Question

integrate 6cos^3(3x)

Answer 1

\[\int\limits\limits{6(\cos^3 (3x)) dx}=6\int\limits\limits{(\cos^2(3x)*\cos(3x) dx}=6\int\limits\limits{(1-\sin^2(3x))(\cos (3x) dx}=6\int\limits\limits{\cos(3x)-\sin^2(3x)\cos(3x) dx}=6[\frac{ \sin(3x) }{ 3 } -\int\limits{\sin^2(3x)\cos(3x) dx]=2\sin (3x)-6[-\frac{\sin^3(3x)}{9}]+C=2\sin(3x)+\frac{2}{3}\sin^2(3x)+C}\]

Answer 2

\[\int\limits\limits\limits{6(\cos^3 (3x)) dx}=6\int\limits\limits\limits{(\cos^2(3x)*\cos(3x) dx}\]
\[=6\int\limits\limits\limits\limits\limits{(1-\sin^2(3x))(\cos (3x) dx}\]

\[=6\int\limits\limits\limits{\cos(3x)-\sin^2(3x)\cos(3x) dx}\]

\[=6[\frac{ \sin(3x) }{ 3 } -\int\limits\limits{\sin^2(3x)\cos(3x) dx]}\]

\[=2\sin (3x)-6[-\frac{\sin^3(3x)}{9}]+C\]

\[=2\sin(3x)+\frac{2}{3}\sin^2(3x)+C\]

Answer 3

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Answer 4

2⋅(sin(3x)−(sin(3x))^3/3)