# Integrating sec x dx

1. Oct 28, 2008

### JFonseka

1. The problem statement, all variables and given/known data
By multiplying the integrand sec x dx by $$\frac{tan x + sec x}{tan x + sec x}$$ find the integral of sec x dx

2. Relevant equations

d/dx sec x = tan x.sec x
d/dx tan x = sec^2 x

3. The attempt at a solution

sec x dx($$\frac{tan x + sec x}{tan x + sec x}$$) =>

$$\frac{tan x.sec x + sec^2 x}{tan x + sec x}$$dx

Just noticed the numerator is the derivative of the denominator, so =>

$$\frac{d(sec x + tan x)}{sec x + tan x}$$dx

Not sure what to do from here...

Last edited: Oct 28, 2008
2. Oct 28, 2008

### Dick

That's integral of du/u where u=sec(x)+tan(x). What's integral of du/u?

3. Oct 29, 2008

### JFonseka

Well...integrating the derivative would just return the original function wouldn't it? But in this case it's the reciprocal so it would be 1/sec x + tan x ???

4. Oct 29, 2008

### Dick

Nooo. Integral of du/u is log(u), isn't it?

5. Oct 29, 2008

### JFonseka

Oh yeaaaaa...I got confused. I always think of it as 1/x, not dx/x.
Thanks Richard!