Find the integrals of the functions `cos2xcos4xcos6x`

Explanation

Given,`∫cos2xcos4xcos6xdx`

As `cos2xcos4x=1/2*(cos6x+cos2x)dx`
[by using property that cos A cos B=cos(A+B)/2+cos(A-B)/2]

so the equation becomes

`∫cos2xcos4xcos6xdx`

=`∫1/2*(cos6x+cos2x)cos6xdx`

=`1/2∫(cos6x+cos2x)cos6xdx`

=`1/2∫(cos6x.cos6x+cos2x.cos6x)dx`

As=`∫cos6x.cos6x=1/2[cos(12x)+1]dx`

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