# How does this integral become zero?

1. Feb 17, 2010

### Peon666

I=1/2 [(integral sign) cos(n+m)wt dt + (integral sing) cos (n-m)wt dt]

"Because cos wt executes one complete cycle during any interval of duration T, cos (n+m)wt executes (n+m) complete cycles during any interval of duration T. Therefore the first integral in the above equation, which represents the area under n+m complete cycles of a sinusoid, equals zero. The same argument shows that the second integral in the above equation is also zero, except when n=m. Hence I in the above equation is zero for all n is not equal to m."

My question is: How do these integrals become zero?

Last edited: Feb 17, 2010
2. Feb 17, 2010

### CompuChip

It's just what the book says: if you integrate sin(x) or cos(x) over an integer number of periods, the integral is zero. If you let ,+{x,+0,+2pi}]]Wolfram Alpha calculate the integral, you can graphically see why in the plot: the blue and pink areas are equal (because of the symmetry around the x-axis), and because they are on different sides of the axis, they cancel.

If you want, you can work it out explicitly: the period of cos(w x) is 2 pi / w, so if you integrate over a complete period
$$\int_0^{2\pi/w} \cos(wt) \, dt$$
you will find 0. You can even make it completely general and integrate
$$\int_a^{a + k \cdot 2\pi/w} \cos(wt) \, dt$$
with a an arbitrary number and k an integer (k = 0, 1, 2, ..., -1, -2, ...).

3. Feb 17, 2010

### Peon666

Got it! Thanks a lot.