Evaluate the following integrals:

Given I $=\int \frac{1}{\cos x(\sin x+2 \cos x)} d x$

$\Rightarrow I=\int \frac{1}{\cos x(\sin x+2 \cos x)} d x=\int \frac{1}{\cos x \sin x+2 \cos ^{2} x} d x$

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

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

Putting $\tan x+2=\mathrm{t}$ so that $\sec ^{2} x \mathrm{dx}=\mathrm{dt}$,

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

We know that $\int \frac{1}{x} d x=\log |x|+c$

$\Rightarrow \int \frac{1}{t} d t=\log |t|+c$

$=\log |\tan x+2|+x$

$\therefore \mathrm{I}=\int \frac{1}{\cos \mathrm{x}(\sin \mathrm{x}+2 \cos \mathrm{x})} \mathrm{dx}=\log |\tan \mathrm{x}+2|+\mathrm{c}$