Probably, there are various ways to find the antiderivative(s) of #f(x)=sin(4x) cos(2x)#. I suggest integration by parts.
In fact, when trying and integrate the product of functions that exhibit cyclic derivatives, integration by parts is often a very powerful technique.
Recall that integration by parts comes from the chain rule and states the following: given #h(x)# and #k(x)# continuously differentiable functions, then
#int h(x) k'(x) dx = h(x) k(x) - int h'(x) k(x) dx#
In our case, we choose #h(x)=sin(4x)# and #k'(x)=cos(2x)#:
#int f(x) dx=sin(4x) sin(2x) / 2-int 4 cos(4x) sin(2x) / 2 dx=1/2 sin(4x) sin(2x) -2 int cos(4x)sin(2x) dx#
and by parts again on #2 int cos(4x)sin(2x) dx#, choosing #h(x)=cos(4x)# and #k'(x)=2sin(2x)#:
#int cos(4x)2sin(2x) dx=cos(4x) [-cos(2x)]-int [-4sin(4x)][- cos(2x)]dx=-cos(4x)cos(2x)-4 int sin(4x)cos(2x) dx=-cos(4x)cos(2x)-4 int f(x) dx#
Finally, substituting the expression obtained for #2 int cos(4x)sin(2x) dx# into the expression calculated for #int f(x) dx#, we get:
#int f(x) dx=1/2 sin(4x) sin(2x)+cos(4x)cos(2x)+4*int f(x) dx#
#3 int f(x) dx=-1/2 sin(4x) sin(2x)-cos(4x)cos(2x)+tilde{C}#
#int f(x) dx=-1/6 sin(4x) sin(2x)-1/3 cos(4x)cos(2x) + C#
This result can be rewritten in many ways using trigonometric identities.
