Question #ec169

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Cem Sentin
Feb 14, 2018

#int_(-2)^2 dx/sqrt(4-x^2)=pi#

Explanation:

#int_(-2)^2 dx/sqrt(4-x^2)#

After using #x=2sinu# and #dx=2cosu*du# transforms, this integral became

#int_(-pi/2)^(pi/2) (2cosu*du)/(2cosu)#

=#int_(-pi/2)^(pi/2) du#

=#[u]_(-pi/2)^(pi/2)#

=#pi#

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