1∫ sin 4x cos 2x dx as we don't have any ready made formulae for product of function in integration, let's split it into sum or difference.
sin A cos B = 1/2 {sin (A + B ) + sin ( A - B )}
A= 4x, B = 2x. ( A + B ) =( 4x + 2x )= 6x. ( A - B ) = ( 4x - 2x )=2x
∴ ∫ sin 4x cos 2x dx =1/2 { ∫ sin 6 x + sin 2x dx }
=1/2{ ∫ sin 6x dx + ∫ sin 2x dx }
= 1/2 {( -1/6 cos 6x ) +(- 1/2 cos 2x )} + C. ( ∫ sin ax dx= - 1/a cos ax + C )
= - 1/2 { 1/6 cos 6x +1/2 cos 2x } + C
= - 1/2• 1/6 { cos 6x + 3 cos 2x } + C
= - 1/12 { cos 6 x + 3 cos 2x } + C
