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)

?

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- Apr 21st 2009, 06:37 PMcoolguy99Very 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)

? - Apr 21st 2009, 07:03 PMCalculus26
No

Let u = 3x du = 3dx

1/3integral(cos^2(u)) = 1/3 (u/2 +1/4sin(u))= x/2 +1/12[sin(3x)] - Apr 22nd 2009, 11:14 AMgammaman
$\displaystyle \frac{1}{2}\int[1+cos6x]$

- Apr 22nd 2009, 12:56 PMCalculus26
Of course my answer should have been 1/3 (u/2 +1/4 sin(2u) )=

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