# What is int tan^2(2x) sec^4(2x) dx?

Apr 8, 2018

$I = \frac{1}{30} \left(3 {\tan}^{5} \left(2 x\right) + 5 {\tan}^{3} \left(2 x\right)\right) + C$

#### Explanation:

We want to solve

$I = \int {\tan}^{2} \left(2 x\right) {\sec}^{4} \left(2 x\right) \mathrm{dx}$

Rewrite the integrand using the trig identity

color(blue)(sec^2(a)=1+tan^2(a)

Thus

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

Make a substitution $u = \tan \left(2 x\right) \implies \mathrm{du} = 2 {\sec}^{2} \left(2 x\right) \mathrm{dx}$

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

$\textcolor{w h i t e}{I} = \frac{1}{2} \int {u}^{4} + {u}^{2} \mathrm{du}$

$\textcolor{w h i t e}{I} = \frac{1}{2} \left(\frac{1}{5} {u}^{5} + \frac{1}{3} {u}^{3}\right) + C$

$\textcolor{w h i t e}{I} = \frac{1}{30} \left(3 {u}^{5} + 5 {u}^{3}\right) + C$

Substitute back $u = \tan \left(2 x\right)$

$I = \frac{1}{30} \left(3 {\tan}^{5} \left(2 x\right) + 5 {\tan}^{3} \left(2 x\right)\right) + C$

Apr 8, 2018

$\int {\tan}^{2} \left(2 x\right) {\sec}^{4} \left(2 x\right) \mathrm{dx} = {\tan}^{3} \frac{2 x}{6} + {\tan}^{5} \frac{2 x}{10} + C$

#### Explanation:

$\int {\tan}^{2} \left(2 x\right) {\sec}^{4} \left(2 x\right) \mathrm{dx}$

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

if $u = \tan \left(2 x\right)$ then $\mathrm{du} = 2 {\sec}^{2} \left(2 x\right) \mathrm{dx} = 2 \left(1 + {u}^{2}\right) \mathrm{dx}$

or $\mathrm{dx} = \frac{\mathrm{du}}{2 \left(1 + {u}^{2}\right)}$

and $\int {\tan}^{2} \left(2 x\right) {\left(1 + {\tan}^{2} \left(2 x\right)\right)}^{2} \mathrm{dx}$

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

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

= $\frac{1}{2} \int \left({u}^{2} + {u}^{4}\right) \mathrm{du}$

= ${u}^{3} / 6 + {u}^{5} / 10 + C$

= ${\tan}^{3} \frac{2 x}{6} + {\tan}^{5} \frac{2 x}{10} + C$