How to evaluate sin(t-tau)cos(tau) wrt tau(0..t)?

[lkj-17]

How to integrate sin(t-tau)cos(tau)d(tau) where tau = 0..t

Maple gives answer (1/2)*t*sin(t)

How can we compute it by hand?

[HallsofIvy]

I would start with a trig identity. sin(t- tau)= sin(t)cos(tau)- cos(t)sin(tau). Now the integral is
\int_0^t (sin(t)cos^2(tau)- cos(t)cos(tau)sin(tau)) dtau
= sin(t)\int_0^t cos2(tau)dtau- cos(t)\int_0^t cos(tau)sin(tau)dtau
You can integrate the first by using cos2(tau)= (1/2(1+ cos(2tau)) and in the second, let u= sin(tau).

[lkj-17]

I would start with a trig identity. sin(t- tau)= sin(t)cos(tau)- cos(t)sin(tau). Now the integral is
\int_0^t (sin(t)cos^2(tau)- cos(t)cos(tau)sin(tau)) dtau[/iex]
[itex]= sin(t)\int_0^t cos2(tau)dtau- cos(t)\int_0^t cos(tau)sin(tau)dtau
You can integrate the first by using cos2(tau)= (1/2(1+ cos(2tau)) and in the second, let u= sin(tau).

Any simple way to do it?

If this happened in the exam, it will be a disaster!

[Ben Niehoff]

You should learn the trig identities well enough so that they ARE simple to you. Then you can apply them in exams.

If you forget the sum-of-angles formula, you can easily derive it from Euler's formula:

e^{i(\theta + \phi)} = \cos (\theta + \phi) + i \sin (\theta + \phi)

e^{i\theta} e^{i\phi} = \cos (\theta + \phi) + i \sin (\theta + \phi)

(\cos \theta + i \sin \theta) (\cos \phi + i \sin \phi) = \cos (\theta + \phi) + i \sin (\theta + \phi)

(\cos \theta \cos \phi - \sin \theta \sin \phi) + i (\cos \theta \sin \phi + \sin \theta \cos \phi) = \cos (\theta + \phi) + i \sin (\theta + \phi)

from which you can read off the two identities:

\cos (\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi

\sin (\theta + \phi) = \cos \theta \sin \phi + \sin \theta \cos \phi

[lkj-17]

You should learn the trig identities well enough so that they ARE simple to you. Then you can apply them in exams.

If you forget the sum-of-angles formula, you can easily derive it from Euler's formula:

e^{i(\theta + \phi)} = \cos (\theta + \phi) + i \sin (\theta + \phi)

e^{i\theta} e^{i\phi} = \cos (\theta + \phi) + i \sin (\theta + \phi)

(\cos \theta + i \sin \theta) (\cos \phi + i \sin \phi) = \cos (\theta + \phi) + i \sin (\theta + \phi)

(\cos \theta \cos \phi - \sin \theta \sin \phi) + i (\cos \theta \sin \phi + \sin \theta \cos \phi) = \cos (\theta + \phi) + i \sin (\theta + \phi)

from which you can read off the two identities:

\cos (\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi

\sin (\theta + \phi) = \cos \theta \sin \phi + \sin \theta \cos \phi

Unable to obtain (1/2)*t*sin(t) from this method.
Would you please show a little steps?

[NoMoreExams]

Which integral are you stuck on? Show your work.

[tiny-tim]

I would start with a trig identity. sin(t- tau)= sin(t)cos(tau)- cos(t)sin(tau).

Hi lkj-17 and HallsofIvy! :smile:
Any simple way to do it?

If this happened in the exam, it will be a disaster!

Do it in one go with sinA cosB = (sin(A+B) + sin(A-B))/2:

∫sin(t-tau)cos(tau)dtau = ∫(sint + sin(t-2tau))dtau/2. :smile:

[HallsofIvy]

Yes, that was what was originally suggested. Apparently ikj-17 feels he is not capable of doing that.

[tiny-tim]

Yes, that was what was originally suggested.

No … the original suggestion was sin(A-B) = sinAcosB - cosAsinB, which is different, and only dealt with half the expression.

sinA cosB = (sin(A+B) + sin(A-B))/2 deals with the whole thing in one go, and avoids all those nasty squares. :biggrin:

ikj-17 may even be able to integrate ∫(sint + sin(t-2tau))dtau/2 from 0 to t just by looking at it! :wink: