# Integral of (1/2)sin2x dx

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1. Apr 28, 2015

### J_M_R

The result of the integral of (1/2)sin2x dx with:

upper limit x = arccos((R-h)/R)
lower limit x = 0

is (-h^2+2Rh)/(2R^2)

I can not seem to get this exact answer my workings yield:

let u = 2x, du/dx = 2 therefore dx = du/2

Integral becomes (1/4) ∫ sinu du (with the same upper and lower limit)

= -1/4 (cos(2x)) (evaluated at the limits)

using cos2x = 2cos^2x-1 it becomes:

= -1/4 (2cos^2x-1) (evaluated at the limits)

= -1/4 ((2(cos*arccos ((R-h)/R))^2 - 1) - (2(cos*arccos(0))^2 -1)

= -1/4 (2((R-h)/R)^2 - 1 + 1)

= -1/2 ((R-h)/R)^2 = -1/2 - (h^2)/(2R^2) + h/R = (-h^2 + 2Rh - R^2)/(2R^2)

As you can see, my answer has an extra R^2 term in the answer and I cannot see where I have gone wrong...

2. Apr 28, 2015

### SteamKing

Staff Emeritus
Since your indefinite integral evaluates to -(1/4) * {2 cos2(x) - 1}, applying the limits of integration should be written as follows:

I = -(1/4) * {2 cos2 (arccos ((R-h)/R)) - 1 - [2 cos2 (arccos (0)) - 1]}

{Note: there is no '*' between the cos and the arccos. These are both functions which take arguments.}

3. Apr 28, 2015

### Staff: Mentor

The last +1 in the second line should be -1. There is some bracket error in the first line I think.