# How do you integrate int tan^5(2x)sec^2(2x)?

$\setminus \frac{\setminus {\tan}^{6} \left(2 x\right)}{12} + C$

#### Explanation:

Let $\setminus \tan \left(2 x\right) = u \setminus \implies 2 \setminus {\sec}^{2} \left(2 x\right) \setminus \mathrm{dx} = \mathrm{du}$

$\setminus \therefore \setminus \int \setminus {\tan}^{5} \left(2 x\right) \setminus {\sec}^{2} \left(2 x\right) \setminus \mathrm{dx}$

$= \setminus \int {u}^{5} \setminus \frac{\mathrm{du}}{2}$

$= \frac{1}{2} \setminus \int {u}^{5} \setminus \mathrm{du}$

$= \frac{1}{2} \setminus \cdot {u}^{6} / 6 + C$

$= \setminus \frac{\setminus {\tan}^{6} \left(2 x\right)}{12} + C$

Jul 27, 2018

${\tan}^{6} \frac{2 x}{12} + c$

#### Explanation:

We'll have to make a trigonometric substitution here, and we have tangents and secants, so we're looking for either a tangent with an odd power or a secant with an event power.

Since we have both in this case, it does not matter which one we choose for substitution. In this particular case, it seems like using $\tan \left(2 x\right)$ would be easier because we already have a ${\sec}^{2} \left(2 x\right)$.

So:

color(red)(u = tan(2x)
color(blue)(du = 2sec^2(2x)dx

Plugging that back into the original integral:

$\int {\tan}^{5} \left(2 x\right) {\sec}^{2} \left(2 x\right) \mathrm{dx}$

$\int \frac{2}{2} \cdot {\tan}^{5} \left(2 x\right) {\sec}^{2} \left(2 x\right) \mathrm{dx}$

int1/2*color(red)(tan)^5color(red)((2x))*color(blue)(2sec^2(2x)dx

int1/2*color(red)(u)^5color(blue)(du

$\frac{1}{12} {u}^{6} + c$

Substitute $u$ back in:

$\frac{1}{12} {\tan}^{6} \left(2 x\right) + c$ or ${\tan}^{6} \frac{2 x}{12} + c$