Use integration by parts. Let:
\[ u = \sec x, dv = \sec^2 x \, dx \Rightarrow du = \sec x \tan x \, dx, v = \tan x. \] \[ \int \sec^3 x \, dx = \sec x \tan x - \int \tan x \cdot \sec x \tan x \, dx = \sec x \tan x - \int \sec x \tan^2 x \, dx. \] Rewrite: \( \tan^2 x = \sec^2 x - 1 \).
\[ \int \sec x \tan^2 x \, dx = \int \sec x (\sec^2 x - 1) \, dx = \int \sec^3 x \, dx - \int \sec x \, dx. \] \[ \int \sec^3 x \, dx = \sec x \tan x - \left( \int \sec^3 x \, dx - \int \sec x \, dx \right). \] \[ 2 \int \sec^3 x \, dx = \sec x \tan x + \int \sec x \, dx. \] \[ \int \sec x \, dx = \ln |\sec x + \tan x| + c. \] \[ \int \sec^3 x \, dx = \frac{1}{2} \left( \sec x \tan x + \ln |\sec x + \tan x| \right) + c. \] Answer: \( \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln |\sec x + \tan x| + c \).
Let \( f: \mathbb{R} \to \mathbb{R} \) \(\text{ be any function defined as }\) \[ f(x) = \begin{cases} x^\alpha \sin \left( \frac{1}{x^\beta} \right) & \text{for } x \neq 0, \\ 0 & \text{for } x = 0, \end{cases} \] where \( \alpha, \beta \in \mathbb{R} \). Which of the following is true? \( \mathbb{R} \) denotes the set of all real numbers.