Integrating sec x dx: Multiply by \frac{tan x + sec x}{tan x + sec x}

[JFonseka]

Homework Statement

By multiplying the integrand sec x dx by \frac{tan x + sec x}{tan x + sec x} find the integral of sec x dx

Homework Equations

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

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...

 

[Dick]

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

 

[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 ?

 

[Dick]

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

 

[JFonseka]

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