Evaluate the following integrals:

Solution:

Given $I=\int \frac{\cos x}{\cos 3 x} d x$

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

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

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

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

Replacing $\sec ^{2} x$ by $1+\tan ^{2} x$ in denominator,

$\Rightarrow \int \frac{\sec ^{2} x}{4-3 \sec ^{2} x} d x=\int \frac{\sec ^{2} x}{4-3-3 \tan ^{2} x} d x$

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

Putting $\tan x=t$ and $\sec ^{2} x d x=d t$, we get

$I=\int \frac{\mathrm{dt}}{1-3 \mathrm{t}^{2}}=\frac{1}{3} \int \frac{1}{\frac{1}{3}-\mathrm{t}^{2}} \mathrm{dt}$

We know that $\int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c$

$\Rightarrow \frac{1}{3} \int \frac{1}{\frac{1}{3}-\mathrm{t}^{2}} \mathrm{dt}=\frac{1}{3} \times \frac{1}{2 \sqrt{3}} \log \left|\frac{\frac{1}{\sqrt{3}}+\mathrm{t}}{\frac{1}{\sqrt{3}}-\mathrm{t}}\right|+\mathrm{c}$

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

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

$\therefore I=\int \frac{\cos x}{\cos 3 x} d x=\frac{1}{6 \sqrt{3}} \log \left|\frac{1+\sqrt{3} \tan x}{1-\sqrt{3} \tan x}\right|+c$