Question

Consider the integral$\int se{c}^{2}x{\tan }^{3}xdx$

(a) Solve this integral by using integration by substitution with u = tan x.

(b) Solve this integral another way, by using integration by substitution with u = sec x.

Short answer

(a)$\begin{array}{rcl}& & \frac{{\tan }^{4}x}{4}+C\end{array}$

(b)$\begin{array}{rcl}& & \frac{{\tan }^{4}x}{4}+C\end{array}$

Step-by-step solution

Step 1. Given information

Integral is$\int se{c}^{2}x{\tan }^{3}xdx$

Part (a) Step 1. Explanation

Let $u=\tan x$

$\begin{array}{rcl}\int se{c}^{2}x{\tan }^{3}xdx& =& \int u^{3}du\\ & =& \frac{u^4}{4}+C\\ & =& \frac{{\tan }^{4}x}{4}+C\end{array}$

Part (b) Step 1. Explanation

Let $u=\sec x$.

$du=secx\tan xdx$

$\begin{array}{rcl}\int se{c}^{2}x{\tan }^{3}xdx& =& \int u\left(u^2-1\right)du\\ & =& \int u^3-udu\\ & =& \frac{u^4}{4}-\frac{u^2}{2}+C\\ & =& \frac{se{c}^{4}x}{4}-\frac{se{c}^{2}x}{2}+C\\ & =& \frac{{\tan }^{4}x}{4}+C\\ & & \end{array}$