f'(x)=1-2(sinθ/2)^2,求原函数f(x)
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f''(x)=1-2(sinθ/2)^2,求原函数f(x)f''(x)=1-2(sinθ/2)^2,求原函数f(x)f''(x)=1-2(sinθ/2)^2,求原函数f(x)不妨用两倍角公式先把f''(x)改
f'(x)=1-2(sinθ/2)^2,求原函数f(x)
f'(x)=1-2(sinθ/2)^2,求原函数f(x)
f'(x)=1-2(sinθ/2)^2,求原函数f(x)
不妨用两倍角公式先把f'(x)改写成f'(x)=cosx,那么f(x)=sinx+C
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f'(x)=1-2(sinθ/2)^2,求原函数f(x)
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