Question

Solve the following integral.

$\int secx{\tan }^{3}xdx$

Short answer

Answer is$\frac{se{c}^{3}x}{3}-secx+C$

Step-by-step solution

Step 1. Given information

An integral is$\int secx{\tan }^{3}xdx$

Step 2. Explanation

$\int secx{\tan }^{3}xdx=\int \left(se{c}^{2}x-1\right)secx\tan xdx$

Let $u=secx$

$\begin{array}{rcl}\int secx{\tan }^{3}xdx& =& \int u^2-1du\\ & =& \frac{u^3}{3}-u+C\\ & =& \frac{se{c}^{3}x}{3}-secx+C\end{array}$