# How do I evaluate the indefinite integral int(tan^2(x)+tan^4(x))^2dx ?

##### 1 Answer
Jul 26, 2014

\int(\tan^2(x)+\tan^4(x))^2 dx=\int((tan^2 x(1+tan^2 x))^2 dx
$= \setminus \int {\tan}^{4} x \left(1 + {\tan}^{2} x\right) \left(1 + {\tan}^{2} x\right) \mathrm{dx}$
$= \setminus \int \left({\tan}^{4} x + {\tan}^{6} x\right) \left(1 + {\tan}^{2} x\right) \mathrm{dx}$
Now, note that $\left(\tan x\right) ' = {\sec}^{2} x = 1 + {\tan}^{2} x$ (see derivatives of trig functions ).

Thus, the substitution $u = \tan x$ yields $\mathrm{du} = \left(1 + {\tan}^{2} x\right) \mathrm{dx}$ and
$\setminus \int {\left(\setminus {\tan}^{2} \left(x\right) + \setminus {\tan}^{4} \left(x\right)\right)}^{2} \mathrm{dx} = \setminus \int {u}^{4} + {u}^{6} \mathrm{du} = {u}^{5} / 5 + {u}^{7} / 7 + C$
$= \frac{\setminus {\tan}^{5} x}{5} + \frac{\setminus {\tan}^{7} x}{7} + C$
using the power rule for integration (see, e.g, antiderivatives )