求（1 sinx)\(1 cosx)與e^xdx的積的不定積分網友廻答：
- 1/(1 cosx)=1/[1 2cos²(x/2)-1]
=sec²(x/2)/2
所以原式=∫e^xsec²(x/2)/2dx ∫e^xsinx*sec²(x/2)/2dx
=∫e^xsec²(x/2)d(x/2) ∫e^x2sin(x/2)cos(x/2)*sec²(x/2)/2dx
=∫e^xdtan(x/2) ∫e^x*sin(x/2)/cos(x/2)dx
=∫e^xdtan(x/2) ∫tan(x/2)de^x
=e^x*tan(x/2)-∫tan(x/2)de^x ∫tan(x/2)de^x
=e^x*tan(x/2) C
admin(1+cosx)與e^xdx的積的不定積分.jpg "求（1+sinx)(1+cosx)與e^xdx的積的不定積分,第1張")
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