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

1. Aug 25, 2008

### lkj-17

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

How can we compute it by hand?

2. Aug 25, 2008

### HallsofIvy

Staff Emeritus
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).

Last edited: Aug 26, 2008
3. Aug 25, 2008

### lkj-17

Any simple way to do it?

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

4. Aug 25, 2008

### 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:

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

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

5. Aug 25, 2008

### lkj-17

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

6. Aug 26, 2008

### NoMoreExams

Which integral are you stuck on? Show your work.

7. Aug 26, 2008

### tiny-tim

Hi lkj-17 and HallsofIvy!
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.

8. Aug 26, 2008

### HallsofIvy

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

9. Aug 26, 2008

### tiny-tim

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.

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