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习题521.试求函数xtdty0sin当x0及4x时的导数.解xtdtdxdyxsinsin0,当x0时,ysin00;当4x时,224siny.2.求由参数表示式tudux0sin,tuduy0cos所给定的函数y对x的导数.解x(t)sint,y(t)cost,ttxtydxdycos)()(.3.求由xyttdtdte000cos所决定的隐函数y对x的导数dxdy.解方程两对x求导得0cosxyey,于是yexdxdycos.4.当x为何值时,函数xtdttexI02)(有极值？解2)(xxexI,令I(x)0,得x0.因为当x0时,I(x)0;当x0时,I(x)0,所以x0是函数I(x)的极小值点.5.计算下列各导数:(1)2021xdttdxd;(2)32411xxdttdxd;(3)xxdttdxdcossin2)cos(.解(1)dxdudttduduxdttdxdux02202112令421221xxxu.(2)323204044111111xxxxdttdxddttdxddttdxd3204041111xxdttdxddttdxd)()(11)()(11343242xxxx12281312xxxx.(3)xxxxdttdxddttdxddttdxdcos02sin02cossin2)cos()cos()cos())(coscoscos())(sinsincos(22xxxx)coscos(sin)sincos(cos22xxxx)sincos(sin)sincos(cos22xxxx)sincos(sin)sincos(cos22xxxx)sincos()cos(sin2xxx6.计算下列各定积分:(1)adxxx02)13(;解aaaxxxdxxxaa230230221|)21()13(.(2)2142)1(dxxx;解852)11(31)22(31|)3131()1(333321332142xxdxxx.(3)94)1(dxxx;解94223942194|)2132()()1(xxdxxxdxxx6145)421432()921932(223223.(4)33121xdx;解66331arctan3arctanarctan13313312xxdx.(5)212121xdx;解3)6(6)21arcsin(21arcsinarcsin1212121212xxdx.(6)axadx3022;解aaaaxaxadxaa30arctan13arctan1arctan1303022.(7)1024xdx;解60arcsin21arcsin2arcsin410102xxdx.(8)dxxxx012241133;解013012201224|)arctan()113(1133xxdxxxdxxxx41)1arctan()1(3.(9)211exdx;解1ln1ln||1|ln12121exxdxee.(10)402tand;解4144tan)(tan)1(sectan40402402dd.(11)dxx20|sin|;解2020sinsin|sin|xdxxdxdxx20coscosxxcoscos0cos2cos4.(12)20)(dxxf,其中12111)(2xxxxxf.解38|)61(|)21(21)1()(2131022121020xxxdxxdxxdxxf.7.设k为正整数.试证下列各题:(1)0coskxdx;(2)0sinkxdx;(3)kxdx2cos;(4)kxdx2sin.证明(1)000)(sin1sin1|sin1coskkkkkxkkxdx.(2))(cos1cos1cos1sinkkkkxkkkxdx0cos1cos1kkkk(3)22|)2sin21(21)2cos1(21cos2kxkxdxkxkxdx.(4)22|)2sin21(21)2cos1(21sin2kxkxdxkxkxdx.8.设k及l为正整数,且kl.试证下列各题:(1)0sincoslxdxkx;(2)0coscoslxdxkx;(3)0sinsinlxdxkx.证明(1)dxxlkxlklxdxkx])sin()[sin(21sincos0])cos()(21[])cos()(21[xlklkxlklk.(2)dxxlkxlklxdxkx])cos()[cos(21coscos0])sin()(21[])sin()(21[xlklkxlklk.(3)dxxlkxlklxdxkx])cos()[cos(21sinsin.0])sin()(21[])sin()(21[xlklkxlklk.9.求下列极限:(1)xdttxx020coslim;(2)xtxtxdttedte0220022)(lim.解(1)11coslimcoslim20020xxdttxxx.(2)22222200002200)(2lim)(limxxtxtxxtxtxxedtedtedttedte22222002002lim2limxxtxxxxtxxedtexeedte2212lim22lim2020222xexeexxxxx.10.设]2,1[]1,0[)(2xxxxxf.求xdttfx0)()(在[0,2]上的表达式,并讨论(x)在(0,2)内的连续性.解当0x1时,302031)()(xdttdttfxxx;当1x2时,6121212131)()(2211020xxtdtdttdttfxxx.因此2161211031)(23xxxxx.因为31)1(,3131lim)(lim30101xxxx,316121)6121(lim)(lim20101xxxx,所以(x)在x1处连续,从而在(0,2)内连续.11.设xxxxxf或000sin21)(.求xdttfx0)()(在(,)内的表达式.解当x0时,00)()(00xxdtdttfx;当0x时,21cos21|cos21sin21)()(000xttdtdttfxxxx;当x时,000|cos210sin21)()(tdttdtdttfxxx10cos21cos21.因此xxxxx10)cos1(2100)(.12.设f(x)在[a,b]上连续,在(a,b)内可导且f(x)0,xadttfaxxF)(1)(.证明在(a,b)内有F(x)0.证明根据积分中值定理,存在[a,x],使))(()(axfdttfxa.于是有)(1)()(1)(2xfaxdttfaxxFxa))(()(1)(12axfaxxfax)]()([1fxfax.由f(x)0可知f(x)在[a,b]上是单调减少的,而ax,所以f(x)f()0.又在(a,b)内,xa0,所以在(a,b)内0)]()([1)(fxfaxxF.