Integration Assistence Please?

[Archaeopteryx]

Hello boards...

So, it's about 12 at night over here, and I'm being killed by an indefinite integral and just about out of options. Does anyone know what Int((e^(sin(.5x)))-1) dx would be?

Sine is in radians. Thanks so much!

[Tirian]

The online integration calculator I just checked suggests that most likely does not have a closed form.

[Archaeopteryx]

Hmmm, ok, so that means... what exactly? :smallfrown:

[Tirian]

Essentially that it doesn't have a solution. Are you sure that's the function you want to integrate?

[Archaeopteryx]

Sigh... yeah, I'm sure. Ok, thanks for your help. I'm going to call it a night.

[drakir_nosslin]

Wolframalpha (http://www.wolframalpha.com/input/?i=int%28%28e^%28sin%28.5x%29%29%29-1%29+dx) is your friend when it comes to problems like this :smallsmile:

And it does seem like Tirian was right, it only has definite solutions.

[Generic Archer]

Well, wolfram major (Mathematica) also agrees that it can't integrate it. Are you sure that we have the right equation here?

[Tirian]

I hope that the problem looks better in daylight. :smallsmile:

For the record, I was also checking Mathematica's online integrator. So even though we're three different votes, we're all relying on Wolfram's knowledge of mathematics. (Of course, if you wanted to trust a single online source for mathematical knowledge, Wolfram would be a very very fine choice.)

[Jack Squat]

I can't guarantee that this is right, since I've been out of calc for a couple years, but I worked out an answer. Just had to apply the law of logs first.

f`(x) = esin(.5x) - 1
esin(.5x) - 1 = 0
esin(.5x) = 1
ln (esin(.5x)) = ln (1)
sin(.5x) = 0

f`(x) = ∫ sin(.5x) dx
f(x) = -2cos(.5x) + C

[Tirian]

I'm afraid not. Setting f'(x)=0 just means that you've found the points at which f has a local maximum or minimum, which you then went on to show is everywhere that x=2k for some integer k. While it is true that cos(x/2) and any of its integer multiples (plus a constant) has that same quality, that's all that function would have in common with the actual indefinite integral of x.

[Jack Squat]

I figured there was something wrong. They don't really make us business majors keep up on any of the fun math.

[nerd-7i+42e]

I can't guarantee that this is right, since I've been out of calc for a couple years, but I worked out an answer. Just had to apply the law of logs first.

f`(x) = esin(.5x) - 1
esin(.5x) - 1 = 0
esin(.5x) = 1
ln (esin(.5x)) = ln (1)
sin(.5x) = 0

f`(x) = ∫ sin(.5x) dx
f(x) = -2cos(.5x) + C

Maybe I just haven't learned this in calc, but why are you setting f'=0?

Edit: Ninja'd!