#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 4 maths Textbook Solution.

Answer:  $\frac{1}{2 \sqrt{3}} \log \left|\frac{1+\sqrt{3} \tan x}{1-\sqrt{3} \tan x}\right|+c$

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

Hint: Use the formula of $\cos 3 x$

Solution:

$\int \frac{\cos x}{\cos 3 x} d x$

$=\int \frac{\cos x}{4 \cos ^{3} x-3 \cos x} d x$

$\left(\because \cos 3 x=4 \cos ^{3} x-3 \cos x\right)$

$=\int \frac{1}{4 \cos ^{2} x-3} d x$

Dividing numerator and denominator by $\cos ^{2} x$

\begin{aligned} &=\int \frac{\sec ^{2} x}{4-3 \sec ^{2} x} d x \\ &=\int \frac{\sec ^{2} x}{4-3\left(1+\tan ^{2} x\right)} d x \end{aligned}

$\left(\because \sec ^{2} x=1+\tan ^{2} x\right)$

$=\int \frac{\sec ^{2} x}{1-3 \tan ^{2} x} d x$

Let

\begin{aligned} &\tan x=t \\ &\sec ^{2} x d x=d t \end{aligned}                                                             (Differentiating w.r.t x)

\begin{aligned} &=\int \frac{1}{1-3 t^{2}} d t \\ &=\frac{1}{3} \int \frac{1}{\left(\frac{1}{\sqrt{3}}\right)^{2}-t^{2}} d t \end{aligned}

$=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{\frac{1}{\sqrt{3}}} \log \left|\frac{\frac{1}{\sqrt{3}}+t}{\frac{1}{\sqrt{3}}-t}\right|+c$                            $\left[\int \frac{d t}{a^{2}-t^{2}}=\frac{1}{2 a} \log \left|\frac{a+t}{a-t}\right|+c\right]$

$=\frac{1}{6} \times \sqrt{3} \log \left|\frac{1+\sqrt{3} t}{1-\sqrt{3} t}\right|+c$

$=\frac{1}{2 \sqrt{3}} \log \left|\frac{1+\sqrt{3} \tan x}{1-\sqrt{3} \tan x}\right|+c$