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
