Using the linearity of the integral:
#int_(-1)^1 sqrt(2-x^2)-x^2dx = int_(-1)^1 sqrt(2-x^2)dx - int_(-1)^1 x^2dx#
The second integral can be solved directly:
# int_(-1)^1 x^2dx = [x^3/3]_(-1)^1 = 2/3#
For the first, solve the indefinite integral substituting #x= sqrt2 sint#. As the integrand is defined only for #x in [-sqrt2,sqrt2]# #t# varies in #[-pi/2,pi/2]#
#int sqrt(2-x^2)dx = int sqrt(2-2sin^2t) d(sqrt2sint)#
#int sqrt(2-x^2)dx = 2int sqrt(1-sin^2t)costdt#
As for #t in [-pi/2,pi/2]# #cost > 0# then: #sqrt(1-sin^2t) = cost#, so:
#int sqrt(2-x^2)dx = 2 int cos^2tdt#
and using the trigonometric identity: #2cos^2alpha = (1+cos 2 alpha)#
#int sqrt(2-x^2)dx = int (1+cos 2t) dt = t +(sin 2t)/2 = t+sintcost#
undoing the substitution:
#t = arcsin(x/sqrt2)#
#sint = x/sqrt2#
#cost = sqrt(1-(x/sqrt2)^2)#
#int sqrt(2-x^2)dx =arcsin(x/sqrt2)+x/sqrt2sqrt(1-(x/sqrt2)^2)#
#int sqrt(2-x^2)dx =arcsin(x/sqrt2)+(xsqrt(2-x^2))/2#
So:
#int_(-1)^1 sqrt(2-x^2)dx =arcsin(1/sqrt2)+(sqrt(2-1))/2 - arcsin(-1/sqrt2)-((-1)sqrt(2-1))/2#
#int_(-1)^1 sqrt(2-x^2)dx =pi/4+1/2 +pi/4+1/2 = 1+pi/2#
And finally:
#int_(-1)^1 sqrt(2-x^2)-x^2dx = 1+pi/2-2/3 = (2+3pi)/6#
