Integral of Secant: Solving for \int \frac{1}{cosx} dx | Step-by-step Solution

[Sisyphus]

Homework Statement

Basically, I have to find

<br /> \int \frac{1}{cosx} dx <br />

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

<br /> let u = sin(x)du = cos(x)dx<br />

The Attempt at a Solution

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

 

[StatusX]

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)

 

[cristo]

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

 

[cristo]

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]!

 

[cristo]

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]!

That's the derivative of sec(x)