Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 24

Determine the integral $\displaystyle \int (\tan^2 x + \tan^4 x) dx$



$\begin{equation} \begin{aligned} \int (\tan^2 x + \tan^4 x) dx =& \int \tan^2 x (1 + \tan^2 x) dx \qquad \text{Apply Trigonometric Identity } \sec^2 x = \tan^2 x + 1 \\ \\ \int (\tan^2 x + \tan^4 x) dx =& \int \tan^2 x \sec^2 x dx \end{aligned} \end{equation}$


Let $u = \tan x$, then $du = \sec^2 x dx$. Thus,


$\begin{equation} \begin{aligned} \int \tan^2 x \sec^2 x dx =& \int u^2 du \\ \\ \int \tan^2 x \sec^2 x dx =& \frac{u^{2 + 1}}{2 + 1} + c \\ \\ \int \tan^2 x \sec^2 x dx =& \frac{u^3}{3} + c \\ \\ \int \tan^2 x \sec^2 x dx =& \frac{(\tan x)^3}{3} + c \\ \\ \int \tan^2 x \sec^2 x dx =& \frac{\tan^3x}{3} + c \end{aligned} \end{equation}$