Integration prob.

[JazzMan]

How do you integrate cos3x. Is there a substitution I should know?

[DaveD]

How do you integrate cos3x. Is there a substitution I should know?

Use the identity
cos2A ≡2cosA-1
cosA ≡(1+cos2A)/2
then ∫cos3x dx = 1/2 ∫(1+cos6x) dx
= 1/2 [ x + sin6x/6]+c

ok!

[JazzMan]

Thanks DaveD. I was wayout on what i was doing