Evaluating Integrals?

Question

Evaluate ∫2cos^2 x dx

I've gotten an answer, but I'm not 100% sure about it. Any help is appreciated :)

Answer 1

cos(2x) = 2cos^2 x - 1
2cos^2 x = 1 + cos(2x)
∫2cos^2 x dx = ∫1 + cos(2x)dx
= x + sin(2x)/2 + c
= x + sin(x)cos(x) + c

Answer 2

cos 2x = 2 cos ² x - 1
2.cos ² x = cos 2x + 1
I = ∫ (cos 2x + 1) dx
I = sin 2x / 2 + x + C

Answer 3

It is cos(x) * sin(x) + x

integral of cos^2(x) dx = (1/2)(x) + (1/4)sin(2x)
so
integral 2*cos^2(x) dx
2((1/2)(x) + (1/4)sin(2x)
x + (1/2)sin(2x)

and sin(2x) = 2 sin(x) cos(x), so it can be simplified to:
x + (1/2) sin(2x)
x + (2/2) sin(x) cos(x)
x + sin(x) cos(x)