Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 34

Determine the integral $\displaystyle \int \tan^2 x \sec x dx$


$
\begin{equation}
\begin{aligned}

\int \tan^2 x \sec x dx =& \int (\sec^2 x - 1) \sec x dx
\qquad \text{Apply Trigonometric Identity } \sec^2 x = \tan^2 x + 1
\\
\\
\int \tan^2 x \sec x dx =& \int (\sec^3 x - \sec x) dx
\\
\\
\int \tan^2 x \sec x dx =& \int \sec^3 x dx - \int \sec x dx

\end{aligned}
\end{equation}
$


We integrate the equation term by term

@ 1st term, using integration by parts

$ \int \sec^3 x dx = \int udv$

where


$
\begin{equation}
\begin{aligned}

u =& \sec x
\\
\\
du =& \sec x \tan x dx
\\
\\
v =& \tan x
\\
\\
dv =& \sec^2 x dx

\end{aligned}
\end{equation}
$


then


$
\begin{equation}
\begin{aligned}

\int \sec^3 x dx =& \int udv
\\
\\
\int \sec^3 x dx =& uv - \int v du
\\
\\
\int \sec^3 x dx =& \sec x \tan x - \int \tan x \cdot \sec x \tan x dx
\\
\\
\int \sec^3 x dx =& \sec x \tan x - \int \sec x \tan^2 x dx
\qquad \text{Apply Trigonometric Identity } \sec^2 x = \tan^2 x + 1
\\
\\
\int \sec^3 x dx =& \sec x \tan x - \int \sec x (\sec^2 x - 1) dx
\\
\\
\int \sec^3 x dx =& \sec x \tan x - \int (\sec^3 x - \sec x) dx
\\
\\
\int \sec^3 x dx =& \sec x \tan x - \int \sec^2 x dx + \int \sec x dx
\qquad \text{Combine like terms}


\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

\int \sec^3 x dx + \int \sec^2 x dx =& \sec x \tan x + \int \sec x dx
\\
\\
2 \int \sec^3 x dx =& \sec x \tan x + \int \sec x dx
\\
\\
2 \int \sec^3 x dx =& \sec x \tan x + \ln (\sec x + \tan x) + c
\\
\\
\int \sec^3 x dx =& \frac{\sec x \tan x + \ln (\sec x + \tan x)}{2} + c


\end{aligned}
\end{equation}
$


@ 2nd term

$\int \sec x dx = \ln (\sec x + \tan x) + c$

Combine the results of the integration term by term


$
\begin{equation}
\begin{aligned}

\int \tan^2 x \sec x dx =& \frac{\sec x \tan x + \ln(\sec x + \tan x)}{2} - \ln (\sec x + \tan x) + c
\\
\\
\int \tan^2 x \sec x dx =& \frac{\sec x \tan x + \ln(\sec x + \tan x) - 2 \ln (\sec x + \tan x)}{2} + c
\\
\\
\int \tan^2 x \sec x dx =& \frac{\sec x \tan x - \ln (\sec x + \tan x)}{2} + c
\\
\\
\text{ or} &
\\
\\
\int \tan^2 x \sec x dx =& \frac{1}{2} (\sec x \tan x - \ln (\sec x + \tan x)) + c


\end{aligned}
\end{equation}
$