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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No Let u = 3x du = 3dx 1/3integral(cos^2(u)) = 1/3 (u/2 +1/4sin(u))= x/2 +1/12[sin(3x)]
$\displaystyle \frac{1}{2}\int[1+cos6x]$
Of course my answer should have been 1/3 (u/2 +1/4 sin(2u) )= x/2 +1/12sin(6x)
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