integral of sin^3 3x cos 3x dx?

Answer 1

integral of sin^3 3x cos 3x dx
= integral of (1/3)sin^3 3x d sin 3x
= (1/12)sin^4(3x) + c
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Reason: Used mental substitution because cos 3x dx = (1/3) d sin 3x.

Answer 2

I'll use a double substitution (although it's not necessary):
∫ sin^3 3x cos 3x dx
u = 3x
du = 3 dx (or dx = du/3)

∫ sin^3 3x cos 3x dx = (1/3) ∫ sin^3 u cos u du
v = sin u
dv = cos u du

(1/3) ∫ sin^3 u cos u du = (1/3) ∫ v^3 dv
(1/3)(v^4)/4 + C
(1/12)v^4 + C

now resubstitute:
(1/12)(sin^4 u) + C
(1/12)(sin^4 3x) + C

Answer 3

Let u = sin(3x)
then du = 3cos(3x)dx
cos(3x)dx = (1/3)du
and
∫sin^3(3x)cos(3x)dx
becomes
(1/3)∫u^3du =
(1/12)u^4 + C
Substituting back,
∫sin^3(3x)cos(3x)dx =
(1/12)sin^4(x) + C

Answer 4

Just use the substitution
u = sin(3x)
It will work itself out.

Answer 5

Int[Sin^2 3x. Sin 3x. Cos 3x ]dx = 0.5 Int[Sin^2 3x.Sin 6x]dx
=0.25 Int [(1-Cos 6x)Sin 6x]dx
=0.25Int[Sin6x]dx - 0.125 Int[Sin12x]dx
=-{0.25. Cos 6x}/6 + {0.125 Cos 12x}/12 + C