求f（x） f²（x）＝∫f（t）sint/（2+cost）dt上限x下限0

f²（x）=∫(0->x) f(t)sint/(2+cost) dt
2f(x) f'(x) = f(x) sinx/(2+cosx)
f'(x) = (1/2) sinx/(2+cosx)
f(x) = (1/2)∫ (0->x) [sint/(2+cost)] dt
= -(1/2) [ ln(2+cost)] (0->x)
= -(1/2) ( ln(2+cosx) - ln3)