# Integral of secant

Sisyphus

## Homework Statement

Basically, I have to find

$$\int \frac{1}{cosx} dx$$

by multiplying the integrand by $$\frac{cosx}{cosx}$$

I go through and arrive at a solution, but when I differentiate it,
I get -tan(x)

something's clearly wrong, but I can't see what it is that I'm doing wrong here...

## Homework Equations

$$let u = sin(x)$$

$$du = cos(x)dx$$

## The Attempt at a Solution

$$\int \frac{1}{cosx} dx = \int \frac{cosx}{cos^2x} dx\\ = \int \frac{cosx}{1-sin^2x} dx\\ =\int \frac{du}{1-u^2} \\ =\int \frac{du}{(1-u)*(1+u)} \\ =\frac{1}{2} * \int \frac {1}{1+u} + \frac {1}{1-u} du\\ = \frac{1}{2} * (ln(1-u^2}}) =\frac{1}{2} * (ln(cos^2))$$

Last edited:

Homework Helper
Your second to last step (where you actually perform the integration) is wrong. In the second term, remember that u has a minus sign.

Sisyphus
hmm..

I've never run into anything like this before

so why does u having a minus sign in front of it pose a problem with what I did in my original solution?

(thanks for the help)

Staff Emeritus
Well, the value of your integral will be ln(1+u)-ln(1-u)

Sisyphus
if I kept the minus sign where it was, which is what I did in my original solution, my integral would've been
$$ln(1+u)+ln(1-u)=ln(1-u^2)$$

StatusX told me to watch out for the negative sign in front of the u, so I factored it out, made sure that u was positive, and then integrated it, which gave me the correct solution.

the thing is I'm not sure why I couldn't proceed as usual with the minus sign in front of the u

Staff Emeritus
StatusX said to note the minus sign in front of the u. You can proceed as normal, but noting that $$\int\frac{1}{1-u}du=-ln(1-u)$$. In general $$\int\frac{1}{f(u)}du=\frac{ln[f(u)]}{df/du}$$. In this case, f(u)=1-u, and so df/du=-1

Sisyphus
ah, ok

thank you!

silver-rose i didn't know we would utilizie such a method to do this integral.

i've always thought the integral of secant was just sec[x]tan[x]!

Staff Emeritus i didn't know we would utilizie such a method to do this integral.