The result of the integral of (1/2)sin2x dx with:(adsbygoogle = window.adsbygoogle || []).push({});

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...

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# Integral of (1/2)sin2x dx

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