不定积分03-巩固习题.pdf

不定积分巩固练习不定积分巩固练习1.2d23xxx_.2.4d1x xx_.3.d4xx x_.4.计算下列不定积分：（1）2sind2cosxxx.（2）2cosd1cosxxx.5.计算22d1xxx.6.计算21d1xxx.7.计算arctanedexxx.8.计算ln 1dxx.9.求32d368xxxx.10.求2d11xxx x.11.求arctandxxx.12.求d4e1xx.13.求2arctand1xxxx.14.求4dsin cosxxx.15.求arctandxx x.16.求arctan 1dxx.17.求2lnsindsinxxx.18.求e de1xxxx.19.求21arcsin dxx x.20.求lnd2xxx.1.【解】232233d 23d 23113dd 2344234232323xxxxxxxxxx13ln 2344 23xCx.2.【解】22242dd11arctan1221xx xxCxx.3.【解】22dd22ln442xxxxCx xx.4.【解】（1）222d cossin1cosdarctan2cos222cosxxxxCxx .（2）22coscos1 11ddcos1d1cos1cos1cosxxxxxxxxx 21cossindsincotcscsinxxxxxxxxCx.5.【解】22222211ddd1d111xxxxxxxxxx21arcsinarcsin122xxxxC21arcsin122xxxC.6.【解】221sectanddcsc dtan sec1txtxtt tttxx211ln csccotlnxttCCxx.7.【解】22arctanearctanearctanedd edeexxxxxxttxtt21arctan1arctan dd1ttttttt 22arctan1arctan1dlnln 112ttttttCtttt 22arctane1elne21exxxxC.8.【解】令2xt，则 222ln 1dln 1dln 1d1txxtttttt21ln 11d1ttttt 22ln 1ln12tttttC 1 ln 12xxxxC.9.【解】令22211368243xxxxxx22433ABCDxxxx，由224323234241A xxB xxC xxxD xx得0108903321266036182481ABCABCDABCDABCD，解得11,0,122ABCD ，故222d111d22243683xxxxxxxx141ln223xCxx.10.【解】2dsec tan1secddarccossec tan1xttxttttCCttxx x.11.【解】arctand2arctand2arctan dxxtxxxt tx222 arctan2d2 arctanln 11tttttttCt2arctanln 1xxxC.12.【解】2222d eded24e1e4e4xxxxxxx 22ln ee4xxC.13.【解】22arctandarctan d11xxxxxx222221d1arctand1arctan11xxxxxxxxx221arctanln1xxxxC.14.【解】22224444dsincossincosdddsin cossin cossin cossin cosxxxxxxxxxxxxxxxx224232d cos11sincosddcossin cos3cossin cosxxxxxxxxxxx 32d cos1csc d3coscosxx xxx311ln csccot3coscosxxCxx.15.【解】341arctand2arctan darctan d2xtxx xtt ttt444422111111arctandarctand221221tttttttttt422111arctan1d221ttttt 4311 1arctanarctan22 3tttttC 231111 arctan262xxxxC.16.【解】令xt，则 2222arctan 1darctan 1darctan 1d11txxtttttt2222arctan 11d22tttttt22arctan 1ln22tttttC arctan 1ln22xxxxxC.17.【解】2lnsincosdlnsin d cotcot lnsincotdsinsinxxxxxxxxxxx 2cot lnsincsc1 dcot lnsincotxxxxxxxxC .18.【解】令e1xt，则222ln 1,dd1txtxtt，则2221ln 1e d2d1e1xxttxxtttt222222ln 1d2 ln 12d1ttttttt22212 ln 141d2 ln 144arctan1tttttttCt2e14 e14arctane1xxxxC 19.【解】221sin1arcsin dcosd1cos2d2xtxx xtt tttt2211d sin2sin2sin2 d4444ttttttt t2111sin2cos2448ttttC222111arcsin1arcsin424xxxxxC.20.【解】ln2d2ln d222ln2d2xxxxxxxxxx因为222222d2d21d22xtxtxttxtt422arctan222 2arctan222txtCxC，所以ln2d22ln424 2arctan22xxxxxxCx.