解法二：
dxcosxdxdtanx???sinxcosx?sinxcos2x?tanx?ln|tanx|?C
dx(sin2x?cos2x)dx?解法三：?
sinxcosx?sinxcosx??(sinxcosx?)dx?ln|sinx|?ln|cosx|?C cosxsinxdxexdxdexx???arctane?C (23) ?x?x2x2x??e?ee?1e?12x?3d(x2?3x?8)dx??2?ln|x2?3x?8|?C (24) ?2x?3x?8x?3x?8x2?2(x?1)2?2(x?1)?3dx??dx(25) ?33(x?1)(x?1)
12323??(??)dx?ln|x?1|???C232x?1(x?1)x?12(x?1)(x?1)(26)
?dxdxx?a22
解 令x?atant, 则
?asec2tdt???ln|sect?tant|?C1?ln|x?x2?a2|?C
asectx2?a2dx1x1x?d??(x2?a2)32a2?(x2?a2)12a2(x2?a2)12?C
(27)
解法2 令x?atant, 则
dxasec2tdt11x??costdt?sint?C??C 2?(x2?a2)32?a3sec3ta2?222aax?a(28)
?x51?x2dx
解 令x?sint, 则
157
?sin5tcostdx??dt??sin5tdt???(1?cos2t)2dcostcost1?x2123252x5
2121??cost?cos3t?cos5t?C??(1?x2)?(1?x2)?(1?x2)?C3535(29)
?1?16x3xdx
解 令x?t, 则x?t, dx?6t
65t3?t5dt(t2)4?1?1(t2?1)(t6?t4?t2?1)?1dt2?1?3xdx?6?1?t2?6?1?t2dt?6?1?t
7531ttt6t?1?6?(?(t6?t4?t2?1)?)dt??6(???t)?ln||?C27532t?11?tx其中t?x (30)
16?x?1?1x?1?1dx
2解 令x?1?t, 则x?1?t, dx?2tdt,
x?1?1t?124tdx?2tdt?(1?)2tdt?(2t?)dt?x?1?1?t?1?t?1?t?14??(2t?4?)dt?t2?4t?4ln|t?1|?C1
t?1?x?1?4x?1?4ln|x?1?1|?C1?x?4x?1?4ln|x?1?1|?C
2．应用分部积分法求下列不定积分：
⑴
?arcsinxdx?xarcsinx??x1?x2dx?xarcsinx?1?x2?C
⑵
1lnxdx?xlnx?x???xdx?xlnx?x?C
158
x⑶ ?2cosxdx??x2dsinx?x2sinx??2xsinxdx?x2sinx??2xdcosx?x2sinx?2xcosx?2?cosxdx ?x2sinx?2xcosx?2sinx?C⑷ ⑸
lnx?11?1lnx11lnx1dx?lnxd??dx????C 22322?x3??22x2xx2x4x?(lnx)2dx?x(lnx)2?2?lnxdx?x(lnx)2?2xlnx?2x?C
1121x22dx ⑹ ?xarctanxdx??arctanxdx?xarctanx??2221?x21211121xarctanx??(1?)dx?xarctanx?(x?arctanx)?C 222221?x11?(x2?1)arctanx?x?C 2211]dx??ln(lnx)dx??dx ⑺ ?[ln(lnx)?lnxlnx11?xln(lnx)??x?dx??dx?xln(lnx)?C
xlnxlnx?⑻
22(arcsinx)dx?x(arcsinx)???2xarcsinx1?x2dx
?x(arcsinx)2?2?arcsinxd1?x2
?x(arcsinx)2?21?x2arcsinx?2?1?x211?x2dx
?x(arcsinx)2?21?x2arcsinx?2x?C
⑼
32secxdx?secxdtanx?secxtanx?secxtanxdx ????secxtanx??secx(sec2x?1)dx?secxtanx??sec3xdx??secxdx ?secxtanx??sec3xdx?ln|secx?tanx|
所以
3sec?xdx?1secxtanx?ln|secx?tanx|)?C 2159
⑽
?x2?a2dx?xx2?a2??x?2222xx?a22dx
?xx?a??(x?a?2222a2x?a22)dx
?xx?a??x?adx??a2x?a22dx
?xx2?a2??x2?a2dx?a2ln(x?x2?a2)
所以
?x2?a2dx?1(xx2?a2?a2ln(x?x2?a2))?C 2类似地可得
?x2?a2dx?1(xx2?a2?a2ln(x?x2?a2))?C 2
3．求下列不定积分：
⑴ [f(x)]f?(x)dx?[f(x)]df(x)?⑵
?a?a1[f(x)]a?1?C a?1f?(x)1dx??1?[f(x)]2?1?[f(x)]2df(x)?arctanf(x)?C
⑶
??f?(x)df(x)dx???ln|f(x)|?C f(x)f(x)f(x)f?(x)dx?ef(x)df(x)?ef(x)?C ⑷ e?
4．证明：
n⑴ 若In?tanxdx，n?2,3,?，则In??1tann?1x?In?2 n?1n?2x(sec2x?1)dx?tann?2xsec2xdx?tann?2xdx 证： In?tan?????tann?2xdtanx?In?2.
n?2xdtanx?tann?1x?(n?2)tann?2xdtanx， 因为tan?? 160
所以tan从而In??n?2xdtanx?1tann?1x. n?11tann?1x?In?2. n?1mn⑵ 若I(m,n)?cosxsinxdx，则当m?n?0时，
?cosm?1xsinn?1xm?1I(m,n)??I(m?2,n)
m?nm?ncosm?1xsinn?1xn?1???I(m,n?2)，n,m?2,3,?
m?nm?n证： I(m,n)?cos?mxsinnxdx?1m?1n?1cosxdsinx ?n?11[cosm?1xsinn?1x?(m?1)?cosm?2xsinn?2xdx] n?11?[cosm?1xsinn?1x?(m?1)?cosm?2xsinnx(1?cos2x)dx] n?11?[cosm?1xsinn?1x?(m?1)(I(m?2,n)?I(m,n))] n?1?cosm?1xsinn?1xm?1?I(m?2,n)， 所以I(m,n)?m?nm?ncosm?1xsinn?1xn?1?I(m,n?2) 同理可得I(m,n)??m?nm?n
5．利用上题的递推公式计算：
(1)?tan3xdx;
(2)?tan4xdx;(3)?cos2xsin4xdx.解：(1)
3?tanxdx?11tan2x??tanxdx?tan2x?lncosx?C 22 161