=1?x2?2arctg四.计算积分?解法一：?sinxcosxdx .
sinx?cosx1?x?c 1?x1sinxcosxdx=??sinx?cosx2sin2x2sin(x?)4?dx?122??cos(2x?)2dx sin(x?)4??=
121??1sin2(x?)?42dx
sin(x?)4?=
??1??sin(x?)d(x?)?csc(x?)d(x?) ??444422212cos(x? =??4)?x?lntg(?)?c
2822111x?=(sinx?cosx)?lntg(?)?c . 22822解法二：?sinxcosx12sinxcosx?1?1dx=?dx
sinx?cosx2sinx?cosx1(sinx?cosx)211 =?dx??dx 2sinx?cosx2sinx?cosx1=?(sinx?cosx)dx??2d(x??4)22sin(x?)4?
11??=(sinx?cosx)?csc(x?)d(x?) ?2442211x?=(sinx?cosx)?lntg(?)?c 228221xdx亦可用万能代换法（令：u?tg)求取.
sinx?cosx2x?sinx五．计算积分?dx .
1?sinxx?sinxxsinx解：?dx=?dx+?dx
1?sinx1?sinx1?sinxxxdxxx?=dx?2d(?) ?1?sinx???x?241?cos(x?)2cos2(?)224x?x?x? =?xdtg(?)?xtg(?)??tg(?)dx
242424注：解法二中积分?x?x? =xtg(?)?2lncos(?)?c1
2424sinxsinx?1?1=dx?1?sinx?1?sinxdx?x??x??x?tg(?)?c2 ?241?cos(x?)2dxsinx(1?sinx)sinxsin2xsinx或：?dx??dx??dx dx=?1?sinxcos2xcos2xcos2x =所以：?
11??(sec2x?1)dx??x?tgx?c2 cosxcosxx?x?x?sinxdx=x?(x?1)tg(?)?2lncos(?)?c
24241?sinx一元微积分学题库（24）广义积分
一．
填空题：
d100?1．?sinx2dx? 0 dx0db2．?sinx2dx??sina2
daadx3dt?3．?2x4dx1?t3x21?x12?2x1?x8
?x?tulnududy??1t4．已知?>，则0,t?1??t 12dxy??ulnudu?t?5．?6．?3aadx? ?2212aa?x1dx4?x20?? 67．?二．
dx??1
?e?11?x计算题：
?21．求?sinxdx
02?解：原式???02?sindx??sindx??cos??cos0??4
2???x?1,x?12?2．设f(x)??12 求?f(x)dx
0x,x?1??2解：?f(x)dx??(x?1)dx??002121?x2?121xdx???x??x32?2?06121?8 3三．
xnsin3x求lim?dx n??01?sin3x13xnsin3xn解：x??0,1?,时sinx≥0， 0≤≤ x31?sinx1xnsin3x1n≤ dx?0≤?xdx??001?sin3xn?11n31xsinx1由于limdx?0 ?0, 故lim?n??01?sin3xn??n?1注：利用介值定理或定积分中值定理也可求得结果. 四．
设f(x)在?a,b?上连续，且单调增加，证明?tf(t)dt≥
aba?bbf(t)dt ?a2证：令F(x)?2?tf(t)dt?(a?x)?f(t)dt
aaxx?F?(x)?2xf(x)??f(t)dt?(a?x)f(x)?(x?a)f(x)??f(t)dtaaxx??f(x)dt??f(t)dt??[f(x)?f(t)]dtaaaxxx
由于f(x)单调增，当a
?F?(x)>0， F(x)在?a,b?单调增，故F(b)>F(a)?0
即 ?tf(t)dt≥
aba?bbf(t)dt ?a2一元微积分学题库（26）定积分的分部积分法
一．填空题：
1．设f(0)?1,f(2)?3,f?(2)?5,则?xf??(2x)dx?2
012．已知f(x)??x12tdt,则1f(x)dx?1(e?1?1) e?02?二．计算下列定积分： 1．?xe?xdx
01解：设u?x,dv?e?xdx,.则
du?dx,v??e?x
由分部积分公式，得
?x?x1?xxedx?[?xe]?e0??dx0?1??e?1?[?e?x]10?1?2e1
2．?4lnxx1d
解：设u?lnx,dv?du?dxx则
1dx,v?2x x由分部积分公式，得
?4lnxx14dx?[22lnx]1??42dxx
14?8ln2?[4x]1?4(ln4?1)3．??3xdx 2sinx4?解：设u?x,dv?dx则 2sinxdu?dx,v??cotx
由分部积分公式，得
xdx3??sin2x?[?xcotx]?3???3cotxdx4441313133?(?)??[lnsinx]??(?)??ln4949224e1e????
e4．?1lnxdx.解：原式=?1?lnxdx??lnxdx??[xlnx?x]11?[xlnx?x]1?2?ee1ee 25.
e?sin(lnx)dx
1eee1esin(lnx)dx?[xsin(lnx)]?xcos(lnx)dx1?1?1xe?esin1?[xcos(lnx)]1??sin(lnx)dx1
?esin1?ecos1?1??sin(lnx)dx1e??sin(lnx)dx?1e1(esin1?ecos1?1) 2x0三．设f(x)在(??,??)上连续，且对一切x有f(x)?x?2?f(t)dt求f(x) 解：由题设： f?(x)?1?2f(x) 则
??2dx?2dx1f(x)?e?(?1?e?dx?c)?ce2x?
21又f(0)?0?c?
2e2x?1 ?f(x)?2