Re: [積分]

[GayerDior ( 未通過身分認證)]

※ 引述《king911015 (早已放棄愛上你)》之銘言：
: 已知微分式 f'(sin^2 x) = cos^2 x, 原式 f(x) = Ax^2 + Bx + C, 則 A = ?

f'(sin^2 x) = cos^2 x = 1 - sin^2 x


令t=sin^2 x


f'(t) = 1 - t => f(t) = t - t^2/2 + c

同理

f(x) = x - x^2/2 + c

       1
A = - ---
       2

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