Verified answer

∫ sin^5(3x) cos^(1/2)(3x) dx

= ∫ sin^4(3x) * sin(3x) * cos^(1/2)(3x) dx

= ∫ (1 - cos^2(3x))^2 * cos^(1/2)(3x) * (sin(3x) dx)

= ∫ (1 - u^2)^2 * u^(1/2) * (-1/3) du, letting u = cos(3x)

= (-1/3) ∫ [u^(1/2) - 2u^(5/2) + u^(9/2)] du

= (-1/3) [(2/3)u^(3/2) - (4/7)u^(7/2) + (2/11)u^(11/2)] + C

= (-1/3) [(2/3)(cos(3x))^(3/2) - (4/7)(cos(3x))^(7/2) + (2/11)(cos(3x))^(11/2)] + C.

I hope this helps!