Re: [積分]

[chrisjon (布丁狗)]

※ 引述《king911015 (早已放棄愛上你)》之銘言：
: ∫sin^5 xdx
  ∫sin^5 xdx
= -∫sin^4 xdcosx
= -∫(1-cos^2 x)^2 xdcosx
= -∫1 - 2cos^2 x + cos^4 x dcosx
= -cosx + (2/3)cos^3 x -(1/5)cos^5 x + c

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