#### Provide Solution For  RD Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.23 Question 10 Maths Textbook Solution.

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

Hint: To solve this equation we have to change cosx into tanx form

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

Solution :$\int \frac{1}{5-4 \cos x}dx$

$\operatorname{Cos} x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

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

\begin{aligned} &=\int \frac{\sec ^{2} \frac{x}{2}}{1+9 \tan ^{2} \frac{x}{2}} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{1+\left(3 \tan \frac{x}{2}\right)^{2}} d x \end{aligned}

\begin{aligned} &\text { Let } 3 \tan \frac{x}{2}=t \\ &=3 \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t \end{aligned}

\begin{aligned} &=\frac{2}{3} \int \frac{d t}{1+t^{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left\langle\int\frac{1}{a^{2}+t^{2}}dt=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)\right\rangle \\ &=\frac{2}{3} \tan ^{-1}(t)+C \\ &=\frac{2}{3} \tan ^{-1}\left(3 \tan \frac{x}{2}\right)+C \end{aligned}