Efficiently Solve Tricky Integral: cos(x)/(1+e^x) with Odd Function Property

[Elysian]

Homework Statement

Find the integral of

\int\frac{cos(x)}{1+e^{x}}

Homework Equations

Given that

\frac{1}{1+e^{x}}-\frac{1}{2} is an odd function

The Attempt at a Solution

I tried integration by parts, with both u = cos(x) and u = 1+e^x, and both only complicated it even more. I've not learned infinite series and sequences so I'm not sure that's the way.

I did try writing cos(x) as \frac{e^{ix}+e^{-ix}}{2}, but I'm not sure what to do after that. I can see that there'd be some hyperbolic stuff in here but again, I'm not sure where exactly to start.

 

[LCKurtz]

Homework Statement

Find the integral of

\int\frac{cos(x)}{1+e^{x}}

Homework Equations

Given that

\frac{1}{1+e^{x}}-\frac{1}{2} is an odd function

I don't see how that is relevant unless you are doing a definite integral on an interval like [-a,a].

The Attempt at a Solution

I tried integration by parts, with both u = cos(x) and u = 1+e^x, and both only complicated it even more. I've not learned infinite series and sequences so I'm not sure that's the way.

I did try writing cos(x) as \frac{e^{ix}+e^{-ix}}{2}, but I'm not sure what to do after that. I can see that there'd be some hyperbolic stuff in here but again, I'm not sure where exactly to start.

I don't think there is an elementary antiderivative.

 

[Elysian]

I don't see how that is relevant unless you are doing a definite integral on an interval like [-a,a].


I don't think there is an elementary antiderivative.

Oh jeez I forgot to put the interval. its from -pi/2 to pi/2. I'm really sorry about that

 

[LCKurtz]

Oh jeez I forgot to put the interval. its from -pi/2 to pi/2. I'm really sorry about that

If $f(x)$ is odd what do you know about $\int_{-a}^a f(x)\, dx$?

 

[SammyS]

So, what is \displaystyle <br /> \int_{-\pi/2}^{\pi/2}\left(\frac{1}{1+e^{x}}-\frac{1}{2}\right)\cos(x)\,dx\ ?

 

[Elysian]

If $f(x)$ is odd what do you know about $\int_{-a}^a f(x)\, dx$?

It would equal 0. So as SammyS put it, it would be \displaystyle <br /> \int_{-\pi/2}^{\pi/2}\left(\frac{1}{1+e^{x}}-\frac{1}{2}\right)\cos(x)\,dx\ ?, where the first term would go to zero and then I'm left to evaluate the last bit which would be -\int_{-\pi/2}^{\pi/2}\frac{cos(x)}{2}?

So, what is \displaystyle <br /> \int_{-\pi/2}^{\pi/2}\left(\frac{1}{1+e^{x}}-\frac{1}{2}\right)\cos(x)\,dx

Is this just the integral of the right side then? So it's just -1?

 

[LCKurtz]

It would equal 0. So as SammyS put it, it would be \displaystyle <br /> \int_{-\pi/2}^{\pi/2}\left(\frac{1}{1+e^{x}}-\frac{1}{2}\right)\cos(x)\,dx\ ?, where the first term would go to zero and then I'm left to evaluate the last bit which would be -\int_{-\pi/2}^{\pi/2}\frac{cos(x)}{2}?

No. You don't evaluate an integral in pieces like that. You have $\frac{1}{1+e^{x}}-\frac{1}{2}$ multiplied by $\cos x$. Is $\cos x$ even, odd, or neither? What about its product with $\frac{1}{1+e^{x}}-\frac{1}{2}$?

 

[SammyS]

It would equal 0. So as SammyS put it, it would be \displaystyle <br /> \int_{-\pi/2}^{\pi/2}\left(\frac{1}{1+e^{x}}-\frac{1}{2}\right)\cos(x)\,dx\ ?, where the first term would go to zero and then I'm left to evaluate the last bit which would be -\int_{-\pi/2}^{\pi/2}\frac{cos(x)}{2}?

Remember, the hint says that \displaystyle \frac{1}{1+e^{x}}-\frac{1}{2}\ is odd (that entire expression).

How do you get \displaystyle \frac{1}{1+e^{x}}-\frac{1}{2}\ from \displaystyle \frac{1}{1+e^{x}}\ ?