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ammu123

  • 2 years ago

integrate 6cos^3(3x)

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

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

  3. MrWho
    • 2 years ago
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    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
    • 2 years ago
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    2⋅(sin(3x)−(sin(3x))^3/3)

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