# What's the integral of int (tanx)^8 ?

Jun 7, 2016

${\tan}^{7} \frac{x}{7} - {\tan}^{5} \frac{x}{5} + {\tan}^{3} \frac{x}{3} - \tan x + x + K$

#### Explanation:

First, to prevent loss in presentation marks, we always tag on the $\mathrm{dx}$ on the end of the integral. That is, our goal is to find $\int \left({\tan}^{8} x\right) \mathrm{dx}$.

Notice that ${\tan}^{8} x = {\tan}^{6} x {\tan}^{2} x = {\tan}^{6} x \left({\sec}^{2} x - 1\right) = {\tan}^{6} x {\sec}^{2} x - {\tan}^{6} x$

Then $\int \left({\tan}^{8} x\right) \mathrm{dx} = \int \left({\tan}^{6} x\right) \left({\sec}^{2} x\right) \mathrm{dx} - \int \left({\tan}^{6} x\right) \mathrm{dx}$

To solve the first integral, we let $u = \tan x$ which implies that $\mathrm{du} = {\sec}^{2} x \mathrm{dx}$, thus the integral is simplified to

$\int {u}^{6} \mathrm{du} = {u}^{7} / 7 + K = {\tan}^{7} \frac{x}{7} + K$.

Using similar methods to find $\int \left({\tan}^{6} x\right) \mathrm{dx}$, $\int \left({\tan}^{4} x\right) \mathrm{dx}$ and $\int \left({\tan}^{2} x\right) \mathrm{dx}$, we derive that

$\int \left({\tan}^{8} x\right) \mathrm{dx} = {\tan}^{7} \frac{x}{7} - {\tan}^{5} \frac{x}{5} + {\tan}^{3} \frac{x}{3} - \tan x + x + K$

For a slightly shorter method, we use the method of reduction formulae. Consider the general integral ${I}_{n} = \int \left({\tan}^{n} x\right) \mathrm{dx}$. Using the same trigonometric identity, we find that

$\int \left({\tan}^{n} x\right) \mathrm{dx} = \int \left({\tan}^{n - 2} x\right) \left({\sec}^{2} x\right) \mathrm{dx} - \int \left({\tan}^{n - 2} x\right) \mathrm{dx}$

which stands to reason that ${I}_{n} = {\tan}^{n - 1} \frac{x}{n - 1} - {I}_{n - 2}$.

All we now need to do is find ${I}_{8}$, which in turn requires us to find ${I}_{6}$, ${I}_{4}$, ${I}_{2}$ and ${I}_{0}$ respectively.

${I}_{0}$ is fairly easy, since $\int \left({\tan}^{0} x\right) \mathrm{dx} = \int 1 \mathrm{dx} = x + K$.

Using the formulae given $\left({I}_{n} = {\tan}^{n - 1} \frac{x}{n - 1} - {I}_{n - 2}\right)$, we can find everything else.

${I}_{2} = \tan x - {I}_{0} = \tan x - x - K$

${I}_{4} = {\tan}^{3} \frac{x}{3} - {I}_{2} = {\tan}^{3} \frac{x}{3} - \tan x + x + K$

${I}_{6} = {\tan}^{5} \frac{x}{5} - {I}_{4} = {\tan}^{5} \frac{x}{5} - {\tan}^{3} \frac{x}{3} + \tan x - x - K$

${I}_{8} = {\tan}^{7} \frac{x}{7} - {I}_{6} = {\tan}^{7} \frac{x}{7} - {\tan}^{5} \frac{x}{5} + {\tan}^{3} \frac{x}{3} - \tan x + x + K$

Therefore, $\int \left({\tan}^{8} x\right) \mathrm{dx} = {\tan}^{7} \frac{x}{7} - {\tan}^{5} \frac{x}{5} + {\tan}^{3} \frac{x}{3} - \tan x + x + K$