Question

Solve the following integral.

$\int {\tan }^{6}xdx$

Short answer

Answer is$\begin{array}{rcl}& & \frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}+\tan x-x+C\end{array}$

Step-by-step solution

Step 1. Given information

An integral is$\int {\tan }^{6}xdx$

Step 2. Explanation

Let $\tan x=u$

$\begin{array}{rcl}\int {\tan }^{6}xdx& =& \int \frac{u^6}{u^2+1}du\\ & =& \int u^4-u^2+1-\frac{1}{u^2+1}du\\ & =& \frac{u^5}{5}-\frac{u^3}{3}+u-{\tan }^{-1}u+C\\ & =& \frac{{\tan }^{5}x}{5}-\frac{{\tan }^{3}x}{3}+\tan x-x+C\end{array}$