#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.3 Question 13 Maths Textbook Solution.

Answer: $\frac{1}{3}\left[\frac{1-\cos 3 x}{\sin 3 x}\right]+C$

Hint: $\text { We will add } 1-\cos 3 x \text { to the equation }$

Given: $\int \frac{1}{1+\cos 3 x} d x$

Solution: $\int \frac{1}{1+\cos 3 x} d x$

$\text { Multiplying } 1-\cos 3 x \text { in numerator and denominator }$

\begin{aligned} &\int \frac{1-\cos 3 x}{(1+\cos 3 x)(1-\cos 3 x)} d x \\ &=\int \frac{1-\cos 3 x}{1-\cos ^{2} 3 x} d x \\ &=\int \frac{1-\cos 3 x}{\sin ^{2} 3 x} d x \end{aligned}

\begin{aligned} &=\int \frac{1}{\sin ^{2} 3 x} d x-\int \frac{\cos 3 x}{\sin 3 x \sin 3 x} d x \\ &=\int \operatorname{cosec}^{2} 3 x d x-\int \cot 3 x \operatorname{cosec} 3 x d x \\ &=\frac{-\cot 3 x}{3}+\frac{\operatorname{cosec} 3 x}{3}+c \end{aligned}

\begin{aligned} &=\frac{1}{3}[\operatorname{cosec} 3 x-\cot 3 x]+c \\ &=\frac{1}{3}\left[\frac{1}{\sin 3 x}-\frac{\cos 3 x}{\sin 3 x}\right]+C \\ &=\frac{1}{3}\left[\frac{1-\cos 3 x}{\sin 3 x}\right]+C \end{aligned}