- #1
PrudensOptimus
- 641
- 0
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?
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?