{[3sin2x+4(cosx^2]}/2?

Question

how do you find the definate integral of {[3sin2x+4(cosx^2]}/2 with limits o to pi/6

Answer 1

I'm not sure whether you mean 4 (cos x)^2 or 4 cos (x^2). I'll assume the first because it's much easier (and therefore I think much more likely to be the one).

∫(0 to π/6) [3 sin 2x + 4 cos^2 x]/2 dx
= (3/2) ∫(0 to π/6) sin 2x dx + 2 ∫(0 to π/6) cos^2 x dx
= (3/2) [-cos 2x / 2] [0 to π/6] + 2 ∫(0 to π/6) (1 + cos 2x) / 2 dx
= (3/4) [-cos π/3 + cos 0] + [x + sin 2x / 2] [0 to π/6]
= (3/4) [-1/2 + 1] + [(π/6 + (1/2) sin π/3) - (0 + (1/2) sin 0)]
= 3/8 + π/6 + √3 / 4