Question #c71e4

1 Answer
Aug 7, 2017

#int_(-1)^1 sqrt(2-x^2)-x^2dx = (2+3pi)/6#

Explanation:

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#