Re: [積分] 基本的三角積分

作者: Honor1984 (希望願望成真)   2014-04-09 16:57:09
※ 引述《kstmasa (雞排)》之銘言:
: http://ppt.cc/T1yt
: 如圖
: (抱歉,用打的我不太會)
∫[sin(x)]^2 dx
= ∫ [1 - cos(2x)]/2 dx
= x/2 - sin(2x)/4
∫ [sin(x/6)]^3 dx
= ∫[-1 + cos(x/6)]/2 * 6 dcos(x/6)
= -3cos(x/6) + 3/2 * [cos(x/6)]^2

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