# Very quick trig identity check

• April 21st 2009, 06:37 PM
coolguy99
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)

?
• April 21st 2009, 07:03 PM
Calculus26
No

Let u = 3x du = 3dx

1/3integral(cos^2(u)) = 1/3 (u/2 +1/4sin(u))= x/2 +1/12[sin(3x)]
• April 22nd 2009, 11:14 AM
gammaman
$\frac{1}{2}\int[1+cos6x]$
• April 22nd 2009, 12:56 PM
Calculus26
Of course my answer should have been 1/3 (u/2 +1/4 sin(2u) )=

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