2022年题课教案3.docx

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1、精选学习资料 - - - - - - - - - 学习必备 欢迎下载高等数学习题课作业（不定积分）答案名师归纳总结 13x1 dx13x1 3x213x1 dxx2x11 dx3x53x4xC. 第 1 页,共 4 页3333x3x5421xlnxdx1dxsinlnxdxx1xlnxdlnx11ln2xC.1x 2cosx 2d x x xx2x xdxx3lnsinC. xsinxsinx4x1xdxxdxx2d1x41xC. x11x5arcsinxdxarcsinxdx2arcsinxdxx2 xx1x1x262arcsinxdarcsinx4arcsinx3C.231lnxdxdxl

2、nx1xC. xlnx2xlnx2xln7x41x2dx11112dx113arctanxC. 1x4x2xx3x88 xxdx 114 x2 dx112 x112 x1 12 dx1ln2 x112 ar c t anC. 241 x42 2 82 x149xx2dx x122x11 d x1x1982 x199 x1100dx . 100 1x11001097x11 9749x119890x11 99Csin x cos xdx x2 costan xdx1tan xd tan x1d2 tanx112 arctantanxC. 4 sinx4 cos4 x tanx4 tanx122 t

3、anx 2211sinx13dx xsin2xcos2xdxtanx2 secxdxsin1xdxcossinxcos3xxcostanxdtanxdtanx1tan2xlntanxC 12tanx21sinxxdxsinx 1sinx dxtanxsec x2 secx1 dxsecxtanxxC. sin2 cosx- - - - - - -精选学习资料 - - - - - - - - - 132 x8dx11x2 xdx9学习必备x 2欢迎下载d 1x2 x9d x1 2 12 1dx1xx 2x 21x1xx 25 22 x2名师归纳总结 21xx29arcsin2x51. 4tdtC

4、 .第 2 页,共 4 页14xx12xdx1xx2dx. 令xx2t,就x122,dx2x2t1t22原式dttCxx2C. （也可以令x1）1 C. t15x3x3xdx. 令6xt,就x6t,dx6 t5dtx原式6tdt1 61t11dt6 lntlnt1 Clnx6ln6xtt161x edx. 令1ext, 就xln2t1 ,dx2 tdt,1C. 2t1原式2t2dt21t211dt2 tlnt1C21x eln1ext21t11x e117arcsinxx2dxxarcsinxx21xxx 221x2xx2xdxx22arctanx 2xarcsinxx21xxdxxarcsi

5、nxx22xx2dx22xarcsinxx221x222 2 dxxarcsinxx22 x218exsin2xdx1ex1cos2xdxex1excos2xdx. 222x ecos 2xdx而excos 2 xdxx ecos 2 x2x esin2 xdxx ecos 2 xx 2 esin2x4于是,excos2xdxexcos2x2sin2xC,5所以,原式exexcos2x2sin2x C. 21019xln1xdxdxln1x d1lnxln1xx dxxx2xxx1- - - - - - -精选学习资料 - - - - - - - - - lnxln 1x 111 xdxlnx

6、学习必备欢迎下载ln 1xC11ln 1x C）ln 1x lnxxxxx名师归纳总结 20xlnx2dx1lnxd1122lnx211dxx. ln1x 2C .第 3 页,共 4 页1x22x 1x2x2lnx2111x dxlnx21lnx1ln 12 xClnx12 1x2x2 x12x242 12 x42 x21xsinxdx2xdxx 2sinxdxxdtanxd1cosx1cosxcos21cosx21cosxC 1ln2）xtanxtanxdxln1cosxxtanx2lncosxln1cosxC12222xtanxln 1cosx ln2ln 1cosx C 1xtanxC.

7、 （其中C2222如函数fx有连续导数,且fxln 1x,求fx . x解： 由fxln 1x2,有fxln12x2,所以x221xdxxfx fx dxl n 2x2dxl n xx2x2x2dxx1xl n xx221dx2l n xx22ar c t a nC.x23如xfx dxarcsinxC,求fdx. x解： 所给等式两边对x 求导,有xfx11x2,于是fx x1x2,所以1fdxx1x2dx11x2d 12 x11x23C.2x 2324如fsin2xxx,求1xxfx dx. sin解： 由fsin2xxxarcsinsin2x,得fxarcsinx,于是sinsin2xx

8、1xxfx dxarcsinxxdx21xarcsinx21x1x11- - - - - - -精选学习资料 - - - - - - - - - 21xarcsinx1dx学习必备x欢迎下载x2xC. 21arcsinx名师归纳总结 25已知函数Fx 是连续函数fx的一个原函数,当x0时,Fxfxsin22x,且第 4 页,共 4 页F0 1,Fx0,求fx. ,有解： 由F x fx sin22x,即FxFxsin22x1 1cos4x 21F2x FxFx dx1 1c o s 4 xdx1x1s i n 4x1C.222820,有于是,F2x x1sin4 xC .由F0 1,得C1,再由Fx4Fx 14xs i n 4x4,所以fxFx41cos4x4. 2xsin4x- - - - - - -