integrate sin^43x dx
sstarica the question is∫sin4 3x dxsubstitute(cos x)^2 = 1 - (sin x)^2∫(1−cos^2(3x))^2 dx = ∫(1−2cos^2u + cos^2u) du1/15 ∫ 5(3)sin^4 (3x) dx= -cos^5 (3x)/15 + C
