# Integtrating secant and tangent

1. Feb 5, 2007

### Aerosion

1. The problem statement, all variables and given/known data

(integrate)sec^3(2x)*tan(2x)dx

2. Relevant equations

3. The attempt at a solution

Okay, so first I tried to separate secant and tangent, to get (integrate)1/cos^3(2x) * sin(2x)/cos(2x)dx. The u would be cos(2x), the du would be -du=sin(2x)dx.

I subsitute these in the aforementioned equation, whcih leaves (integrate)1/u^3 x 1/u. The problem is, I don't think this is right.

2. Feb 5, 2007

### IMDerek

u = 2x
du = 2 dx

$\frac 12 \int \sec ^3 u \tan u du$

Now let $v = \sec ^2 u$ What is dv?

3. Feb 6, 2007

### dextercioby

There's the mistake.

4. Feb 6, 2007

### Gib Z

Slight change to IMDereks post. I dont know wat dv is, cant be bothered working it out, i think my methods quicker. let v= sec u, split the integrand into sec^2 u (sec u tan u), that way the things in the brackets it dv/du, the integral is now integral: v^2 dv. simpler i think