Question

integration of sin^4 2x

Answer 1

sin4(2x)dxSubstitute u=2x ⟶ dx=12du (Steps):=12∫sin4(u)du
∫sin4(u)duApply reduction formula:∫sinn(u)du=n−1n∫sinn−2(u)du−cos(u)sinn−1(u)nwith n=4:=−cos(u)sin3(u)4+34∫sin2(u)du
apply product rule
∫sin2(u)duApply the last reduction formula again with n=2:=−cos(u)sin(u)2+12∫1du… or choose an alternative:
Apply product-to-sum formulasNow solving:∫1duApply constant rule:=uPlug in solved integrals:−cos(u)sin(u)2+12∫1du=u2−cos(u)sin(u)2Plug in solved integrals:−cos(u)sin3(u)4+34∫sin2(u)du=−cos(u)sin3(u)4−3cos(u)sin(u)8+3u8Plug in solved integrals:12∫sin4(u)du=−cos(u)sin3(u)8−3cos(u)sin(u)16+3u16Undo substitution u=2x:=−cos(2x)sin3(2x)8−3cos(2x)sin(2x)16+3x8The problem is solved:∫sin4(2x)dx=−cos(2x)sin3(2x)8−3cos(2x)sin(2x)16+3x8+CRewrite/simplify:=sin(8x)−8sin(4x)+24x64+C

i guess that helps you