Verified answer

Since the power of cosine is even, we need to use the power-reduction formulas to express cos^6(3x) in terms of the sums of cosines/sines with a power of 1.

We can use:

cos^2(x) = [1 + cos(2x)]/2.

Then:

cos^6(3x)

= [cos^2(3x)]^3

= {[1 + cos(6x)]/2}^3

= [1 + cos(6x)]^3/8

= [cos^3(6x) + 3cos^2(6x) + 3cos(6x) + 1]/8, by binomial expansion

= (1/8)cos^3(6x) + (3/8)cos^2(6x) + (3/8)cos(6x) + 1/8

= (1/8)cos^3(6x) + (3/16)[1 + cos(12x)] + (3/8)cos(6x) + 1/8

= (1/8)cos^3(6x) + (3/16)cos(12x) + (3/8)cos(6x) + 5/16.

To finish this off, we can use:

cos(A)cos(B) = [cos(A + B) + cos(A - B)]/2.

Then:

(1/8)cos^3(6x)

= 1/8 * cos^2(6x) * cos(6x)

= {[1 + cos(12x)] * cos(6x)}/16

= [cos(6x) + cos(12x)cos(6x)]/16

= [cos(6x) + (1/2)cos(12x + 6x) + (1/2)cos(12x - 6x)]/16

= (3/32)cos(6x) + (1/32)cos(18x).

Finally:

cos^6(3x)

= (1/8)cos^3(6x) + (3/16)cos(12x) + (3/8)cos(6x) + 5/16

= [(3/32)cos(6x) + (1/32)cos(18x)] + (3/16)cos(12x) + (3/8)cos(6x) + 5/16

= (1/32)cos(18x) + (3/16)cos(12x) + (15/32)cos(6x) + 5/16.

Integrating now yields:

∫ cos^6(3x) = (1/576)sin(18x) + (1/64)sin(12x) + (5/64)sin(6x) + (5/16)x + C.

I hope this helps!

∫cos^6(3x) dx

1/8*∫(1 + cos(6x))³ dx

1/8*∫(1 + 2*cos(6x) + cos²(6x))(1 + cos(6x)) dx

1/8*∫(1 + cos(6x) + 2*cos(6x) + 2*cos²(6x) + cos²(6x) + cos³(6x)) dx

1/8[x + 1/2*sin(6x) + 3/2*[x + 1/12*sin(12x] + sin(6x)/6 - sin³(6x)/18] + C