B Integrating sec(x): Understanding the Answer

[ForceBoy]

Hello.

I integrated $\sec(x)$ and got an answer. I differentiated it to verify it and it came out well. Later when I was looking in a table of integrals I saw the solution $\ln| \sec(x) +\tan(x) | + c$. This was completely different than my solution. I do not think I made a mistake (unless I'm wrongly convinced of something) and I can't see how my answer is related to the given one.

My work:

$I = \int\sec(x) dx$

$\sec x = \frac{1}{\cos x}$ and $\cos(x) = \frac{e^{ix}+e^{-ix}}{2}$ so

$I = \int \frac{2dx}{e^{ix}+e^{-ix}}$

$I = 2 \int \frac{dx}{e^{ix}+e^{-ix}}$

Multiplication by $\frac{e^{ix}}{e^{ix}}$ yields

$I = 2 \int \frac{e^{ix}dx}{e^{2ix}+1}$

Let $u = e^{ix}$
then $du = i u dx$
so $\frac{du}{iu} = dx$

The integral now:
$I = 2\int \frac{u}{u^{2}+1} \frac{du}{iu}$

$I = -2i\int \frac{1}{u^{2}+1}$

We know $\int \frac{1}{u^{2}+1} = arctan(x)+c$

So it follows that
$I = -2i \arctan(u) + c$

and finally that

$I = -2i \arctan(e^{ix}) + c$

Differentiation:

$\frac{d}{dx}(-2i \arctan(e^{ix}))$

$-2i \frac{d}{dx}(\arctan(e^{ix}))$

$t = ix$
$u = e^{t}$
$v = \arctan(u)$

so

$-2i ( \frac{dv}{du} * \frac{du}{dt} *\frac{dt}{dx})$

$-2i(\frac{1}{u^{2}+1}*u*i)$

$-2i( \frac{i e^{ix}}{e^{2ix}+1})$

$-2i(\frac{i e^{ix}}{e^{ix} (e^{ix} +e^{-ix})})$

$-2i(\frac{i}{e^{ix} +e^{-ix}})$

$\frac{2}{e^{ix} +e^{-ix}}$

$\sec(x)$

This is then clearly different than the mentioned solution. I don't see where I could've made a mistake. If I'm right, my solution would require more simplification. I don't see how this could be done.

Any thoughts would be appreciated. Thanks.

 

[mathman]

$\int \frac{1}{1+u^2}du=arctan(u)+c$ holds only for real u. Here $u=e^{ix}$ is complex.

 

[ForceBoy]

I understand. Would my next step then be to evaluate $\int \frac{1}{z^{2} +1} dz$ where z is a complex number?

 

[ForceBoy]

Also a question.

$\int \frac{1}{1+u^2}du=arctan(u)+c$ holds only for real u. Here $u=e^{ix}$ is complex.

Why is this? In the differentiation of arctan(x), where is it that it is assumed that x is real?

 

[mathman]

Technically you are correct. However arctan(u) when u is complex is not very useful. It is more useful with the form given (ln|...|)..

 

[ForceBoy]

I now saw how to evaluate $\int \frac{du}{u^2+1}$ for complex u. It's $\frac{i}{2}\ln|\frac{u+i}{u-i}|$ if I'm not wrong. If I use this instead of $arctan(u)$ I get:

$I = \ln|\frac{e^{ix}+i}{e^{ix}-i}|$

When looking at this, something tells me it can quickly be expressed as $\ln|\sec(x)+\tan(x)|$. Is this right?

 

[ForceBoy]

I just found out how to make it $\ln|\sec(x)+\tan(x)|$. I now understand this better and know how to integrate sec(x). Thank you for your help and time.