∫dx/(sinx^3+cosx^3) sinx+cosx=√2cos(x-π/4)
=∫dx/[(sinx+cosx)(1-sinxcosx)] 1-sinxcosx=1-[(sinx+cosx)^2-1]/2=(1/2)-cos(x-π/4)^2
=∫dx/[√2cos(x-π/4)[1/2-cos(x-π/4)^2]
=(1/√2)∫cos(x-π/4)dx/cos(x-π/4)^2[(1/2-cos(x-π/4)^2]
=(1/√2)∫dsin(x-π/4)/[1-sin(x-π/4)^2][sin(x-π/4)^2-1/2]
sin(x-π/4)=u
=(1/√2)∫du/(1-u^2)(u^2-1/2)
=(-1/√2)(2)∫[(1-u^2)+(u^2-1/2)]du/[(1-u^2)(u^2-1/2)]
=(-√2)∫du/(u^2-1/2)+(-√2)∫du/(1-u^2)
= -∫d(√2u)/(2u^2-1) +(-√2/2)ln|(1+u)/(1-u)|
=(1/2)ln|(1+√2u)/(1-√2u)|+(-√2/2)ln|(1+u)/(1-u)|+C
=(1/2)ln|(1+√2sin(x-π/4))/(1-√2sin(x-π/4)|+(-√2/2)ln|(1+sin(x-π/4))/(1-sin(x-π/4))|+C
追问首先，十分感谢您抽出时间帮我解答问题。
这确实是好方法，但是变换过程中有一个代换有点问题：
1-sinxcosx=1-[(sinx+cosx)^2-1]/2=(1/2)-cos(x-π/4)^2
右边应改为(3/2)-cos(x-π/4)^2
不过我按照您的思维还是继续写出来了，谢谢！
我很喜欢学数学，想必您也对数学很感兴趣
我QQ是1181158196，以后多多交流学习
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