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Solution to a trigonometric problem requested by one student

14/01/2017 0 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


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Karthik | 14/01/2017

Neatly explained.

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