Integral of \( \sec^3 x \)

Derivation and Evaluation

The method of integration by parts is used to find the integral:

\[ \int \sec^3 x \, dx \]

The integration by parts formula is expressed as follows:

\[ \int u' v \, dx = uv - \int u v' \, dx \]

Let \( v = \sec x \) and \( u' = \sec^2 x \); hence \( u = \displaystyle \int \sec^2 x \, dx = \tan x \) and \( v' = \sec x \tan x \). Applying this:

\[ \begin{aligned} \int \sec^3 x \, dx &= \int \sec^2 x \cdot \sec x \, dx \\[6pt] &= \tan x \sec x - \int \tan x \sec x \cdot \tan x \, dx \\[6pt] &= \tan x \sec x - \int \tan^2 x \sec x \, dx \qquad (I) \end{aligned} \]

Use the trigonometric identity \( \tan^2 x = \sec^2 x - 1 \) to rewrite the integral:

\[ \begin{aligned} \int \tan^2 x \sec x \, dx &= \int (\sec^2 x - 1)\sec x \, dx \\[6pt] &= \int \sec^3 x \, dx - \int \sec x \, dx \end{aligned} \]

Substitute this back into equation (I):

\[ \int \sec^3 x \, dx = \tan x \sec x - \left( \int \sec^3 x \, dx - \int \sec x \, dx \right) \] \[ \int \sec^3 x \, dx = \tan x \sec x - \int \sec^3 x \, dx + \int \sec x \, dx \]

Add \( \displaystyle \int \sec^3 x \, dx \) to both sides of the equation and simplify:

\[ 2 \int \sec^3 x \, dx = \tan x \sec x + \int \sec x \, dx \]

Use the standard integral formula \( \displaystyle \int \sec x \, dx = \ln|\tan x + \sec x| \):

\[ 2 \int \sec^3 x \, dx = \tan x \sec x + \ln|\tan x + \sec x| \]

Dividing all terms by \( 2 \) and adding the constant of integration \( c \), we obtain the final answer: