Calculus
posted by Lee .
Evaluate£º
¡Ò dy/(rad£¨4y^2))=

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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