∫(0,π/2)[f(cosx)cosx-f'(cosx)sin^2x]dx=
∫(0,π/2)d[sinxf(cosx)]
=sinxf(cosx)|(0,π/2)
=1*f(0)-0*f(1)
=f(0)
=1
设f(x)导数在【-1,1】上连续,且f(0)=1,计算∫【f（cosx）cosx-f‘（cosx）sin^2x】dx（
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