# How do I find inttan^4x dx?

Mar 7, 2015

Well, this one is quite difficult;

$\int {\tan}^{4} \left(x\right) \mathrm{dx} = \int {\tan}^{2} \left(x\right) {\tan}^{2} \left(x\right) \mathrm{dx} =$
$= \int {\sin}^{2} \frac{x}{\cos} ^ 2 \left(x\right) {\tan}^{2} \left(x\right) \mathrm{dx} =$
$= \int \frac{1 - {\cos}^{2} \left(x\right)}{\cos} ^ 2 \left(x\right) {\tan}^{2} \left(x\right) \mathrm{dx} =$
$= \int \left[\frac{1}{\cos} ^ 2 \left(x\right) - 1\right] {\tan}^{2} \left(x\right) \mathrm{dx} =$
$= \int \left[{\tan}^{2} \frac{x}{\cos} ^ 2 \left(x\right) - {\tan}^{2} \left(x\right)\right] \mathrm{dx} =$
But $d \left[\tan \left(x\right)\right] = \frac{1}{\cos} ^ 2 \left(x\right) \mathrm{dx}$

$= \int {\tan}^{2} \left(x\right) d \left[\tan \left(x\right)\right] - \int {\tan}^{2} \left(x\right) \mathrm{dx} =$
$= {\tan}^{3} \frac{x}{3} - \int {\tan}^{2} \left(x\right) \mathrm{dx} =$
$= {\tan}^{3} \frac{x}{3} - \int {\sin}^{2} \frac{x}{\cos} ^ 2 \left(x\right) \mathrm{dx} =$
$= {\tan}^{3} \frac{x}{3} - \int \left[\frac{1 - {\cos}^{2} \left(x\right)}{\cos} ^ 2 \left(x\right)\right] \mathrm{dx} =$
$= {\tan}^{3} \frac{x}{3} - \left\{\int \frac{1}{\cos} ^ 2 \left(x\right) \mathrm{dx} - \int \mathrm{dx}\right\} =$
$= {\tan}^{3} \frac{x}{3} - \tan \left(x\right) - x + c$