1 不定积分换元法例题 2009-12-18 【不定积分的第一类换元法】
已知 fuduFuC()(),,,
求 【凑微分】 gxdxfxxdxfxdx()(())'()(())(),,,,,,,,,
【做变换,令,再积分】 ,,,fuduFuC()()ux,,(),
【变量还原,】 ,,FxC(()),ux,,()
【求不定积分的第一换元法的具体步骤如下:】 gxdx(),
(1)变换被积函数的积分形式:gxfxx()(())'()dx,,,dx ,,
(2)凑微分:gx()(())((dxfxfx,,,,,,'(()x)dxdx)) ,,,
(3)作变量代换得:gxdxfxx()(())'()(),,,,dxfd,,()()xx ,f()uduux,,(),,,,(4)利用基本积分MATCH_
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_1714149374542_0fuduFuC()(),,求出原函数: ,
gxdxfxxdxfxdx()(())'()(())(),,,,,,,,,fuF()()duuC ,,,,(5)将代入上面的结果,回到原来的积分变量得: ux,,()x
gxdxfxxdxfxdx()(())'()(())(),,,,,,,,,fuduFuC()() ,,FxC(()),,,,,【注】熟悉上述步骤后,也可以不引入中间变量,省略(3)(4)步骤,这与复合函数的求导法则类似。 ux,,()__________________________________________________________________________________________
【第一换元法例题】
1199991、 (57)(57)(57xdxx,,,,,,,,,,xxdxdxd)(57)(57)(57),,x,,,,55
111191010 ,,,,,,,(57)(57)(57)xx,,,dCx(57)x,C,555010
1【注】 (57)'5,(57)5,(57)xdxdxdxdx,,,,,,,,5
lnx12、 dxxx,,,,lnlndxdxln,,,xx
1122 ,,,,,,lnln(ln)xxxdCxC(ln),22
111【注】 (ln)',(ln),(ln)xdxdxdxdx,,,,,xxx
sinxsincoscosxdxd,xdx3(1) tanxdxdx,,,,,,,,,,coscoxscoscosxxx
dcosx ,,,,,,,,ln|cos|xCxCln|cos|,cosx
2 不定积分换元法例题 2009-12-18
【注】 (cos)'sin,(cos)sin,sin(cos)xxdxxdxxdxdx,,,,,,,,
cosxcosxdxdxsin3(2) cotxdxdx,,,,,,,sinsinsinxxx
dsinx ,,,,,ln|sin|xCxCln|sin|,sinx
【注】 (sin)'cos,(sin)cos,cos(sin)xxdxxdxxdxdx,,,,,
1114(1) dx,,,,dxdx()a,,,,axa,,,xax
1 ,,,,,,dCax(axax,,)ln||ln||,C,ax,
【注】 ()'1,(),()axdaxdxdxdax,,,,,,,,
1114(2) dx,,,,dxda()x,,,,xax,,,axa
1 ,,,,,,dCxa(xaxa,,)ln||ln||,C,xa,
【注】 ()'1,(),()xadxadxdxdxa,,,,,,,,
11111111,,,,4(3) dxdx,,,,,dxdxdx,,,,2222,,,,,x,aaxaaxa,,,2xa2xaxa,,,,,,
11xa, ,,,,,,,ln||ln||lnxaxaCC,,22aaxa,
2sec()xsectansecsectanxxxxx,,5(1) secxdxdx,,,dx,,,sectanxx,sectanxx,
dxxxx()()tansectansec,d, ,,,,,ln|sectan|xxC,,sectanxx,secxx,tan
1cosxcossinxdxdx,5(2) secxdxdxdx,,,,222,,,,,coscxxoscos1sinxx,
dxsinx1111sin111sin,,x,, ,,,,,,,,dCCsinxlnln,,2,,1,,,,,sinxx2112sin121ssinsinxxxin,,
2csc()xcsccotcsccsccotxxxxx,,6(1) cscxdxdx,,,dx,,,csccotxx,csccotxx,
dxxxx()(),,,cotcsccsccotd ,,,,,,,ln|csccot|xxC,,csccxx,otcscxx,cot
2csc()xcsccotcsccsccotxxxxx,,6(2) cscxdxdx,,,dx,,,csccotxx,csccotxx,
3 不定积分换元法例题 2009-12-18 dxxxx()(,,,cotcsccsccodt) ,,,,,ln|cscxxCcot|,,csccotx,xcsccoxx,t
1dx7(1) dxxC,,,arcsin,,2211,,xx
xx,,,,dd,,,,1dxxdxaa,,,,7(2) dxC,,,,,,arcsin,,,,,2222222aa,,xaxxxx,,,,,,a111,,,,,,,,,aaa,,,,,,1dx8(1) dxxC,,,arctan22,,11,,xx
xx,,,,dd,,,,1111dxxdxaa,,,,8(2),() a,0dxC,,,,,,arctan222222,,,,,2axaxaaa,,a,,xxx,,,,,,211,,a1,,,,,,,,,aaa,,,,,,,,,,
3525259(1)sincossincossinxxdxxxxx,,,,sicnxdxdx,cosos ,,,
86coscosxx2575 ,,,,,,,,,,,(1cos)coscos(coscos)cosxxdxxxdxC,,86353434sincossincossincosxxdxxxx,,,,cossinxdxdxx9(2) ,,,
468sinsinsinxxx322357 ,,,,,,,,,,,sin(1sin)sin(sin2sinsin)sinxxdxxxxdxC,,438
dx111110(1) ,,,,,,,,,dxdxlndxClnxlnln,,,,xxxxlnllnnxlnx
dx1111110(2) ,,,,,,,,,,dxxdlndClnx2222,,,,xxxxlnlnlnlnxlnxx
222()xdxd2xdxdxx,1211(1) ,,,,,,arctan(xC1)42424222,,,,xxxxx,,,,,,222)22(x21,x,1
22xdxd12xdxd1x1()x,111(2) ,,,,,,,42424222xxxx,,,,,,,25222xx52524()x,1
2,,x,1d,,2221(1)111dxx,,,, ,,,,arctan()C22,,228442,,,,xx,,1111,,,,,,22,,,,
4 不定积分换元法例题 2009-12-18 sinx1112、 dxxx,,,,,,,,sinsdxdx22insinxdx,,,,xxx2
,,,,,,,,2sin2cos2cosxxxdCxC,
111222xxx2x13、 edxedxde,,,,exC22,,,222
4sinx333314、 sincossinsinxxdxxxdC,,,,,,,,cossinxdxdxsinsnxxi,,,,4
1110010010010015、 (25)xdx,,,,,,(25)(25)(25)xxxdxd,,,,(25)(25)xdx,,,,,,22
1111100101101 ,,,,,,,(25)(25)(25)xxx,,,dCxC(25),,22202101
111222222216、 xxdxxsinsins,,,,,,,,,xdxdxinsincosxdCxxx,,,,222
lnlnln(1ln)1xxxx1,,17、 dxdx,,,,,,dxdlnxln,,,,xxxxxx1ln1ln1ln1ln,,,,
1,,,,,1lnlnlnxdxdx,,1ln,x
1,,,,,,,1ln(1ln)(1ln)xdxdx ,,1ln,x
31222,,,,,(1ln)2(1ln)xxC3
arctanx1earctanarctanarcxxxarctanxtan18、 arctanarctandxeeedeC,,,,,,,,dxdxx,,,,2211,x,x
x1112219、 dx,,,,,,xddxxxd,,(1),,,,22221111,,,,xxx22x
122 ,,,,,,dx(1),x1,C,221,x
31,,sin11x22,,,,coscosxdxx2cosC20、 dx,,,sincosxdxd,,x,,,,333coscoscosxxx
5 不定积分换元法例题 2009-12-18 xe111xxxx21、 dxdeC,,,,,,,,edxde,e()ln(22)xxxx,,,,,,,eee,e2222
23lnlnxx122222、 dxx,,,,,,,,lnldxdxnxdClnlnlnxx,,,,xx3
dxdxxdxdx(1,)(1)1,,23、 ,,,,,arcsinC,,,,222222122(1)2(1),,,,,xxxx,(2)(1,,x)
11dx(),dx(),dxdx2224、 ,,,2,,,,171722xx,,21722()()xx,,,,()x,,()242422
11dx()x,,2221x,22 ,,arctanarctan,,CC,,17777722()x,,()222
sincosxx2225、计算, ab,dx,2222axbxsincos,
22222222【
分析
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】因为: (sincos)'2sincos2cos(sin)2()sincosaxbxaxxbxxabxx,,,,,,
222222 所以: daxbxabxxdx(sincos)2()sincos,,,
12222 sincos(sincos)xxdxdaxbx,,,222()ab,
2222sincossincos(sixxxxdxda1ncos)xbx,【解答】 dx,,,,,22222222222222ab,axbxaxbxsincossinc,,os2sincosaxbx,
22221()1daxbxsincos,2222 ,,,,asincosxbxC,22222222abab,,2siaxbxncos,
6 不定积分换元法例题 2009-12-18 【不定积分的第二类换元法】
已知 ftdtFtC()(),,,
求 【做变换,令,再求微分】 gxdxgtdtgttdt()(())()(())'(),,,,,,xt,,(),,,
【求积分】 ,,,ftdtFtC()(),
,1,1 【变量还原,】 ,,FxC(()),tx,,()__________________________________________________________________________________________
【第二换元法例题】
令xt,sinsinsinxtt2dxtdt,,,,,,,,,,,,dttdt22sin1、 ,,,,2xt,ttx
变量还原
,,,,,2cos2costCxC,,,,,,, tx,
令xt,1111t,,22221dxdtdt,,,,,,,,dttd,,,,,,t,,2(1) ,,,,,2xt,1111,,,,tttt1,x,,
变量还原
,,,,,,,2ln|1|2ln|1|ttCxxC,,,,,,,,,,, tx,
令1+xt,,11111t,,2,,,,,,,,,,,,,,,,dxdtdtd(1ttdt)2(1)221,,2(2) ,,,,,2xt(1),,tttt,1x,,
变量还原
,,,,,,,2ln||21ln|1|ttCxxC,,,,,,,,,,, tx,,1
34令,,xt134111343321,,,,xdxtt,,,,,,,,dtttdt(1)4(1)3,,,,,3、 32,,,3434xt,,(1)t(1),xt(1),
334474,,74变量还原(1)(1)xx,,,,tt63,,,,,,,,12,,,C,,,,,12()12ttdtC,, 3,4,,74741tx,,,,,,
令xt,111124、 dxdt,,,,,,,,,dttd,,,t22222,,,,2xt,ttttt,,,(1)(1)1xx,(1)
变量还原
,,,2arctan2arctantCxC,,,,,,,tx,
7 不定积分换元法例题 2009-12-18 xet令,1111111,,5、 dxdtdt,,,,,,,,,,,,,,ddttln,,x,,,,,lnxt,,,,,,ettttttt111(1)1,,
x变量还原te ,ln||ln|1|lnln,,,,,,,,,,,,,ttCCCxx,te11,,te
62令xt,dxt111,,656、 ,,,,,,,,,,,dttt6d,,,661dtdt3,,,,,,,2323226xt,,,,,(1)(ttttt1)11t(1),xx,,
变量还原66 ,,,,6(arctan)6(arctan)ttCxxC,,,,,,,,6tx,
mnk
xx,xt,【注】被积函数中出现了两个根式时,可令,其中为的最小公倍数。 mn,k
322令xt,,2,,dxtt
,,,,,,33ln|1|,,,,,dtttC3,,7(1) ,,3xt,,212,t,,x12,,
32,,变量还原,(2)x33
,,,,,,,32ln|12|,,,,,,xxC6 ,,2tx,,,
1,x,t令2xt211,x,,,,,,,,,,,,,,222ln|1|ln|1|dttttCdx2,7(2) 1,t,1x,xx2t,1
变量还原111xxx,,,
22ln|1|ln|1|C,,,,,,,,,,, 1,xxxxt,x
naxb,n
axb,t【注】被积函数中含有简单根式或时,可令这个简单根式为 ,即可消去根式。 cxd,
111令,tddt,8dx2xt1,,642tt,,,,,,,,,,,,,dttttdt18(1) 82,,,22,,,,1111111tt,,,,,,xx(1),x,,,t11,,,,,,8282tttt,,,,
753变量还原ttt11111 ,,,,,,,,,,,,,,,,,,ttCCarctanarctan7531753753xxxxxt,x
8 不定积分换元法例题 2009-12-18
111,t令,,1ln1lnxxt,,1ln1ln11ttdxdtddt,,,,,,,,,,22222,,,,8(2) 1xxtt,(ln)x,tt,1ln1111,,,,,,t,,lnln,,,,tttt,,,,
111,,,,,,,,,,tdtdttC(1ln)(1ln)22,,,tt1ln,,tttt1ln1ln,,,,变量还原x1 ,,,,,,,CC,111x,xlnt,,1lnx
xx
【注】当被积函数中分母的次数较高时,可以试一试倒变换。
22ttx令tan,t11,,2221sin,2xtdt11,,tt2arctandx,,,,,,d,,,2229、 ,,,xt,2arctan2121,,ttttsin(1cos),21,xxt(1)(1),,22221,,,111,tttt
2111t,,tdtttC2ln||,,,,,,,,,,242t,,
x2tan 变量还原xx12tanln|tan|C,,,,,,,,14222t,x
【注】对三角函数有理式的被积函数,可以用万能公式变换,化为有理分式函数的积分问题。
,令,xatt,,sin||22222222(1)10 axdxaatdatatdt,,,,,,,,,,,sinsincos,,,xt,arcsnia
,,,令,xattsin||变量还原2dxdatxsin ,,,,,,,,,,,,,,,,,,dttCCarcsin,,,xx22222a,,tarcsintarcsin,,axaatsinaa
221cos2sin2,taat,,2,,,,,,adttdttC(1cos2),,,,2222,, 2变量还原ax122,,,,,,,arcsin,xaxC,,x22a,arcsinta
,令,xatt,,tan||2dxdattan10(2)secln|sectan|tdtttC ,,,,,,,,,,,,,,x22222t,arctanaxaat,,tana
22变量还原xax,22,,,,,,,ln||ln||,,,,CxaxC,, xaat,arctana
9 不定积分换元法例题 2009-12-18
2a2222因为: ()'2xaxax,,,,,22ax,
2a2222所以: ()'2xaxdxaxdxdx,,,,,,,,22ax,
,,1122222axdxxaxdxadx,,,,,()',,即: ,,,222ax,,,
21a2222,,,,,,xaxxaxCln|| 22
,令,xatt,,,sec02dxdatsec10(3) secln|sectan|tdtttC,,,,,,,,,,,,,,22222atasec,xa,
22变量还原xxa,22 ,,,,,,,,,,,,,CxxaCln||ln||xat,secaa
2a2222因为: ()'2xxaxa,,,,,22xa,
2a2222所以: ()'2xxadxxadxdx,,,,,,,,22xa,
,,1122222xadxxxadxadx,,,,,()',,即: ,,,222xa,,,
21a2222,,,,,,xxaxxaCln|| 22
222222【注】当被积函数中出现因子时,可以用三角变换,化为三角函数的积分问题。 axaxxa,,,,,
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10 不定积分换元法例题 2009-12-18 【附加】【应用题】
Cx()100已知生产单位的某种产品,边际单位成本是,产量为 1 个单位时,成本为102,xCx'()()',,,2xx又知边际收益为,且, Rxx'()120.1,,R(0)0,
求:(1)利润函数; Lx()
(2)利润最大时的产量;
(3)利润最大时的平均价格。
【解答】
Cx()100(1)因为: Cx'()()',,,2xx
100100,由得:, , 所以:Cxx()1002,,C,2C(1)102,,,CxC(),,Cx()2,,111xx
2 又已知:,, Rxx'()120.1,,R(0)0,Rxxx()120.05,,,
2 于是: LxRxCxxx()()()100.05100,,,,,
(2)令 得: Lxx'()100.10,,,x,100
因为:,所以当时利润最大, LL'(100)0,"(100)0,,L(100)400,x,100max
R(100)700(3)利润最大时的平均价格为: P,,,7100100