求不定积分问题
2011-09-08 07:11:39独***
求不定积分问题求不定积分问题：1/(1+cosx)=1/[1+2cos²(x/2)-1]
=sec²(x/2)/2
所以原式=∫e^xsec²?
最佳回答
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
2011-09-08 09:50:16
e^xtan(x/2)+C
C为任意常数
2011-09-08 11:32:54
解:
∫[(1+sinx)/(1+cosx)]e^xdx
=∫[(1+2sin(x/2)cos(x/2))/(2cos(x/2))^2]e^xdx
=∫[1/2(sec(x/2))^2+tan(x/2)]e^xdx
=∫∫e^x[tan(x/2)]^2d(x/2)+∫e^xtan(x/2)dx
=∫e^xd(tan(x/2))+∫e^xtan(x/2)dx
=e^xtan(x/2)-∫e^xtan(x/2)dx+∫e^xtan(x/2)dx
=e^xtan(x/2)+C.
2011-09-08 10:27:03
很赞哦！ (154)
