【2017年整理】第五章不定积分习题课参考
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第五章 不定积分
x,例1 为的一个原函数~则, 。 e,cosxf(x)f(x)
xxx,解: 由已知为的导函数~即 e,cosxf(x)fxexex()(cos)sin,,,,
xx,,所以~ fxexex()(sin)cos,,,,
12,例2 设为的原函数~求:。 x,xxf(x)dxf(x),0
22,解: 由已知为的导函数~即 x,xf(x)xxfx,,()
1111123243,xfxxxxxxxxxxxC()d()d()d,,,,,,,所以~ ,,,00043
例3 求下列不定积分:
231,2x(x,1)dx?, ?( dx2,,1,xx
分析:用分项积分法~
5311,332,,(1)331xxxx,,,,2222解: ?, dd33dxxxxxxx,,,,,,,1,,,x,,2x
75313(1)26x,2222故 d22xxxxxC,,,,,,75x
221232(1)3,,,xx,,? dd2d3arctan2xxxxxC,,,,,,,,,222,,111,,,xxx,,
1例4 计算不定积分:dx。 ,x(2x,1)
1(21)21211xx,,,,,解: ddddd(21)xxxxx,,,,,,,,,,,,,,xxxxxxxx(21)(21)2121,,,,,,
1故 dlnln21xxxC,,,,,xx(21),
f(x)dx,F(x),Cf(sinx)cosxdx例5 设~则, 。 ,,
fxxxfxxFxC(sin)cosd(sin)d(sin)(sin),,,解: ,,
例6 求下列不定积分:
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dxxdx11?, ?, ?, sindx22,,,2x1,xxx1,lnx凑微分求不定积分~必须牢记基本积分公式类型~这样就不会被复杂的式子所迷
惑~同时为提高凑微分技巧~应熟悉常见的微分公式( 常用的凑微分积分类型:
1nn,1nn?, f(ax,b)xdx,f(ax,b)d(ax,b),,an
?, f(sinx)cosxdx,f(sinx)dsinx,,
2?, f(tanx)secxdx,f(tanx)dtanx,,
1?, f(arctanx)dx,f(arctanx),darctanx2,,1,x
1xxxxxxxx?f(a)adx,f(a)da,?, f(e)edx,f(e)de,,,,lna
1?f(lnx)dx,f(lnx)dlnx( ,,x
2xxxd1d(1)1,2解: ? , ,,,,ln(1)xC,,221212,,xx
11111? , sindsind()cosxC,,,,2,,xxxxx
ddlnxx? ,,,arcsin(ln)xC,,22xxx1ln1ln,,
2xedx例7 计算:, x,1,e
2xxxeee(1)11,,xxxxxxddddd(1+)ln(1)xeeeeeeC,,,,,,,,解: ,,,,,xxxx1111,,,,eeee
32例8 计算:x,1,xdx。 ,
4133322223: xxxxxxC,,,,,,,,1d1d(1)(1)解 ,,28
1dx例9 计算。 ,xlnx
11dd(ln)lnlnxxxC,,,解: ,,xxxlnln
例10 求下列不定积分:
dxdx1,lnx,xdx1,edx?, ?, ?, ?( ,,2,,222(x,lnx)xx,1x4,x
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11,xx,2tant分析:?令或~?令~?令~?令( x,x,t,1,ex,secttt
第二类换元法常用代换有:根式代换~三角代换~倒代换~其中三角代换可使被
积函数消去根号而有理化~尤为多用(使用第二类换元法求出原函数后一定要将
变量回代(常用第二类换元法积分类型:
22nnx,asint?~令,?~令, t,ax,bf(x,ax,b)dxf(x,a,x)dx,,
2222x,atant?~令,?~令, f(x,a,x)dxf(x,x,a)dxx,asect,,
1?倒代换~常用于消去被积函数分母中的因子( x,xt
,,d1d111xx,,解: ?, ,,,,d()arcsinC2,2xxx,,11xx,1211(),,,,2xx
x,0法, 当时~
1令x,,tdd11xtt,,, ,,,,,,,,,darcsinarcsinttCC,,222tx,1,xxt,,11,1,2t
x,0 当时~同理。
,,,令xtt2tan, (,),,222dsecdsecd1cosdxtttttt?,,,,,,,,, 22,,,,22228tan8sintt,,xxtt44tan44tan
d111x,,,,21 ,,,,,sind(sin)sintttCC,,22888sintxx4,
2xd4xx,, 由于~所以~。 sint,,,C,2228xx,4xx4,
,x22令则text,,,,-1)1,ln(,,,2(22)2tt,x? 1ddd,,,,,,,,,,,exttt,,,22-1-1tt
,x2111te,,,,,,x ,,,,,,,,,,,,2d2ln21lnttCeC 2,,,,x-1tt,1,,11,,e
1令 x,t1ln1ln11ln,,,,xtt?( dddxtt,,,,,222,,,1(ln)(1ln)xxttt,,2(ln),tt
11x ,,,,,,,d(1ln)ttCC2,(1ln)(1ln)(ln),,,ttttxx
24,xdx例11 计算:。 ,
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,,令 xtt,,,2sin, [,]22222解: 4d44sin2cosd4cosd,,,,,,,,,,xxttttt,,,
1cos2,t,,,,,,4d2dcos2d22sin2ttttttC,,,2 2xxxxx4,2,,,,,,,,,2arcsin22arcsin4CxC22222
例12 求下列不定积分:
xarcsinx2axdx?, ?, ?, ?( xlnxdxxsin3xdxxedx,,,,21,x一般被积函数为两类函数的乘积时~考虑用分部积分法~适当选取函数~ u,v
22222xxxxx1解: ?, xxxxxxxClndlnd()lndln,,,,,,,,,22224x
,,cos3cos3cos3cos3sin3xxxxxx?, xxxxxxCsin3dd()d,,,,,,,,,,,33339
axaxaxaxaxeeeee2222axdd()2d2d, ?xexxxxxxx,,,,,2,,,,aaaaa
axaxaxaxaxax2,,eeexexee222,,,,,,,xxxC22d ,,2223,aaaaaa,,
xxarcsin222?( darcsind(1)1arcsin1d(arcsin)xxxxxxx,,,,,,,,,,,21,x
1222 ,,,,,,,,,,1arcsin1d1arcsinxxxxxxxC,21,x
2xcosxdx例13 计算:, ,
21cos2cos2sin2,xxxxx2解: xxxxxxxxxcosddddd(),,,,, ,,,,,22244
22xxxxxxxxsin2sin2sin2cos2,,,,,,,dxC ,444448
例14 求下列不定积分:
4xdxx,12,3xdxdx?, ?, ?( 32,,6,x,3x,2,,5xxx,1
x122,,,分析:?~?去分母的一次项~ 329(x,1)9(x,2)x,3x,23(x,1)
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42x,11x?计算有理函数的积分可分为两步进行~第一步:用待定,,626x,1x,1x,1
系数法或赋值法~将有理分式化为简单分式之和,第二步:对各简单分式分别积分(其中把被积函数变成部分分式是关键(
对有些有理函数的积分~应分析表达式的具体特点~采用一定技巧~如分项积分~变量代换等~简化计算过程(
x122解: ?因为 ,,,329(x,1)9(x,2)x,3x,23(x,1)
,,xxd122所以~, ,,,dx32,,,,xxxxx,,,,,323(1)9(1)9(2),,
122,,,,,,,,,d(1)d(1)d(2)xxx,,,23(1)9(1)9(2)xxx,,,,,,
,122,,,,,,ln1ln2xxC3(1)99x,
11令tx,,t,,2()32xxt,,,2323242? xxtt,,,,,dddd,,,,21191919xx,,5222xtt,,,,()2444
211982tt,,2,,,,,,d4dlnarctantttC,,,,19194221919,,tt,, , 44
821x,2,,,,,ln5arctanxxC,,1919
42x,11x,,?因为 626x,1x,1x,1
42xx,1111,3,,所以~dddarctandxxxxx,,,,( ,,,,6266,,xxxx,,,,11131,
4x,113darctanarctanxxxC,,, ,6x,13
例15 求下列不定积分:
31,,x,1dx?, ?( dx,,34x(1,x)x,1
分析:通过适当的变换将无理函数有理化(
323x,1令txxt,,, ,,t,11232432解: ?, dxtdttttdttttC,,,,,,,,,,,,22,,,,,t,123x,1
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3x,1,,122所以~ dxxxxxC,,,,,23x,1
44令txxt,,,11(1)1t,,3?( d4d4dxttt,,,,,,,,,,23334ttt(1)(1),,xx(1),
,,,,2312,, ,,,,,,,,,,,4(1)(1)d(1)4(1)2(1)tttttC,,,
124所以~ dxC,,,,32444xxxx(1)(1)1,,,
附 三角函数相关公式:
(1) ; . sin()sincoscossin,,,,,,,,,sin()sincoscossin,,,,,,,,,
(2) ; . cos()coscossinsin,,,,,,,,,sin()sincoscossin,,,,,,,,,(3) ;. . 2sincossin()sin(),,,,,,,,,,2cossinsin()sin(),,,,,,,,,,
;. . 2coscoscos()cos(),,,,,,,,,,,,,,,2sinsincos()cos(),,,,,,
222sincossin2,,,,2cos1cos2,,,,2sin1cos2,,,,(4) ;..; .
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