不定积分练习题04【10题】
1.excosxdx【分部积分法】
解:原式=exdsinxexsinxsinxdexexsinxexsinxdx
=exsinxexdxcosxexsinxexcosxcosxdexexsinxcosxexcosxdx 令excosxdxA,则有:2A=exsinxcosx
exsinxcosxC 综上:A=2
x02.e2xsindx【分部积分法】
2解:原式= =
=
x12xx1x1xesine2xcose2xsind(x)【令Ae2xsindx】
22242221x12xxx12xx1xsind(e2x)esine2xd(sin)esine2xcosd(x) 22222222212xx11x2x)1e2xsinx1e2xcosxe2xd(cosx) esincosd(e2222224222 整理后的:A=
12xx1x112xx1x1esine2xcosA esine2xcosA224222282161712xx1xAesine2xcos 16228282xx2x2xxesine2xcosCe2x4sincosC 1721721722 A
03.x2arctanxdx【分部积分法】
111111dx 解:原式=arctanxdx3x3arctanxx3darctanxx3arctanxx323331x33131x(1x2)x131xdxxarctanxx =xarctanxdx
331x2331x2131x2113x2122ln(1x)Cxarctanxln(1x)C =xarctanx3322366
x04.xcosdx【分部积分法】
2xxxxx解:原式=x2d(sin)2xsin2sindx2xsin4cosC
22222
05.xtan2xdx【分部积分法】
x2xd(tanx) 解:原式=x(secx1)dxxsecxxdxxsecxdxxdx2222x2x2xtanxtanxd(x)xtanxlncosxC =2206.xsinxcosxdx【分部积分法】
1111xcos(2x)cos(2x)dx 解:原式=xsin2xdxxdcos2x2224 =
07.te2tdt【分部积分法】 解:原式=t1112t12tdte2tte2te2tdteC te22221111xcos(2x)sin(2x)Cxcos(2x)sin(2x)C 4248 =
12t1etC 2208.ln2xdx【分部积分法】
1解:原式=xln2xxd(ln2x)xln2xx2lnxd(x)xln2x2lnxd(x)
x1 =xln2x2xlnxxd(lnx)xln2x2xlnxxd(x)
x =xln2x2xlnx2dxxln2x2xlnx2xC
09.x2cosxdx【分部积分法】
解:原式=x2d(sinx)x2sinxsinxd(x2)x2sinx2xsinxd(x)
=x2sinx2xsinxd(x)x2sinx2xd(cosx)x2sinx2xcosx2cosxd(x) =x2sinx2xcosx2sinxC
x10.x2cos2dx【分部积分法】
2xxxx1cosx解:∵cosxcos2sin22cos21∴cos2
22222cosx1cosx111 原式=x2dxx2dxx2dxx2dxx2d(sinx)
22222x3x21x3x212 =sinxsinxd(x)sinx2xsinxd(x)
622622x3x2x3x2 =sinxxsinxd(x)sinxxd(cosx)
6262x3x2x3x2 =sinxxcosxcosxd(x)sinxxcosxsinxC
6262