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RupaBose
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits{6(\cos^3 (3x)) dx}=6\int\limits\limits{(\cos^2(3x)*\cos(3x) dx}=6\int\limits\limits{(1\sin^2(3x))(\cos (3x) dx}=6\int\limits\limits{\cos(3x)\sin^2(3x)\cos(3x) dx}=6[\frac{ \sin(3x) }{ 3 } \int\limits{\sin^2(3x)\cos(3x) dx]=2\sin (3x)6[\frac{\sin^3(3x)}{9}]+C=2\sin(3x)+\frac{2}{3}\sin^2(3x)+C}\]

RupaBose
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits\limits{6(\cos^3 (3x)) dx}=6\int\limits\limits\limits{(\cos^2(3x)*\cos(3x) dx}\] \[=6\int\limits\limits\limits\limits\limits{(1\sin^2(3x))(\cos (3x) dx}\] \[=6\int\limits\limits\limits{\cos(3x)\sin^2(3x)\cos(3x) dx}\] \[=6[\frac{ \sin(3x) }{ 3 } \int\limits\limits{\sin^2(3x)\cos(3x) dx]}\] \[=2\sin (3x)6[\frac{\sin^3(3x)}{9}]+C\] \[=2\sin(3x)+\frac{2}{3}\sin^2(3x)+C\]

MrWho
 one year ago
Best ResponseYou've already chosen the best response.0If you couldn't find the integration just use the online integration solver and see the solution to have full grasp of the integration techniques.

sashankvilla
 one year ago
Best ResponseYou've already chosen the best response.02⋅(sin(3x)−(sin(3x))^3/3)
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