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intergrate cos^2 (3x) dx
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rewrite it as \[\frac{1}{2}\cos(6x)+1\]
Thanks, i appreciate it.
yw
wow i made a typo sorry
forgot parentheses, it should be \[\frac{1}{2}\left(\cos(6x)+1\right)\]
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using the identity \[\cos^2(x)=\frac{1}{2}\left(\cos(2x)+1\right)\] only with \(3x\) instead of \(x\)
you do sine the same way
don't worry i understood what you meant
k good
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