Integrate sin^43x dx

Sstarica the question is ∫ sin4 3x dx

substitute (cos x) ^2 = 1 - (sin x) ^2

∫ (1 - cos^2 (3x)) ^2 dx = ∫ (1-2 cos^2 u + cos^2 u) d u

1/15 ∫ 5 (3) sin^4 (3x) d x

= - cos^5 (3x) / 15 + C