We have to remember a few formulas. Here, we will need #2sin(theta)cos(theta) = sin(2theta)#. We can make it appear easily because we're dealing with the squares of #sin(x)# and #cos(x)# and we're multiplying them by an even number.
#16sin^2(x)cos^2(x) = 4(4cos^2(x)sin^2(x)) = 4(2sin(x)cos(x))^2 = 4(sin(2x))^2#.
So #int16sin^2(x)cos^2(x)dx = 4intsin^2(2x)dx#.
And we know that #sin^2(theta) = (1-cos(2theta))/2# because #cos(2theta) = 1-2sin^2(theta)#, so #sin^2(2x) = (1 - cos(4x))/2#.
Hence the final result : #4intsin^2(2x) = 4int(1 - cos(4x))/2dx = 4intdx/2 - 4intcos(4x)/2dx = 2x - 2intcos(4x)dx = 2x + c - 2sin(4x)/4 + a# with #a,c in RR#. Let's say #k = a + c#, hence the final answer.
