Let the original function of (Sint+Sint 2) be F(t), then f' (t) = Sint+Sint 2, so ∫(-x versus x) (Sint+Sint 2) dt = f (x)-f (-x) takes its derivative as f' (x+sinx2). When the limit ratio of x tends to 0 is 1, the limit lim(x tends to 0)[∫(-x vs. x) (Sint+Sint [(4xcosx2)/(AK (k-1) x (k-2))] (x except the following) =

k=3