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习题531.计算下列定积分:(1)2)3sin(dxx;解0212132cos34cos)3cos()3sin(22xdxx.(2)123)511(xdx;解51251110116101)511(2151)511(22122123xxdx.(3)203cossind;解203203sincoscossindsd410cos412cos41cos4144204.(4)03)sin1(d;解02002003cos)cos1(cossin)sin1(dddd34)cos31(cos03.(5)262cosudu;解2626262622sin4121)2cos1(21cosuuduuudu836)3sin(sin41)62(21.(6)dxx2022;解dtttdtttxdxx2020202)2cos1(cos2cos2sin22令2)2sin21(20tt.(7)dyy22228;解44222222cos2cos22sin24228xdxxxydyydyy令)2(2)2sin21(22)2cos1(224444yxdxx.(8)121221dxxx;解41)cot()1sin1(cossincossin12424224212122ttdtttdttttxdxxx令.(9)adxxax0222;解2024202202222sin4coscossinsintdtatdtatatataxdxxaxa令164sin328)4cos1(84204204204atatadtta.(10)31221xxdx;解34223122secsectan1tan1tdttttxxxdx令3322sin1sincos34342tdttt.(11)1145xxdx;解61)315(81)5(81454513133211uuduuuxxxdx令.(12)411xdx;解)32ln1(2|)1|ln(2)111(2211121212141uuduuuduuuxxdx令.(13)14311xdx;解2ln21|)1|ln(2)111(2)2(11111210210021143uuduuduuuuxxdx令.(14)axaxdx20223;解)13(3)3(3121320202222222022axaxadxaxaxdxaaa.(15)dttet1022;解2110102221021)2(222eetdedttettt.(16)21ln1exxdx;解)13(2ln12lnln11ln1222111eeexxdxxxdx.(17)02222xxdx;解2)1arctan(1arctan)1arctan()1(112202022022xdxxxxdx.(18)222coscosxdxx;解32)sin32(sinsin)sin21(2coscos22322222xxxdxxdxx.(19)223coscosdxxx;解222223cos1coscoscosdxxxdxxx34cos32cos32sincos)sin(cos202302232002xxxdxxdxxx(20)02cos1dxx.解22cos2sin22cos1000xxdxdxx.2.利用函数的奇偶性计算下列积分:(1)xdxxsin4;解因为x4sinx在区间[,]上是奇函数,所以0sin4xdxx.(2)224cos4d;解202204224)22cos1(8cos42cos4dxdd20202)4cos212cos223(2)2cos2cos21(2dxxdxx23)4sin412sin23(20xx.(3)2121221)(arcsindxxx;解210221022212122)(arcsin)(arcsin21)(arcsin21)(arcsinxdxdxxxdxxx324)(arcsin3232103x.(4)55242312sindxxxxx.解因为函数12sin2423xxxx是奇函数,所以012sin552423dxxxxx.3.证明:aaadxxdxx022)(2)(,其中(u)为连续函数.证明因为被积函数(x2)是x的偶函数,且积分区间[a,a]关于原点对称,所以有aaadxxdxx022)(2)(.4.设f(x)在[b,b]上连续,证明bbbbdxxfdxxf)()(.证明令xt,则dxdt,当xb时tb,当xb时tb,于是bbbbbbdttfdttfdxxf)()1)(()(,而bbbbdxxfdttf)()(,所以bbbbdxxfdxxf)()(.5.设f(x)在[a,b]上连续.,证明babadxxbafdxxf)()(.证明令xabt,则dxdt,当xa时tb,当xb时ta,于是babaabdttbafdttbafdxxf)()1)(()(,而babadxxbafdttbaf)()(,所以babadxxbafdxxf)()(.6.证明:11122)0(11xxxxdxxdx.证明令tx1,则dttdx21,当xx时xt1,当x1时t1,于是11121122211)1(1111xxxdttdtttxdx,而xxdxxdtt1121121111,所以1112211xxxdxxdx.7.证明:1010)1()1(dxxxdxxxmnnm.证明令1xt,则10100110)1()1()1()1(dxxxdtttdtttdxxxmnnmnmnm,即1010)1()1(dxxxdxxxmnnm.8.证明:020sin2sinxdxxdxnn.证明2020sinsinsinxdxxdxxdxnnn,而2020202sinsin))((sinsinxdxtdtdtttxxdxnnnn令,所以020sin2sinxdxxdxnn.9.设f(x)是以l为周期的连续函数,证明1)(aadxxf的值与a无关.证明已知f(xl)f(x).alalllallaaadxxfdxxfdxxfdxxfdxxfdxxfdxxf00001)()()()()()()(,而aaalaldxxfdxlxfdtltfltxdxxf000)()()()(令,所以laadxxfdxxf01)()(.因此1)(aadxxf的值与a无关.10.若f(t)是连续函数且为奇函数,证明xdttf0)(是偶函数;若f(t)是连续函数且为偶函数,证明xdttf0)(是奇函数.证明设xdttfxF0)()(.若f(t)是连续函数且为奇函数,则f(t)f(t),从而)()()()1)(()()(0000xFdxxfdxufduufutdttfxFxxxx令,即xdttfxF0)()(是偶函数.若f(t)是连续函数且为偶函数,则f(t)f(t),从而)()()()1)(()()(0000xFdxxfdxufduufutdttfxFxxxx令,即xdttfxF0)()(是奇函数.11.计算下列定积分:(1)10dxxex;解11011010101021eeedxexexdedxxexxxxx.(2)exdxx1ln;解)1(414121121ln21ln21ln21220212121exedxxxxxxdxxdxxeeeee.(3)20sintdtt(为常数);解20202020cos1cos1cos1sintdtttttdtdtt220222sin12t.(4)342sindxxx;解34343434342sinln4313cotcotcotsinxxdxxxxxddxxx23ln21)9341(.(5)41lndxxx;解4141414112ln2ln2lndxxxxxxxddxxx)12ln2(442ln8122ln84141xdxx.(6)10arctanxdxx;解xdxxxxxdxxdxx1022102102101121arctan21arctan21arctan214)41(218)arctan(218)111(21810102xxxdx.(7)202cosxdxex;解202202202202sin2sinsincosxdxexexdexdxexxxx202202202202cos42cos4cos2cos2xdxeexdxexeexdeexxxx所以)2(51cos202exdxex,于是(8)212logxdxx;解212212221222122ln121log21log21logdxxxxxxdxxdxx2ln432212ln212212x.(9)02)sin(dxxx;解020302022sin4161)2cos1(21)sin(xdxxdxxxdxxx003000332cos41622sin412sin416xxdxdxxxx462sin81462cos412cos416303003xxdxxx.(10)edxx1)sin(ln;解法一101sinln)sin(lndtettxdxxte令.因为10101010cossinsinsintdtetetdedtettttt101010sincos1sincos1sintdteteetdeettt10sin11cos1sintdteeet,所以)11cos1sin(21sin10eetdtet.因此)11cos1sin(21)sin(ln1eedxxe.解法二eeeedxxedxxxxxxdxx1111)cos(ln1sin1)cos(ln)sin(ln)sin(lneedxxxxxxe111)sin(ln)cos(ln1sinedxxee0)sin(