標題:
∫e^(ax)cos(bx+c)dx=?
發問:
∫e^(ax)cos(bx+c)dx=? 請列詳細步驟。 更新: *** Integrate this by part! 更新 2: To 翻雷滾天 風卷殘雲: Can you please state the process from step 1 to step 2?
最佳解答:
圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Dec11/Crazyint4.jpg aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa 2011-12-06 00:42:01 補充： Step 1 to step 2 is using the rule: d(e^ax)/dx = ae^ax Hence d(e^ax) = ae^ax dx, giving d(e^ax)/a = e^ax dx
其他解答:
∫e^(ax)cos(bx+c)dx=?
發問:
∫e^(ax)cos(bx+c)dx=? 請列詳細步驟。 更新: *** Integrate this by part! 更新 2: To 翻雷滾天 風卷殘雲: Can you please state the process from step 1 to step 2?
最佳解答:
圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Dec11/Crazyint4.jpg aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa 2011-12-06 00:42:01 補充： Step 1 to step 2 is using the rule: d(e^ax)/dx = ae^ax Hence d(e^ax) = ae^ax dx, giving d(e^ax)/a = e^ax dx
其他解答:
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