f'(π/4)是常数
当x=π/4时
f'(π/4)=f'(π/4)cos(π/4)+sin(π/4)
f'(π/4)[1-cos(π/4)]=sin(π/4)
f'(π/4)=1/(√2-1)=√2+1
所以原式两边积分得
f(x)=(√2+1)sinx-cosx+C
f'(x)=f'(丌/4)cosx+sinx则f(x)?
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