- #1

ForceBoy

- 47

- 6

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.