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Mathematics 18 Online
OpenStudy (anonymous):

integral of integral sin^3(6x) cos^3 (6x)

OpenStudy (tkhunny):

\(\sin(2x) = 2\sin(x)\cos(x)\) -- That might come in handy. \(\cos^{2}(x) = 1 - \sin^{2}(x)\) -- That might be another way to go. Seriously, just try something. You won't break it.

OpenStudy (turingtest):

I would try\[\sin^3(6x)\cos^3(6x)=\sin^3(6x)\cos^2(6x)\cos(6x)\]\[=\sin^3(6x)[1-\sin^2(6x)]\cos(6x)=[\sin^3(6x)-\sin^5(6x)]\cos(6x)\]

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