whats integral of...?

Question

1/2(1-cos2x) dx

Answer 1

(-sin(2x) + 2x)/4 + C

Answer 2

write cos(2x) = 1 - 2sin^2 x
The denominator becomes
4*sin^2 x
so now you must integrate
1/4 * csc^2 (x)

That's a standard integral.
answer: -1/4 * cot(x) + constant

*EDIT*
This would of course be correct if hte 1-cos2x is in the denominator. The above answerer would be correct if it were inthe numerator. It's hard to tell from the way you wrote it.

Answer 3

it is really 2 integrals (B/C of addition/subtraction you can seperate them)
int of 1/2dx & int of -1/2 cos2xdx
1/2dx=> 1/2 X
-1/2 cos2xdx=> -1/2 * 1/2 sin2x = -1/4sin2x
which gives us=== 1/2x-1/4sin2x

Answer 4

I = (1/2).∫(1 - cos 2x).dx
I = (1/2).x - (1/2) ∫ cos(2x).dx
let u = 2x
du/dx = 2
dx = du / 2
I = (1/2).x - (1/4).∫ cos u.du
I = (1/2).x - (1/4).sin u + C
I = (1/2).x - (1/4).sin 2x + C

Answer 5

=(1/2)x -(sin2x)/4