{-1}\left[\frac{\tan \frac{x}{2}}{3}\right]+c \end{aligned}