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October 1, 2014

October 1, 2014

Posted by **Lee** on Saturday, January 7, 2012 at 4:52pm.

¡̉ dy/(rad£¨4-y^2))=

- Calculus -
**Lee**, Saturday, January 7, 2012 at 5:19pmEvaluate the indefinite integral

- Calculus -
**Steve**, Saturday, January 7, 2012 at 6:35pmlook 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

- Calculus - oops -
**Steve**, Saturday, January 7, 2012 at 6:37pmoops - too many 1/2's there. integral(du) = u = arcsin(y/2)+C

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