Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 36

Determine the integral $\displaystyle \int \frac{\sin \Phi}{\cos^3 \Phi} d \Phi$


$\begin{equation} \begin{aligned} \int \frac{\sin \Phi}{\cos^3 \Phi} d \Phi =& \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi \end{aligned} \end{equation}$


Let $u = \cos \Phi$, then $du = - \sin \Phi d \Phi$, so $\sin \Phi d \Phi = -du$. Thus


$\begin{equation} \begin{aligned} \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \int \frac{1}{u^3} \cdot -du \\ \\ \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& - \int \frac{1}{u^3} du \\ \\ \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& - \int u^{-3} du \\ \\ \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{-u^{-3 + 1}}{-3 + 1} + c \\ \\ \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{-u^{-2}}{-2} + c \\ \\ \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{u^{-2}}{2} + c \\ \\ \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{1}{2u^2} + c \\ \\ \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{1}{2 (\cos \Phi)^2} + c \\ \\ \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{1}{2 \cos^2 \Phi} + c \\ \\ \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{1}{2} \sec^2 \Phi + c \end{aligned} \end{equation}$


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


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