## anonymous 3 years ago integrate 6cos^3(3x)

1. anonymous

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

2. anonymous

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

3. anonymous

If you couldn't find the integration just use the online integration solver and see the solution to have full grasp of the integration techniques.

4. anonymous

2⋅(sin(3x)−(sin(3x))^3/3)