【例14】 I??解：
x2dx 2x?4I??x2dxdx???2x?4?1?d??1?x?????2x?4x2x?4xt2?4, dx?tdt2dx?2x?4?232??x?2x?4?32tdt1?t?4??t??????????2??2dt322t?4tt?4t????2x?4?t?x?1?11?1?11t?112x?4?dt??+arctan=??arctan?c ?22????2?tt?4?2?t22?222x?442x?41x?2?arctan?c4x42?I?cos2x?sinx【例15】 I??dx sinxcosx(1?cosxe)?解：注意到：?cosxesinx??(cos2x?sinx)esinx
d(cosxesinx)dycosxesinxy?cosxesinxI??sinx???????ln?C sinxsinxecosx(1?cosxe)y(1?y)1?cosxe1?lnxdx21?lnx1x?x?lnx?xdx???d??C 【例16】 ???x?lnx?2??x?lnx?2??(x?lnx)2xx?lnx???????x??x?111??d(x?)x???2x2?11xxx??C dx???arctg?【例17】 ?4dx??21x?12?2?1??x2?2??x??2??x??x??1?
【例18】F?x?为f?x?的原函数，x?0时，有f?x?F?x??sin2x, 且 F?0??1, F?x??0,求f?x?。 解：由F??x??f?x?有：
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??2sin2x2?f?x?F?x??sin2x?F??x?F?x??sin2x??Fx????1?F2?x?=?2sin2xdx?x?sin4x?C41F?0??1?C?1?F?x??x?sin4x?141?cos4xf?x??F??x??12x?sin4x?14
xarctanx ?x?0?，求f?x?。 【例19】设f?x?连续，且f?x???f?t?dt?1?????0?x?1?x? 解：令F?x???f?t?dt?1
0x?F?0??1,F??x??f?x??F??x?F?x??212??arctanx??Fx???2?x?1?x?arctanx ?F?x??2?dx?4?arctanxd?arctanx??2arctanx??x?1?x???2?c
F?0??1?c?1?F?x???1?2arctanx?f?x??F??x???2、回归法 【例20】求I????24arctanx21?2arctanx??2?1??2?1?x?xarctanxx?1?x?1?2arctanx??2cosx?xsinx?x?cosx?2dx
x?cosx???1?sinx?x?dx1??dx??xd 解：I???? 2??x?cosxx?cosx???x?cosx???dxxdxx?????c
x?cosxx?cosxx?cosxx?cosx2tx?x?【例21】设f?x?连续，且lim?1????f?x?t?costdt，求f?x?。
0t???t??x? 解：易知：lim?1???e2x
t???t?2t 114
e??f?x?t?costdtx?t?u?f?u?cos?x?u?du?cosx?f?u?cosudu?sinx?f?u?sinudu2x0000xxxx?2e??sinx?f?u?cosudu?cosx?f?u?sinudu?f?x?2x00xx ?4e2x?f??x??cosx?f?u?cosudu?sinx?f?u?sinudu?f?x?sinxcosx?f?x?sinxcosx
xx00?4e2x?f??x??e2x?f??x??5e2x?f?x??5e2x2?c
【例22】设f?x??x1?x2?1?x2?10f?x?dx，求I??10f?x?dx 解：设?1f?x?dx?A，x0f?x??1?x?1?x2?10f?x?dx两边同时取积分得
?1f?x?dx??1x11001?x2dx?????0f?x?dx???????01?x2dx??? ?A??1x01?x2dx?A?12121??x12?11?01?xdx?A?2ln?1?x?|0?A??1?x?arcsinx|???2??0? 02???A?1?12ln22ln2?A4??0f?x?dx?A?4??3、待定函数法
1?x?1【例23】求I????1?x??ex?x?dx
解：设I????1?x?1?x?1x?1?exdx?F?x?ex?x??c，两边求导得：
??1?x?1?ex?1x?ex?1x?F?x?Fx?1?1?? ?x??????????x2?????1?x?1?1?
x?F??x??F?x???1?x2??上式很容易看出，F?x??x是它的一个特解，根据定积分的定义，故 I?xex?1x? c4、相关积分法
【例24】求I1??e2xsin2xdx, I2??e2xcos2xdx
解：
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1I1?I2??e2xdx?e2x?c21I1?I2??e2xcos2xdx?e2xcos2x??e2xsin2xdx21111 ?e2xcos2x?e2xsin2x??e2xcos2xdx?e2xcos2x?e2xsin2x??I1?I?
22221?I1?I?e2x?cos2x?sin2x?412x12x?I?e?e?cos2x?sin2x??c1??48???I?1e2x?1e2x?cos2x?sin2x??c2?48?5、换元法（注意：凑微分和换元法是计算积分的两大核心而普遍的技术） 5.1 三角换元
● 三角换元?一? ____ 去根式
sec2x?1?tg2xa2?x2?x?asint 或 acost
x2?a2?x?atant(dx?asec2tdt)x2?a2?x?asect(dx?asecttantdt)1?tan2x?sec2x 1?ctan2x?csc2x
●三角换元?二?____万能公式
x2dt?dx?21?t2
2t1?t22tsinx?; cosx?; tanx?221?t1?t1?t2t?tan
注意：三角万能代换只有在没有其他简单方法可用时才使用，实际上三角万能代换后计算
量很大。 如?sinxdx令u?cosx更方便???????，如用三角替换反而繁琐。 2cosx?sinx●三角换元?三?____和差化积或积化和差
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1?sin(???)x?sin(???)x?21?cos(???)x?cos(???)x? sin?xsin?x??
21cos?xcos?x??cos(???)x?cos(???)x?2sin?xcos?x?●三角换元?四?____倍角公式
1sinxcosx?sin2x cos2x?1?2sin2x21122 sinx?(1?cos2x) cosx?(1?cos2x)
22xx1?cosx?2cos2 1?cosx?2sin222评 注：不管引用何种三角替换，其本质是去掉根号和化简，从这个意义上读者根据具体题型要广义使用。
x2【例25】 I??dx 22(1?x)解：三角换元法：
x2tan2tx?tantI??dx ??????sec2tdt??sin2t2222(1?x)(1?tant)
11111x ??(1?cos2t)dt?t?sin2t?C?arctanx??C224221?x2【例26】I??xxdx ?a?0? 2a?x解：三角换元法：令 x?2acos2t?dx??4asintcostdt
xcostdx ??2ac2ots?cos??a4tsint?dt2a?xsitn
1?? ??8a2?co4tdts??a2t?32st?in2t?icn4??s4??I??x115.2 倒换元 x? ?dx??2dt
tt【例27】?dxx2a?x22
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