Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 28

Determine the integral $\displaystyle \int \tan^3 (2x) \sec^5 (2x) dx$

Let $u = 2x$, then $du = 2 dx$, so $\displaystyle dx = \frac{du}{2}$. Thus,


$\begin{equation} \begin{aligned} \int \tan^3 (2x) \sec^5 (2x) dx =& \int \tan^3 u \sec^5 u \cdot \frac{du}{2} \\ \\ \int \tan^3 (2x) \sec^5 (2x) dx =& \frac{1}{2} \int \tan^3 u \sec^5 u du \\ \\ \int \tan^3 (2x) \sec^5 (2x) dx =& \frac{1}{2} \int \tan^2 u \sec^4 u \sec u \tan u du \qquad \text{Apply Trigonometric Identity } \sec^2 u = \tan^2 u + 1 \text{ for } \tan^2 u \\ \\ \int \tan^3 (2x) \sec^5 (2x) dx =& \frac{1}{2} \int (\sec^2 u - 1) \sec^4 u \sec u \tan u du \\ \\ \int \tan^3 (2x) \sec^5 (2x) dx =& \frac{1}{2} \int (\sec^6 u - \sec^4 u) \sec u \tan u du \end{aligned} \end{equation}$


Let $v = \sec u$, then $dv = \sec u \tan u du$. Thus,


$\begin{equation} \begin{aligned} \frac{1}{2} \int (\sec^6 u - \sec^4 u) \sec u \tan u du =& \frac{1}{2} \int (v^6 - v^4) dv \\ \\ \frac{1}{2} \int (\sec^6 u - \sec^4 u) \sec u \tan u du =& \frac{1}{2} \left( \frac{v^{6 + 1}}{6 + 1} - \frac{v^{4 + 1}}{4+1} \right) + c \\ \\ \frac{1}{2} \int (\sec^6 u - \sec^4 u) \sec u \tan u du =& \frac{1}{2} \left( \frac{v^7}{7} - \frac{v^5}{5} \right) + c \\ \\ \frac{1}{2} \int (\sec^6 u - \sec^4 u) \sec u \tan u du =& \frac{v^7}{14} - \frac{v^5}{10} + c \qquad \text{Substitute value of } v \\ \\ \frac{1}{2} \int (\sec^6 u - \sec^4 u) \sec u \tan u du =& \frac{(\sec u)^7}{14} - \frac{(\sec u)^5}{10} + c \\ \\ \frac{1}{2} \int (\sec^6 u - \sec^4 u) \sec u \tan u du =& \frac{\sec^7 u}{14} - \frac{\sec^5 u}{10} + c \qquad \text{Substitute value of } u \\ \\ \frac{1}{2} \int (\sec^6 u - \sec^4 u) \sec u \tan u du =& \frac{\sec^7 (2x)}{14} - \frac{\sec^5 (2x)}{10} + c \end{aligned} \end{equation}$