Why is cos(t)(cos(wt) - jsin(wt)) considered negligible in integration?

[Gliese123]

Homework Statement

Hi,

I'm having a problem comprehending the odd-even trigonometry properties when doing an integration and I hope someone here feel like explaining since I can't seem to find anything of this in my course literature.
I suppose it's more or less of a integration problem.

f(t) = (cost)(Θ(t + \frac{\pi}{2}) - Θ(t - \frac{\pi}{2}))

Homework Equations

The Attempt at a Solution

When doing the integration of:
\int (cost)(coswt - jsinwt) dt from -pi/2 \rightarrow +pi/2
(The period is pi and the function is obviously even)

Why do they then consider cost * -jsinwt to be negligible and equal to 0?

http://www.wolframalpha.com/input/?i=integrate+cost*e^iwt+dt+from+-pi/2+to+pi/2

 

[PeroK]

Homework Statement

Hi,

I'm having a problem comprehending the odd-even trigonometry properties when doing an integration and I hope someone here feel like explaining since I can't seem to find anything of this in my course literature.
I suppose it's more or less of a integration problem.

f(t) = (cost)(Θ(t + \frac{\pi}{2}) - Θ(t - \frac{\pi}{2}))

Are you sure this is correct?

$Θ(t + \frac{\pi}{2}) - Θ(t - \frac{\pi}{2}) = Θ \pi$

 

[Gliese123]

Are you sure this is correct?

$Θ(t + \frac{\pi}{2}) - Θ(t - \frac{\pi}{2}) = Θ \pi$

Yes. The function stretches from -pi/2 --> pi/2
It's cut off

 

[gopher_p]

Homework Statement

Hi,

I'm having a problem comprehending the odd-even trigonometry properties when doing an integration and I hope someone here feel like explaining since I can't seem to find anything of this in my course literature.
I suppose it's more or less of a integration problem.

f(t) = (cost)(Θ(t + \frac{\pi}{2}) - Θ(t - \frac{\pi}{2}))

Homework Equations

The Attempt at a Solution

When doing the integration of:
\int (cost)(coswt - jsinwt) dt from -pi/2 \rightarrow +pi/2
(The period is pi and the function is obviously even)

Why do they then consider cost * -jsinwt to be negligible and equal to 0?

http://www.wolframalpha.com/input/?i=integrate+cost*e^iwt+dt+from+-pi/2+to+pi/2

1. $\cos t(\cos\omega t - j\sin\omega t)$ is neither even nor odd (w.r.t. $t$), and its periodicity (as a function of $t$) depends on $\omega$.

2. $\cos\omega t-i\sin\omega t=e^{-i\omega t}$

3. $\cos t\cdot (-j)\sin\omega t$ is an odd function (w.r.t. $t$), so $\int_{-a}^a \cos t\cdot (-j)\sin\omega t\ dt=0$ for any real $a$. In particular, $\int_{-\pi/2}^{\pi/2} \cos t\cdot (-j)\sin\omega t\ dt=0$.

Are you sure this is correct?

$Θ(t + \frac{\pi}{2}) - Θ(t - \frac{\pi}{2}) = Θ \pi$

I would assume, from context, that $\Theta$ is a function of $t$ here.

 

[Gliese123]

1. $\cos t(\cos\omega t - j\sin\omega t)$ is neither even nor odd (w.r.t. $t$), and its periodicity (as a function of $t$) depends on $\omega$.

2. $\cos\omega t-i\sin\omega t=e^{-i\omega t}$

3. $\cos t\cdot (-j)\sin\omega t$ is an odd function (w.r.t. $t$), so $\int_{-a}^a \cos t\cdot (-j)\sin\omega t\ dt=0$ for any real $a$. In particular, $\int_{-\pi/2}^{\pi/2} \cos t\cdot (-j)\sin\omega t\ dt=0$.



I would assume, from context, that $\Theta$ is a function of $t$ here.

Thanks. Seems logical :)