Posted by Lee on Saturday, January 7, 2012 at 4:52pm.
Evaluate the indefinite integral
look into trig substitutions
let y = 2sinu
then 4 - y^2 = 4 - 4sin^2 u = 4 cos^2 u
and √(4 - y^2) = 2cos u
dy = 2 cosu du
and your integral is just
1/(2 cosu) * 2 cosu du = 1/2 du
integral(1/2) du = u/2 = (1/2) arcsin(y/2) + C
oops - too many 1/2's there. integral(du) = u = arcsin(y/2)+C
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