# Integral Problem

1. Oct 11, 2003

### PrudensOptimus

Take the integral of:

sin3x, with respect to x. [-a,a] - interval

I end up getting 0/3:

= [-cos(3x)/3]

= [-cos(3a) - (- cos (-3a))]/3

= [-cos (3a) + cos (3a)]/3 ---> cos(-3a) = cos 3a

= 0/3 = 0

I think I did something wrong, right?

2. Oct 12, 2003

### Suicidal

No you are correct , sine is an odd function and integrating it over any interval symmetric about the origin will give you 0.