中国人民大学出版社（第四版）高等数学一第4章课后习题详解(9)

??★★(2)

3332arctan(cotx)?C?arctan(tanx)?C. 6263dx?3?cosx

思路：万能代换！

1?t22dtx解：令t?tan，则cosx?,dx?; 2221?t1?t2dt2dxdt1t1?t??????arctan?C22?1?t3?cosx2?t223?

1?t2dx11x???arctan(tan)?C.3?cosx222注：另一种解法是：

xdxdx1dx12dx ????3?cosx???xx2x3?2cos2?121?cos2sec2?12221x1x11x ?dtan??dtan?arctan(tan)?C.

xx22222tan2?2(tan)2?(2)222dx★★(3)?2?sinx

sec2思路：万能代换！ 解：令t?tanx2t2dt，则sinx?,dx?; 21?t21?t22t?12dtd()2dxdtdt23????1?t??2????2t?122t1232?sinxt?t?132?(t?)?1?()2241?t322t?1?arctan()?C33

x2tan?1dx22)?C. ???arctan(2?sinx33dx?1?tanx

思路：利用变换t?tanx！（万能代换也可，但较繁！）

★★(4)

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解：令t?tanx，则x?arctant,dx?dt; 1?t2dt2dxdt ????1?t??21?tanx1?t(1?t)(1?t)?111t?111t1?(?)?(??)2222(1?t)(1?t)21?t1?t21?t1?t1?tdt11t1???(dt?dt?dt)222???(1?t)(1?t)21?t1?t1?t

11?[ln1?t?ln(1?t2)?arctant]?C22dx11???[ln1?tanx?ln(1?tan2x)?x]?C.1?tanx22★★(5)

dx?1?sinx?cosx

思路：万能代换！

2t1?t22dtx解：令t?tan，则sinx?,cosx?,dx?; 22221?t1?t1?t2dt2dtx1?t????ln1?t?C?ln1?tan?C 2?2t1?t1?t21??1?t21?t2dx★★(6)?5?2sinx?cosx

思路：万能代换！

2t1?t22dtx,cosx?,dx?; 解：令t?tan，则sinx?21?t21?t21?t22dtdxdt1?t2 ?????22?5?2sinx?cosx2t1?t3t?2t?25?2?1?t21?t2d(3t?15)?dt1??而?23t?2t?23dt15(t?)2?()233?1?5153t?12()?15arctan(3t?15)?C

x3tan?1dx12)?C. ???arctan(5?2sinx?cosx55 37

★★★★(7)

dx?(5?4sinx)cosx

思路一：万能代换！

2t1?t22dtx解：令t?tan，则sinx?,cosx?,dx?; 22221?t1?t1?t2dt2dx2(1?t2)dt1?t???22(5?4sinx)cosx2t1?t(5t?8t?5)(1?t2)(5?4)

1?t21?t224??(2?2)dt5t?8t?5(5t?8t?5)(t2?1)而

44?，

(5t2?8t?5)(t2?1)(5t2?8t?5)(t?1)(t?1)4At?BCD???，等式右边通分后比较两边分子t的同22(5t?8t?5)(t?1)(t?1)5t?8t?5t?1t?1令

次项的系数得：

?A?5C?5D?0?5A=?B?13C?3D?0???2,解之得：???A?13C?3D?0??B=7???8?B?5C?5D?41?C???16; ??D??9?16??4120t?71191?????? 22(5t?8t?5)(t?1)(t?1)85t?8t?516t?116t?11191110t?891?????2??216t?116t?145t?8t?585t?8t?5dx1191110t?871??(??????2??2)dt(5?4sinx)cosx16t?116t?145t?8t?585t?8t?5dx1191110t?871?????dt??dt??2dt??2dt(5?4sinx)cosx16t?116t?145t?8t?585t?8t?519175t?4??lnt?1?lnt?1?ln(5t2?8t?5)?arctan()?C16164243x5tan?41x9x1xx72??lntan?1?lntan?1?ln(5tan2?8tan?5)?arctan()?C162162422243?思路二：利用代换t?sinx！ 解：令t?sinx，x

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dt2dxdtdt1?t?????????(5?4t)(t2?1)

(5?4sinx)cosx(5?4t)(1?t2)(5?4t)1?t211??(5?4t)(t2?1)(5?4t)(t?1)(t?1)令

1ABC，等式右边通分后比较两边分子t的同次项的系数得： ???2(5?4t)(t?1)5?4tt?1t?116?A??9?A?4B?4C?0?111611111????????? ?9B?C?0解之得:?B?218(5?4t)(t?1)95?4t18t?12t?1??A?5B?5C?1??1?C???2???dt1611111?dt?dt?dt(5?4t)(t2?1)9?5?4t18?t?12?t?1

411?ln5?4t?ln1?t?ln1?t?C9182??dx411??ln5?4sinx?ln1?sinx?ln1?sinx?C.

(5?4sinx)cosx9182注：比较上述两解法可以看出应用万能代换对某些题目可能并不简单！

★★★★(8)

1?sinx?(1?cosx)sinxdx

思路：将被积函数分项得，对两个不定积分分别利用代换t?cosx和万能代换！ 解：?1?sinx11 ??(1?cosx)sinx(1?cosx)sinx1?cosx??1?sinx11dx??dx??dx

(1?cosx)sinx(1?cosx)sinx1?cosxdt1dx??,sinx?1?t2; dx，令，则t?cosx,x?(0,?)?(1?cosx)sinx1?t2?dt对积分

1dtdt1?t2??dx????? 22?2(1?cosx)sinx(1?t)(t?1)(1?t)(t?1)(1?t)1?t令

1ABC???(1?t)2(t?1)t?11?t(1?t)2，等式右边通分后比较两边分子t的同次项的系数得：

39

1?A??4A?B?0??11111111??2A?C?0B?????????解之得：??4(1?t)2(t?1)4t?141?t2(1?t)2?A?B?C?1??1?C???2???1111111dt?dt?dt?dt4?t?14?1?t2?(1?t)2(1?t)2(t?1)

1111?lnt?1?lnt?1???C14421?t

??11111dx?ln1?cosx?ln1?cosx???C1;

(1?cosx)sinx4421?cosx

x1?t22dt1对积分?,dx?dx，令t?tan,cosx?21?t21?t21?cosx2dt2dt221x1?t1?t??dx????dt?t?C?tan?C2;21?t2?1?t2?1?cosx21?1?21?t1?t21?sinx1111x??dx?ln1?cosx?ln1?cosx???tan?C3

(1?cosx)sinx4421?cosx2?1x1xxlntan?tan2?tan?C.22422★★(9)

?1?dx3x?1 思路：变无理式为有理式，变量替换t?31?x。

32解：令t?31?x，则 1?x?t,dx?3tdt;

3t2dtt2dt132????3?3(t?1)dt?3dt?t?3t?3lnt?1?C????31?t1?t1?t21?x?1

3?3(1?x)2?331?x?3ln31?x?1?C.2dx★★(10)

?1?(x)31?xdx

思路：变无理式为有理式，变量替换t?x。

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