Integral of secant function

[BigGlenntheHeavy]

$\displaystyle Speaking \ of \ the \ secant, \ \int sec(x)dx \ = \ ln|sec(x)+tan(x)|+C, \ Prove \ it.$

 

[galactus]

$\displaystyle \int sec(x)dx$

$\displaystyle \int sec(x)\cdot\left(\frac{sec(x)+tan(x)}{sec(x)+tan(x)}\right)dx$

$\displaystyle =\int\frac{sec^{2}(x)+sec(x)tan(x)}{sec(x)+tan(x)}dx$

Let $\displaystyle u=sec(x)+tan(x), \;\ du=(sec^{2}(x)+sec(x)tan(x))dx$

Then, we get:

$\displaystyle \int\frac{1}{u}du$

$\displaystyle =ln(u)+C$

Resub:

$\displaystyle \boxed{ln(sec(x)+tan(x))+C}$

Of course, for the purists, we can throw in the absolute value.

 

[Deleted member 4993]

$\displaystyle Speaking \ of \ the \ secant, \ \int sec(x)dx \ = \ ln|sec(x)+tan(x)|+C, \ Prove \ it.$

It is very difficult (I would say impossible) to deduce that we should multiply by [sec(x)+tan(x)]/[sec(x)+tan(x)] without knowing the expected final result. Of course, in this case you have given the final result - you wanted to prove an identity. So we can attack the RHS and go:

d/dx [ln{|sec(x)+tan(x)}] = 1/[sec(x)+tan(x)] * [sec(x)*tan(x) + sec[sup:1q83zvwe]2[/sup:1q83zvwe](x)] = sec(x) ..... Q.E.D.

 

[galactus]

The previous method is the one usually shown in calc texts. It just uses a clever substitution to hammer it into something we can work with.

Here is another way I just thought up

$\displaystyle \int sec(x)dx$

$\displaystyle \int\frac{1}{cos(x)}dx$

Multiply top and bottom by cos(x):

$\displaystyle =\int\frac{cos(x)}{cos^{2}(x)}dx$

$\displaystyle =\int\frac{cos(x)}{1-sin^{2}(x)}dx$

Now, let $\displaystyle u=sin(x),\;\ du=cos(x)dx$

$\displaystyle \int\frac{1}{1-u^{2}}du$

$\displaystyle \frac{1}{2}\left[\int\frac{1}{1+u}+\int\frac{1}{1-u}\right]du$

$\displaystyle \frac{1}{2}\left[\int\frac{1}{u+1}-\int\frac{1}{u-1}\right]du$

$\displaystyle =\frac{1}{2}ln(\frac{u+1}{u-1})$

$\displaystyle \frac{1}{2}ln(\frac{sin(x)+1}{sin(x)-1})$

Here, I thought I was stuck. But hammering away at the identity, it does indeed work.

Continuing:

$\displaystyle \frac{sin(x)+1}{sin(x)-1}\cdot\frac{sin(x)+1}{sin(x)+1}=\frac{(sin(x)+1)^{2}}{sin^{2}(x)-1}=\frac{(sin(x)+1)^{2}}{-cos^{2}(x)}$

Because of the absolute value, we can shed the negative in the denominator in $\displaystyle -cos^{2}(x)$ and write:

$\displaystyle \frac{1}{2}ln\left(\frac{(sin(x)+1)^{2}}{cos^{2}(x)}\right)$

Property of logs:

$\displaystyle =ln\left(\frac{sin(x)+1}{cos(x)}\right)=ln(tan(x)+sec(x))$

Is that acceptable, SK?.

I was thinkin', perhaps another way would be to note that $\displaystyle sec(x)=\frac{2}{e^{ix}+e^{-ix}}$ and use complex integration.

 

[BigGlenntheHeavy]

Subhotosh Khan, I'm curious, what does this proof have to do with the Oxford English Dictionary (O.E.D.)?

 

[Deleted member 4993]

Subhotosh Khan, I'm curious, what does this proof have to do with the Oxford English Dictionary (OED)?

Actually it is Q[the letter that follows P].E.D.

which comes from (Latin) quod erat demonstrandum - which translates to "which was to be demonstrated".

Curiously, translated to Greek - Q.E.D. - does become OED (???? ???? ?????? ), according to my Greek friend.

 

[BigGlenntheHeavy]

Thank you Subhotosh Khan, I figured it probably was a typo, but then one never knows unless one asks.

By the way galactus, good show on your second proof - never saw that one before.