解答：解法一：万能代换！
令u＝tanx/2，则sinx＝2u/(1＋u²)，cosx＝(1－u²)/(1＋u²)，dx＝2du/(1＋u²)，于是得
∫1/(sinx＋cosx)＝∫2/(1＋2u－u²)du
＝√2/2∫［1/(u－(1－√2))－1/(u－(1＋√2))］du
＝√2/2ln|(u－(1－√2))/(u－(1＋√2))|＋C
＝√2/2ln|(tanx/2－1＋√2)/(tanx/2－1－√2)＋C.
解法二：
∫dx/(sinx＋cosx)＝√2/2∫dx/(√2/2sinx＋√2/2cosx)＝√2/2∫dx/cos(x－π/4)
＝√2/2∫sec(x－π/4)d(x－π/4)
＝√2/2ln|sec(x－π/4)＋tan(x－π/4)|＋C.
令u＝tanx/2，则sinx＝2u/(1＋u²)，cosx＝(1－u²)/(1＋u²)，dx＝2du/(1＋u²)，于是得
∫1/(sinx＋cosx)＝∫2/(1＋2u－u²)du
＝√2/2∫［1/(u－(1－√2))－1/(u－(1＋√2))］du
＝√2/2ln|(u－(1－√2))/(u－(1＋√2))|＋C
＝√2/2ln|(tanx/2－1＋√2)/(tanx/2－1－√2)＋C.
解法二：
∫dx/(sinx＋cosx)＝√2/2∫dx/(√2/2sinx＋√2/2cosx)＝√2/2∫dx/cos(x－π/4)
＝√2/2∫sec(x－π/4)d(x－π/4)
＝√2/2ln|sec(x－π/4)＋tan(x－π/4)|＋C.
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第1个回答 2013-03-06
第2个回答 2019-09-03
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