By using the substitution, x = 2sin(y) find the exact value of integral sqrt(1/3(4-x^2)) dx with limits 0 and 1.

In order to calculate this integral we must use the sustitution provided. x=2siny. Firtsly I will differentiate to find the dx component of the integral, so dx/dy=2cosy hence, dx=2cosydy. Now for the limits of the integral. For x=0 => 2siny=0 so y=0. For x=1 => 2siny=1 so y=arcsin(1/2) so y=pi/6. now substiture x=2siny into sqrt(1/3(4-x^2)) dx => sqrt(1/3(4-4sin^2(y)))(2cosydy) using our trigonometry identities we know cos^2(x)+sin^2(x)=1 => sqrt(4/3cos^2(y)))(2cosydy)= 4/sqrt(3)cos^2(y)dy = 2/sqrt(3)(cos(2y) + 1) dy
Now integrating that function yields 2/sqrt(3)*[1/2sin(2y)+y] from y=0 to y=pi/6 = 2/sqrt(3) [ 1/2 *sqrt(3)/2 + pi/6 - 0 ] =  1/2 + pi/(3sqrt(3))