Using the basis

[1, 2(square root 3)t - (square root 3), 6(square root 5)(t^2 - t + 1/6)]

for R3[t] with the L^2[0, 1] inner product,

find the projection of the function sin(pie(x)) onto R3[t].

I am given hints as to what some integrals equal to facilitate the problem, but they are too complicated for me to type out on a computer... i'll try my best:

integral 0 to 1 of tsin(pie(t))dt = 1/pie

and the integral 0 to 1 of t^2sin(pie(t))dt = (pie^2 - 4)/pie^3

thanks in advance for your help!