Trapezoidal, Midpoint, Simpson's Part II. Am I correct?

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of *n. (Round your answers to six decimal places.) In radians. 6 sin x*^{2} dx, n =4

**T= (1/16) [6sin(0)+2*6sin((1/8)^2)+2*6sin((1/4)^2)+2*6sin((3/8)^2)+6sin((1/2)^2)]**

=.343469 (marked wrong)

EDIT: I forgot to square the last term. My answer is: .256461

M= (1/8)[6sin((1/16)^2)+6sin((3/16)^2)+6sin((5/16)^2)+6sin((7/16)^2)]

=.245097 marked correct

S=(1/24)[6sin(0)+4*6sin((1/8)^2)+2*6sin((1/4)^2)+4*6sin((3/8)^2)+6sin((1/2)^2)]

=.248867 marked correct

Is my work correct? Anyone know of an online program/calculator that would give me radians if I plug this in? I dont have a radian calculator since we don't use calculators during tests...

edit: I used wolfram alpha to calculate the radians. Part 1 (Trapezoidal) was marked incorrect.