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(sin3x)^3

[cos(3x)+isin(3x)]^3 = cos(9x) + isin(9x)

Expand the left side and compare real parts:

cos(9x) = cos^3(3x) - 3cos(3x)sin^2(3x) = -3cos(3x) + 4cos^3(3x) + c

So,

-3 cos(3x)/12 + cos(9x)/36

= (1/36)(-9cos(3x) -3cos(3x) + 4cos^3(3x) + c)

= -(1/3)cos(3x) + (1/9)cos^3(3x) + C

In addition, you can use mental substitution.

∫ (sin(x))^3 dx

= -∫ (sin(x))^2 dcos(x)

= ∫ (cos(x))^2 - 1 dcos(x)

= (1/3)(cos(x))^3 - cos(x)

So,

∫ (sin(3x))^3 dx

= (1/3)∫ (sin(3x))^3 d3x

= (1/9)(cos(3x))^3 - (1/3)cos(3x) + c