The integral can be reduced to the form with the substitution x(0) = - so that, in terms of 0, 64-x k = 64-x x k / sin" (0) sin" (0) cos" (0) d0, The substitution u(0) = reduces the integral to / R(u) du, where the rational function R(u) = The partial fraction expansion of R(u) is The integral [ R(u) du = 0, dx m = and n =