Here's the question you clicked on:

## ammu123 Group Title integrate 6cos^3(3x) one year ago one year ago

• This Question is Closed
1. RupaBose Group Title

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

2. RupaBose Group Title

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

3. MrWho Group Title

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. sashankvilla Group Title

2⋅(sin(3x)−(sin(3x))^3/3)