Evaluate the following integrals:

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

We know that $\cos \mathrm{X}=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

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

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

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

$\Rightarrow \int \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)-4\left(1-\tan ^{2} \frac{x}{2}\right)} d x=\int \frac{\sec ^{2} \frac{x}{2}}{9 \tan ^{2} \frac{x}{2}+1} d x$

Putting $\tan x / 2=t$ and $\sec ^{2}(x / 2) d x=2 d t$,

$\Rightarrow \int \frac{\sec ^{2} \frac{x}{2}}{9 \tan ^{2} \frac{x}{2}+1} d x=\int \frac{2 d t}{9 t^{2}+1}$

$=\frac{2}{9} \int \frac{1}{t^{2}+\frac{1}{9}} d t$

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

$\Rightarrow \frac{2}{9} \int \frac{1}{t^{2}+\frac{1}{9}} d t=\frac{2}{9}\left(\frac{1}{\frac{1}{3}}\right) \tan ^{-1}\left(\frac{t}{\frac{1}{3}}\right)+c$

$=\frac{2}{3} \tan ^{-1}(3 \tan x)+c$

$\therefore \mathrm{I}=\int \frac{1}{5-4 \cos \mathrm{x}} \mathrm{dx}=\frac{2}{3} \tan ^{-1}(3 \tan \mathrm{x})+\mathrm{c}$