# Thread: Very quick trig identity check

1. ## Very quick trig identity check

trying to take the integral of cos^2(3x)

but since I can take that integral, I need to change it up. just need to know if I'm doing this correctly.. I'm getting..

1/2 + 1/2cos(6x)

?

2. No

Let u = 3x du = 3dx

1/3integral(cos^2(u)) = 1/3 (u/2 +1/4sin(u))= x/2 +1/12[sin(3x)]

3. $\frac{1}{2}\int[1+cos6x]$

4. Of course my answer should have been 1/3 (u/2 +1/4 sin(2u) )=

x/2 +1/12sin(6x)