As the integrand function is defined for #x in [-1,1]#, you can substitute:
#x= sint # with #t in [-pi/2,pi/2]#
#dx = costdt#
so the integral becomes:
#int sqrt(1-x^2)dx = int sqrt(1-sin^2t)costdt = int sqrt(cos^2t) cost dt#
In the given interval #cost# is positive, so #sqrt(cos^2t) = cost#:
#int sqrt(1-x^2)dx = int cos^2t dt#
Now we can use the identity:
#cos^2t = (1+cos(2t))/2#
#int sqrt(1-x^2)dx = int (1+cos(2t))/2 dt = int (dt)/2 + 1/4 int cos(2t)d(2t)= 1/2t+1/4sin2t = 1/2(t+sintcost)#
To substitute back #x# we note that:
#x = sint# for #t in [-pi/2,pi/2] => t = arcsinx#
#cost = sqrt(1-sin^2t) = sqrt(1-x^2)#
Finally:
#int sqrt(1-x^2)dx = 1/2(arcsinx + xsqrt(1-x^2)) + C#
