# How do you find the antiderivative of e^(2x)*( tan(e^(2x)) )^2?

Oct 22, 2016

$\frac{1}{2} \tan \left({e}^{2 x}\right) - \frac{1}{2} {e}^{2 x} + C$

#### Explanation:

$I = \int {e}^{2 x} {\tan}^{2} \left({e}^{2 x}\right) \mathrm{dx}$

Let $u = {e}^{2 x}$, so that $\mathrm{du} = 2 {e}^{2 x} \mathrm{dx}$. Thus:

$I = \frac{1}{2} \int {\tan}^{2} \left({e}^{2 x}\right) \left(2 {e}^{2 x} \mathrm{dx}\right)$

$I = \frac{1}{2} \int {\tan}^{2} \left(u\right) \mathrm{du}$

We can integrate this using the identity ${\tan}^{2} \left(u\right) + 1 = {\sec}^{2} \left(u\right)$:

$I = \frac{1}{2} \int \left({\sec}^{2} \left(u\right) - 1\right) \mathrm{du}$

$I = \frac{1}{2} \int {\sec}^{2} \left(u\right) \mathrm{du} - \frac{1}{2} \int \mathrm{du}$

These are both common integrals:

$I = \frac{1}{2} \tan \left(u\right) - \frac{1}{2} u + C$

With $u = {e}^{2 x}$:

$I = \frac{1}{2} \tan \left({e}^{2 x}\right) - \frac{1}{2} {e}^{2 x} + C$