华中师范大学专用
(5)=
sin2xcos3xdx
(0.27)
1
(sin5x sinx)dx211=cosx cos5x+C.2101+lnx
dx
(xlnx)
1
=d(xlnx)
(xlnx)2 1=+C.xlnx
(6)
(0.28)
(7)
1
dx
xππ,(0.29)
<t<π.K§dx=secttantdt,
)µ {1(n {)-x=sect,0<t<√
=|tant|.
1
dx
x
1
=secttantdt
sect|tant|
tant=dt
|tant|
t+C,0<t<π,=
t+C,π<t<π
1
+C,x>1,arccos=1
arccos+C,x< 1.
1
=|arccos|+C.
x
(0.30)
12