Oh i need help finding an integral

cos^4 3xdx .. I'm using the half-angle identity but then i gget lost. :(

Did you mean cos^4 (3x)?

cos^4 (3x)dx

let u = 3x, du / 3 = dx

= (1/3)cos^4 (u) du..........I am going to omit the factor of (1/3) for convenience from now on and put it in at the end

= cos^2(u) cos^2(u) du

= cos^2(u) (1 - sin^2(u) ) du

= cos^2(u) du - cos^2(u) sin^2(u) du

cos^2(u) = (1/2)(1 + cos(2u))
sin^2(u) = (1/2)(1 - cos(2u))

so the above becomes

= (1/2)(1 + cos(2u)) - (1/4)(1 + cos(2u))(1 - cos(2u))

= (1/2)(1 + cos(2u)) - (1/4)(1 + cos^2(2u))

= (1/2)(1 + cos(2u)) - (1/4)(1 + (1/2){1 + cos(4u)})

= (1/2) + (1/2)cos(2u) - (1/4) + (1/8) + (1/8)cos(4u)

integrate

= u / 2 + (1/4)sin(2u) - u / 4 + u / 8 + (1/32)sin(4u) + C

= 3u / 8 + (1/4)sin(2u) + (1/32)sin(4u) + C

Recall I omitted a factor of 1/3 early on, put that back in

= 3u/24 + (1/12)sin(2u) + (1/96)sin(4u) + C

and recall u = 3x,

= (9/24)x + (1/12)sin(6x) + (1/96)sin(12x) + C........[Ans.]

Note, answers may look different because of how easy it is to rewrite answers in one form or another using identities like the one you suggested. If you are not sure if an answer is the same, try and use identities to show one answer is the same as another.