How do you integrate # (1-x^2)^.5#?

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Answer 1 · 2016-07-04T18:26:14

#= 1/2 x sqrt(1-x^2) + arcsin x/2 + C #

#int dx qquad sqrt(1-x^2)#

trig sub #x = sin phi, dx = cos phi dphi#

#implies int dphi qquad cos phi sqrt(1-sin^2 phi)#

#= int dphi qquad cos^2 phi #

double angle # cos 2A = 2 cos^2 A - 1#

#= int dphi qquad (cos 2 phi + 1)/2 #

#= 1/4 sin 2 phi + phi/2 + C #

double angle #sin 2A = 2 sin A cos A#

#= 1/2 x sqrt(1-x^2) + arcsin x/2 + C #