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二、定积分的分部积分法第三节不定积分一、定积分的换元法换元积分法分部积分法定积分换元积分法分部积分法定积分的换元法和分部积分法第五章一、定积分的换元法定理1.设函数函数满足:1),],[)(1Ct2)在],[上;)(,)(ba)(t)(t证:所证等式两边被积函数都连续,因此积分都存在,且它们的原函数也存在.是的原函数,因此有则)()(aFbF)]([F)]([F)(t)(t)(t)(t)(t则说明:1)当,即区间换为,时],[定理1仍成立.2)必需注意换元必换限,原函数中的变量不必代回.3)换元公式也可反过来使用,即))((tx令xxfbad)(或配元)(t)(dt配元不换限)(t)(t)(t)(t)(t)(t例1.计算解:令,sintax则,dcosdttax;0,0tx时当.,2tax时∴原式=2attad)2cos1(2202)2sin21(22tta0220ttdcos222xayxoya且例2.计算解:令,12xt则,dd,212ttxtx,0时当x,4时x.3t∴原式=ttttd231212ttd)3(21312)331(213tt13;1t且例3.证:(1)若aaaxxfxxf0d)(2d)(则xxfaad)((2)若0d)(aaxxf则xxfad)(0xxfad)(0ttfad)(0xxfad)(0xxfxfad])()([0时)()(xfxf时)()(xfxf偶倍奇零tx令例4.计算例5.计算例6.若二、定积分的分部积分法定理2.,],[)(,)(1baCxvxu设则ab证:)()()()(])()([xvxuxvxuxvxu)()(xvxuabxxvxuxxvxubabad)()(d)()()()(xvxuabbaxxvxud)()(上积分两端在],[ba例7.计算解:原式=xxarcsin021210xxxd1212)1(d)1(212022121xx1221)1(2x0211223120dcosttn20dcosxxn例8.证明证:令,22143231nnnnn为偶数n为奇数,2xt则20dsinxxn022d)(sinttn令则,cossin)1(2xxnunxvcos]sincos[1xxInn022022dcossin)1(xxxnn02022dcossin)1(xxxnInn2022d)sin1(sin)1(xxxnn2)1(nIn由此得递推公式21nnnnII于是mI2mm21212mI122mm而0I20dx,2201dsinxxI1故所证结论成立.0I1I22mI2232mm42mI214312mI1222mm32mI3254内容小结基本积分法换元积分法分部积分法换元必换限配元不换限边积边代限思考与练习1.提示:令,txu________d)(sindd0100ttxxx则ttxxd)(sin0100u100sinx100sin2.设解)(3xf3.设求解:(分部积分)解:4.右端,],[)(上有连续的二阶导数在设baxf)(af且试证baxfbxax)(d))((21abxfbxax)())((21xbaxxfbad)2)((21分部积分积分再次分部积分xxfbad)(abxfbax)()2(21=左端,0)(bf
本文标题:二、定积分的分部积分法
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