標題:
2 calculus Q
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00016323/o/701108270058613873449710.jpg __________________________________________________________ 更新: In Q1, is this also correct? -(1/2) cosh[-1] [(2x)/√5] *cosh[-1]=inverse of cosh In Q2, is this also correct? (1/10) tanh[-1] [(2x)/5] *tanh[-1]=inverse of tanh 更新 2: How about by these 2 formula?? ∫dx/√(b2x2-a2)=(1/b) cosh[-1] (bx/a) ∫dx/(a2-b2x2)=[1/(ab)] tanh[-1] (bx/a) **maybe my ans of Q1 is wrong, how about Q2?
最佳解答:
圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Aug11/Crazyint4.jpg aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa 2011-08-27 20:29:59 補充： I don't think so since cosh and tanh involve exponential functions. 2011-08-27 21:35:42 補充： Pls. raise another Q to ask, I think I have completed the mission in this Q
其他解答:
2 calculus Q
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00016323/o/701108270058613873449710.jpg __________________________________________________________ 更新: In Q1, is this also correct? -(1/2) cosh[-1] [(2x)/√5] *cosh[-1]=inverse of cosh In Q2, is this also correct? (1/10) tanh[-1] [(2x)/5] *tanh[-1]=inverse of tanh 更新 2: How about by these 2 formula?? ∫dx/√(b2x2-a2)=(1/b) cosh[-1] (bx/a) ∫dx/(a2-b2x2)=[1/(ab)] tanh[-1] (bx/a) **maybe my ans of Q1 is wrong, how about Q2?
最佳解答:
圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Aug11/Crazyint4.jpg aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa 2011-08-27 20:29:59 補充： I don't think so since cosh and tanh involve exponential functions. 2011-08-27 21:35:42 補充： Pls. raise another Q to ask, I think I have completed the mission in this Q
其他解答:
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