# Integration of Secant Cubed X

In this tutorial we shall discuss the integral of the secant cubed of $X$, and this integration can be evaluated by using integration by parts. But first we break the power of the function from cube into square and one power of the function. Then we can use integration by parts of that function because we know the formula for the integration of secant square of $X$ from previous tutorials.

The integral of the secant cubed of $X$ is of the form
$I = \int {{{\sec }^3}xdx} \,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

First we break ${\sec ^3}x$ into $\sec x$ and ${\sec ^2}x$, and now the integral (i) becomes
$I = \int {\sec x{{\sec }^2}xdx} \,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

$\sec x$ and ${\sec ^2}x$ are the first and second functions that form integration by parts. Using the formula for integration by parts, we have
$\int {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int {g\left( x \right)dx – \int {\left[ {\frac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } }$

Using the formula above, equation (ii) becomes
$\begin{gathered} I = \sec x\int {{{\sec }^2}xdx – \int {\left[ {\frac{d}{{dx}}\sec x\left( {\int {{{\sec }^2}xdx} } \right)} \right]} dx} \\ \Rightarrow I = \sec x\tan x – \int {\left[ {\left( {\sec x\tan x} \right)\tan x} \right]} dx \\ \Rightarrow I = \sec x\tan x – \int {\left[ {\sec x{{\tan }^2}x} \right]} dx \\ \end{gathered}$

Now using trigonometric identity ${\tan ^2}x = {\sec ^2}x – 1$, we have
$\begin{gathered} I = \sec x\tan x – \int {\left[ {\sec x\left( {{{\sec }^2}x – 1} \right)} \right]} dx \\ \Rightarrow I = \sec x\tan x – \int {\left[ {{{\sec }^3}x – \sec x} \right]} dx \\ \Rightarrow I = \sec x\tan x – \int {{{\sec }^3}xdx + \int {\sec xdx} } \\ \end{gathered}$

From our original problem $I = \int {{{\sec }^3}xdx}$, using this value we have
$\begin{gathered} I = \sec x\tan x – I + \ln \left( {\sec x + \tan x} \right) + c \\ \Rightarrow I + I = \sec x\tan x + \ln \left( {\sec x + \tan x} \right) + c \\ \Rightarrow I = \frac{1}{2}\left[ {\sec x\tan x + \ln \left( {\sec x + \tan x} \right)} \right] + c \\ \Rightarrow \int {{{\sec }^3}xdx} = \frac{1}{2}\left[ {\sec x\tan x + \ln \left( {\sec x + \tan x} \right)} \right] + c \\ \end{gathered}$