已知f´(sin^x)=cos2x+tan^x,当0

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已知f´(sin^x)=cos2x+tan^x,当0

数学人气:749 ℃时间:2020-08-29 04:35:34
优质解答
f'(sin^2x)=1-2sin^2x+sin^2x/(1-sin^2x)f'(x)=1-2x+x/(1-x)f(x)=∫[1-2x+x/(1-x)]dx=x-x^2+∫x/(1-x)dx=x-x^2-∫-x/(1-x)dx=x-x^2-∫[1+1/(x-1)]dx=x-x^2-x-ln(1-x)+C=-x^2-ln(1-x)+C
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f'(sin^2x)=1-2sin^2x+sin^2x/(1-sin^2x)f'(x)=1-2x+x/(1-x)f(x)=∫[1-2x+x/(1-x)]dx=x-x^2+∫x/(1-x)dx=x-x^2-∫-x/(1-x)dx=x-x^2-∫[1+1/(x-1)]dx=x-x^2-x-ln(1-x)+C=-x^2-ln(1-x)+C