Find the integral of the function $\frac{1}{\sin x \cos ^{3} x}$.

(N/A) We have $\frac{1}{\sin x \cos ^{3} x} = \frac{\sin ^{2} x + \cos ^{2} x}{\sin x \cos ^{3} x}$.
$= \frac{\sin ^{2} x}{\sin x \cos ^{3} x} + \frac{\cos ^{2} x}{\sin x \cos ^{3} x} = \frac{\sin x}{\cos ^{3} x} + \frac{1}{\sin x \cos x}$.
$= \tan x \sec ^{2} x + \frac{\sec ^{2} x}{\tan x}$.
Therefore,$\int \frac{1}{\sin x \cos ^{3} x} dx = \int \tan x \sec ^{2} x dx + \int \frac{\sec ^{2} x}{\tan x} dx$.
Let $t = \tan x$,then $dt = \sec ^{2} x dx$.
Substituting these into the integral,we get $\int t dt + \int \frac{1}{t} dt$.
$= \frac{t^{2}}{2} + \log |t| + C$.
Substituting $t = \tan x$ back,we get $\frac{1}{2} \tan ^{2} x + \log |\tan x| + C$,where $C$ is an arbitrary constant.