# How do you find the integral of tan^3(2x) sec^100(2x) dx?

Mar 23, 2018

$I = \frac{1}{204} {\sec}^{102} \left(2 x\right) - \frac{1}{200} {\sec}^{100} \left(2 x\right) + C$

#### Explanation:

We want to solve

$I = \int {\tan}^{3} \left(2 x\right) {\sec}^{100} \left(2 x\right) \mathrm{dx}$

Make a substitution $u = 2 x \implies \frac{\mathrm{du}}{\mathrm{dx}} = 2$

$I = \frac{1}{2} \int {\tan}^{3} \left(u\right) {\sec}^{100} \left(u\right) \mathrm{du}$

Use the identity color(green)(tan^2(x)=sec^2(x)-1

$I = \frac{1}{2} \int \tan \left(u\right) \left({\sec}^{2} \left(u\right) - 1\right) {\sec}^{100} \left(u\right) \mathrm{du}$

Make a substitution $s = \sec \left(u\right) \implies \frac{\mathrm{ds}}{\mathrm{du}} = \sec \left(u\right) \tan \left(u\right)$

$I = \frac{1}{2} \int \left({s}^{2} - 1\right) {s}^{99} \mathrm{du}$

$\textcolor{w h i t e}{I} = \frac{1}{2} \int {s}^{101} - {s}^{99} \mathrm{du}$

$\textcolor{w h i t e}{I} = \frac{1}{204} {s}^{102} - \frac{1}{200} {s}^{100}$

Substitute back $s = \sec \left(u\right)$ and $u = 2 x$

$I = \frac{1}{204} {\sec}^{102} \left(2 x\right) - \frac{1}{200} {\sec}^{100} \left(2 x\right) + C$