积分

Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Av ec Yu anh on g Z hi 第三章 Integration §3.1 不定积分和定积分的计算 一、利用不定积分的线性性（Linearity）计算下列不定积分： (1) � p x� x+ x3e3 x3 dx; (2) � ( p x� x)( p x3 � 1)dx; (3) � 2x2 1 + x2 dx; (4) � 2 + 4x2 x2(1 + x2) dx; (5) � 3xexdx; (6) � (2x + 5x)2dx (7) � cos2 �x 2 � dx; (8) � secx � (secx� tanx)dx; (9) � cos(2x) cosx� sinxdx; (10) � cos(2x) cos2 x � sin2 xdx; (11) � cot2 xdx: 解: (1)只需把分式展开即可，但为了便于利用不定积分公式求出原，应尽量 把其中的某些项写成Power function（幂函数）的形式。 原式 = � (x� 5 2 � x�2 + e3)dx = x � 5 2 +1 �52 + 1 � x �2+1 �2 + 1 + e 3x+ C = �2 3 x �3 2 + 1 x + e3x+ C: (2)只需去括号再整理即可： 原式 = � (x2 � x 12 � x 52 + x)dx = x 3 3 � 2x 3 2 3 � 2x 7 2 7 + x2 2 + C: (3)对于有理函数（即可示为两个多项式相除的形式的函数），应该先将之作部 分分式（partial fraction decomposition）分解（详见 elearning上相关内容），但这里可不 必按照常规的分解程序来处理。实际上， 2x2 1 + x2 = 2(x2 + 1� 1) 1 + x2 = 2� 2 �1 + x2��1 Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Av ec Yu anh on g Z hi 22 第三章 Integration 故 原式 = � � 2� 2 �1 + x2��1� dx = 2x� 2 arctanx+ C: (4)此题同 (3)，需要先对被积函数作部分分式分解，然后再利用线性。注意到 2 + 4x2 x2(1 + x2) = 4(x2 + 1)� 2 x2(1 + x2) = 4 x2 � 2 x2(1 + x2) = 4 x2 �2 � 1 x2 � 1 1 + x2 � = 2 x2 + 2 1 + x2 ; 故 原式 = � � 2 x2 + 2 1 + x2 � = 2 arctan (x)� 2x�1 + C: (5) 原式 = � (3e)xdx = (3e)x ln(3e) + C: (6)只需把被积函数按完全平方公式展开即可： 原式 = � (4x+2 � 2x5x+25x)dx = � (4x+2 � 10x+25x)dx = 4 x ln 4 +2 10x ln 10 + 25x ln 25 +C: (7) 原式 降次======= � 1 + cosx 2 dx = 1 2 �� dx+ � cosxdx � = 1 2 (x+ sinx) + C: (8) 原式 去括号========= � (sec2 x�secx�tanx)dx = � sec2 xdx� � secx�tanxdx = tanx�secx+C: (9) 原式 = � cos2 x� sin2 x cosx� sinx dx = � (cosx+ sinx)dx = sinx� cosx+ C: (10) 原式 = � cos2 x� sin2 x cos2 x � sin2 x dx = � (csc2 x� sec2 x)dx = � cotx� tanx+ C: (11) 原式 = � (csc2 x� 1)dx = � cotx� x+ C: School of Mathematics and Statistics, Yunnan University Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Av ec Yu anh on g Z hi 3.1 不定积分和定积分的计算 23 例 3.1.1 用Push forward或者Pull back代换求下列不定积分： (1) � sin(3x)dx; (2) � e�3x+2dx; (3) � 1 3 p 2� 3xdx; (4) � xe�x 2 dx; (5) � x cos(x2)dx; (6) � 3x3 1� x4dx; (7) � sin(lnx) x dx; (8) � 1 x lnx � ln(lnx)dx; (9) � e 1 x x2 dx; (10) � sin(px)p t dt; (11) � asinx cosxdx; (a > 0; a 6= 1); (12) � sinx cos3 xdx; (13) � 1 4 + 9x2 dx; (14) � 1p 16� 25x2dx; (15) � cos3 xdx; (16) � 1 + lnx (x lnx)2dx; (17) � 1 (x+ 2) p 1 + x dx; (18) � 1p x� 1dx; (19) � x+ 1p 1� x2dx; (20) � 1 x2 p 1 + x2 dx; (21) � 1 x p a2 � x2dx; (a > 0); (22) � 1p (x2 + 1)3 dx; (23) � p x2 � 9 x dx; (24) � 1 1 + p 1� x2dx: 解: (1)因为d(3x) = 3dx，故原式 = 1 3 � sin(3x)d(3x) = �1 3 cos(3x) + C: (2)因为dx = �1 3 d(�3x+ 2),所以原式 = �1 3 � e�3x+2d(�3x+ 2) = �1 3 e�3x+2 + C: (3)原式 = � (2� 3x)� 13dx = �1 3 � (2� 3x)� 13d(2� 3x) = �1 3 1 2 3 (2� 3x) 23 + C = �1 2 (2� 3x) 23 + C: (4)原式 = � e�x 2 � (xdx) = �1 2 � e�x 2 d(�x2) = �1 2 e�x 2 + C: (5)利用xdx = 1 2 d(x2)，原式 = 1 2 � cos(x2)dx2 = sin(x2) + C: (6)原式 = 3 � 1 1� x4 � (x 3dx) = 3 �4 � 1 1� x4d(1� x 4) = �3 4 ln ��1� x4��+ C: (7)原式 = � sin(lnx) � � 1 x dx � = � sin(lnx)d lnx = � cos(lnx) + C: (8)原式 = � 1 ln(lnx) � 1 lnx � 1 x dx = � 1 ln(lnx) � 1 lnxd lnx = � 1 ln(lnx)d ln(lnx) = ln �� ln(lnx)��+ C: (9)原式 = � e 1 x � 1 x2 dx = � � e 1 xd � 1 x � = �e 1x + C: (10)原式 = � sin( p t) � 1p t dt = 2 � sin( p t)d p t = �2 cos(pt) + C: Avec Yuanhong Zhi Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Av ec Yu anh on g Z hi 24 第三章 Integration (11)原式 = � asinx � (cosxdx) = � asinxd sinx = a sinx ln a + C: (12)原式 = � 1 cos3 x � sinxdx = � 1 cos3 xd cosx = 1 2 cos�2 x+ C: (13)因为 1 4 + 9x2 dx = 1 4 � 1 + (3x2 ) 2 �dx = 1 6 � 1 1 + (3x2 ) 2 d � 3x 2 � ，所以， 原式 = � 1 6 � 1 1 + (3x2 ) 2 d � 3x 2 � = 1 6 arctan � 3x 2 � + C: (14)因为 1p 16� 25x2dx = 1 4 q 1� (5x4 )2 dx = 1 5 1q 1� (5x4 )2 d � 5x 4 � ; 所以 原式 = � 1 5 1q 1� (5x4 )2 d � 5x 4 � = 1 5 arcsin � 5x 4 � + C: (15)原式 = � cos2 x � cosxdx = � cos2 xd sinx = � (1� sin2 x)d sinx = � (sin0 x� sin2 x)d sinx = sinx� sin 3 x 3 + C: (16)法一: 因为 1 + lnx (x lnx)2dx = 1 (x lnx)2d(x lnx);事实上 d(x lnx) = (1 + lnx)dx; 所以 原式 = � 1 (x lnx)2d(x lnx) = � 1 x lnx + C: 注意，法一不容易想到，若对 (dx lnx) 的微分不熟悉的话。下面的法二就易理解，但此 法要用到分部积分 法二：指数代换+分部 为将微分形式有理化，令 lnx = t;则x = et; dx = et dt; 由此 D := 1 + lnx (x lnx)2dx = 1 + t (ett)2 � etdt = 1 + t t2 e�tdt = t�2e�tdt+ t�1e�tdt 但 t�1e�1dt e �t右移=========== t�1d(�e�t) 分部微分============ d(�t�1e�t)� (�e�t)d(t�1) = d(�t�1e�t)� e�tt�2dt; School of Mathematics and Statistics, Yunnan University Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Av ec Yu anh on g Z hi 3.1 不定积分和定积分的计算 25 由此 D = t�2e�tdt+ d(�t�1e�t)� e�tt�2dt = d(�t�1e�t) 回代======= d ��(x lnx)�1� ; 从而 原式 = �(x lnx)�1 + C: (17)法一：利用根式代换，令px+ 1 = t;则：x = t2 � 1;从而 dx = 2tdt;于是： 原式 = � 1 (t2 + 1)t � 2tdt = 2 � 1 1 + t2 dt = 2 arctan t+ C 回代变元=========== 2 arctan p x+ 1 + C: 此题也可直接使用Push forward 法二： 1 (x+ 2) p 1 + x dx = 1 1 + ( p 1 + x)2 � 1p 1 + x dx = 2 � 1 1 + ( p 1 + x)2 d p 1 + x = 2d arctan p 1 + x; 所以 原式 = 2 arctanp1 + x+ C: (18)利用根式代换，令px = t;则：x = t2;从而 dx = 2tdt;于是： 原式 = � 1 t� 12tdt 部分分式=========== 2 � (1 + 1 t� 1)dt = 2 � dt+ 2 � 1 t� 1dt = 2t+ 2 � 1 t� 1d(t� 1) = 2t+ 2 ln jt� 1j+ C 回代变元=========== 2 p x+ 2 ln jpx� 1j+ C: (19)法一［三角代换］，这里采用正弦代换，用余弦代换也可：令x = sin t; jtj < � 2 ;则： dx = cos tdt;于是： 原式 = � sin t+ 1 cos t cos tdt = � (sin t+ 1)dt = t� cos t+ C = arcsinx� cos t+ C: 为便于回代变元，不妨考虑当0 < t < � 2 时，视 t为 t弧度，在基本直角三角形中考 察，则如图 1.3有： cos t = p 1� x2;从而 原式 = arcsinx� p 1� x2 + C: 法二：注意到 x+ 1p 1� x2dx = xp 1� x2dx+ d arcsinx = �1 2 � (1� x2)� 12d(1� x2) Avec Yuanhong Zhi Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Av ec Yu anh on g Z hi 26 第三章 Integration + d arcsinx = �d(1� x2)1/2 + d arcsinx = d(� p 1� x2 + arcsinx);所以 原式 = � p 1� x2 + arcsinx+ C: (20)利用正切代换以将微分形式有理化，令x = tan t; 0 < jtj < � 2 ;则：dx = sec2 tdt; 于是： 原式 = � sec2 t tan2 t sec tdt = � cos t sin2 t dt = � csc t+ C 回代变量=========== � p 1 + x2 x + C: 注：上述回代变量过程如下：不妨设0 < t < � 2 ，则如图 1.4所示便知。 此题也可用分部积分求解 法二：由于 1 x2 p 1 + x2 dx = p 1 + x2 x2 � 1p 1 + x2 ! dx = p 1 + x2d � �1 x � � 1p 1 + x2 dx; 而 p 1 + x2d � �1 x � 分部微分=========== d � �1 x � p 1 + x2 � + 1 x � xp 1 + x2 dx; 代入后便得 1 x2 p 1 + x2 dx = d � p 1 + x2 x ! ;由此 原式 = � p 1 + x2 x + C: (21)用正弦代换或者余弦代换，这里使用正弦代换，令x = a sin t; jtj < � 2 ;则： dx = a cos t dt;于是： 原式 = � 1 a sin t � a cos ta cos tdt = � 1 a sin tdt = 1 a ln j csc t� cot tj+ C 回代变量=========== 1 a ln �����ax � p a2 � x2 x �����+ C: (变元回代过程见图 1.5) (22)此题同 (20)题类似，用正切代换即可。令x = tan t; 0 < jtj < � 2 ;则：dx = sec2 tdt; 于是 原式 = � sec2 t sec3 tdt = � cos t dt = sin t+ C 回代======= xp 1 + x2 + C: (回代过程参见图 1.4) School of Mathematics and Statistics, Yunnan University Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Av ec Yu anh on g Z hi 3.1 不定积分和定积分的计算 27 (23)易见用正割代换便可有理化 [理由如下：由于被积函数的定义域是 D := fx �� jxj � 3g， 于是由 jxj � 3() j3 x j � 1，故可令 3 x = cos t; 0 � t < � 2 或 � 2 < t � �,于是 x = 3 sec t，其中0 � t < � 2 或 � 2 < t � �]，为此，令 x = 3 sec t;其中0 � t < � 2 或� 2 < t � �;则：dx = 3 sec t tan tdt; 于是换元得： (i)当0 � t < � 2 时（此时对应于x � 3时），原式 = 3 � tan2 tdt = 3 � (sec2 t� 1)dt = 3(tan t� t) + C 回代变量=========== 3 p x2 � 9 3 � arccos � 3 x �! + C = p x2 � 9� 3 arccos � 3 x � + C: 回代变量过程见下图所示。 (ii)当 �2 < t � �（对应于 x � �3）时，类似地有： 原式 = � 3 p tan2 t 3 sec t � 3 sec t tan t dt 此时的回代变量过程见下图所示。 = �3 � tan2 t dt = �3 � (sec2 t� 1) dt = �3(tan t� t) + C = �3 tan t+ 3t+ C 回代======= �3 p x2 � 9 �3 � arccos � 3 x �! = p x2 � 9 + 3 arccos � 3 x � + C: √ x2 − 9x 3 t tan t = √ x 2 −9 3 图 1.1: 例 3.1.1的 23题在 0 � t < � 2 时 回代变元图示请大家请注意，本题在 t � �2（对应于 x � 3 时的情形） 和 �2 < t � � 时 （对应于 x � �3 时的情形）得到的结果并不一样。 实际上，易验证：当 x > 3 时，有� arccos � 3 x ��0 = 3 x p x2�9 ; 而当 x < �3 时，有�� arccos � 3x��0 = �3�x�(�x) 1p1�9/x2 = 3xpx2�9 ; 故 arccos � 3 x � 和 � arccos � 3x� 均是 3xpx2�9 的原函数， 但这两个原函数并不相差一个常数，这和“区间 上的连续函数的任意两个原函数至多相差一个常 数”并不矛盾，因为这两个函数是分别定义在不相 −3 −x O t √ x 2 − 9 tan t = √ x 2 −9 −3 图 1.2: 例 3.1.1的 23题在 � 2 < t � � 时 回代变元图示 Avec Yuanhong Zhi Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Av ec Yu anh on g Z hi 28 第三章 Integration 交的两个区间上的。 (24)法一［三角代换］，这里，我们采用正弦代换，即令x = sin t; jtj � � 2 ，则：dx = cos t dt;于是： 原式 = � cos t 1 + cos tdt = � cos t+ 1� 1 cos t+ 1 dt = � (1� 1 1 + cos t)dt = t� � 1 1 + cos tdt 分母升次=========== t� � 1 2 cos2 � t 2 �dt = t� � sec2� t 2 � d � t 2 � = t� tan � t 2 � + C 回代变量=========== arcsinx� x 1 + p 1� x2 + C: 回代变量过程见图 1.6所示。 法二，利用分母有理化结合分部积分：由于所给的微分形式 1 1 + p 1� x2dx 分母有理化============== 1� p 1� x2 x2 dx = (1� p 1� x2)d � �1 x � = d �� �1 x � � � 1� p 1� x2 �� � � �1 x � d � 1� p 1� x2 � (�) = d �� �1 x � � � 1� p 1� x2 �� � � �1 x � xp 1� x2dx = d �� �1 x � � � 1� p 1� x2 �� + 1p 1� x2dx = d �� �1 x � � � 1� p 1� x2 � + arcsinx � = d � arcsinx� x 1 + p 1� x2 � ; 由此可知 原式 = arcsinx� x 1 + p 1� x2 + C: 注意，上述过程中的 (�) 用到了微分恒等式（本文称为分部微分公式［DBP］） f(x)dg(x) = d(f(x)g(x))� g(x)df(x): 例 3.1.2 请用 Integration By Parts (IBP for short)［分部积分法］求下列不定积分： (1) � x sinx dx; (2) � xe�x dx; (3) � arcsinx dx; (4) � x2ex dx; (5) � x2 lnx dx; (6) � lnx x3 dx; (7) � ln2 x dx; (8) � x sec2 x dx; (9) � ex sinx dx; (10) � cos(lnx) dx; (11) � sec3 x dx; (12) � e 3 p x dx; (13) � x3 cosx dx: School of Mathematics and Statistics, Yunnan University Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Av ec Yu anh on g Z hi 3.1 不定积分和定积分的计算 29 x1 √ 1− x2 t sin t = x cos t = √ 1− x2 图 1.3: 例 3.1.1 的 19题回 代变元图示 x √ 1 + x2 1 t csc t = √ 1+x2 x 图 1.4: 例 3.1.1 的 20题回 代变元图示 xa √ a2 − x2 t sin t = x a cos t = √ a2 − x2 图 1.5: 例 3.1.1 的 21题回 代变元图示 x− α α 1 √ 1− x2 t 2 t 2 Òs �¿ 1Òs�¿š� α x−α = √ 1−x2 1 ⇐⇒ α = x √ 1−x2 1+ √ 1−x2 tan ( t 2 ) = α√ 1−x2 = x 1+ √ 1−x2 图 1.6: 例 3.1.1的 24题回代变元图 示 ⚗ 3.1.1 在使⻗分部积分时，一㯛的䍳则是：如果᰹不定积分ᤑᢾ上䕟为� (ផ1 �ផ2 � � �ផn) dජ; 其中͋ ផ1 �ផ2 � � �ផn Ჶ͌䉚积函数可䈗㋩为 n 个ᆩ⃛ௌ等函数ェࠇ，䦈䭻的ᆩ⃛ௌ 等函数Ჶ：޸䎁函数 Ჶ֞数函数 ៱֞函数 ᖨ֞数函数 ർ֞޸䎁函数，则：在分部积分时， ૷ᖨ䦈ផ䥨的 5大㘪ᆩ⃛ௌ等函数䧉䥶͋ ᰯ⃞上刳处ⶵ͌͋ Ხ到͌͋ ජ 䥨͌虹不ፗ㔯㎟ 为͋ ජ㎪ 虺͌的ࣇ૷利序为：޸>Ჶ>៱>ᖨ>ർ，⫥ල଼䦊䇻 IBP Ⱘ֟௚ჟ，当n=1，且 ͋ផ 1 是͌͋ ᖨ ᯅ͌͋ ർ 时͌，直ᵔ IBPഢ可 ඕ֟ዅ，IBP⃛䛗上是䥿⻗微分ଛᢾ f(x)dg(x) = d � f(x)g(x) �� g(x)df(x) ; ࢔޺不ፗ㎟䖔ଛᢾ为分部微分ଛᢾ (Differentiation By Parts, DBP for short)଼֟ 㦴，ᖨ 于微分ᤑᢾ f(x)dx，如果有్⚄ᖵߺୈᮿ f(x) dx = dF (x) 的ᤑᢾ，则 F (x) ᗠ是微分ᤑᢾ f(x) dx 的一个ൎ函数，ࡽ㦻 � f(x) dx = F (x)+C Ỵ֟ 在◱不定积分虹⻉㮢定积分虺时，有时ᯀ࢛ࠎ直ᵔᖨ微分ᤑᢾ f(x) dx 䧉䥶䦱当的Ὠ⚄ 虹಴ᲛPush forward，Pull back，分部微分等虺ᖵߺୈᮿ dF (x)，䦊㦻८可◱ன㢂果֟ Avec Yuanhong Zhi Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Av ec Yu anh on g Z hi 30 第三章 Integration 解: (1)由于微分形式是 x sinx dx，对应的被积函数正好是“幂 � 三”，所以“三” 应该优先“拿到”右面，于是 x sinx dx sinx右移=========== 幂 �三 xd(� cosx) DBP======== d�x � (� cosx)�� (� cosx)dx = d � x � (� cosx)�+ d sinx = d(�x cosx+ sinx) 从而 原式 = �x cosx+ sinx+ C: (2)由于被积函数是“幂 � 指”形式，故应将“指”（即 e�x）拿到微分运算 d 的右边。于 是 xe�xdx e �x右移=========== xd(�e�x) DBP======== d�x � (�e�x)�� (�e�x)dx = d�� xe�x�+ e�xdx = d �� xe�x�+ d(�e�x) = d�� xe�x � e�x� = d�� e�x(x+ 1)�; 从而 原式 = �e�x(x+ 1) + C: (3)此题的微分形式正好是“反 dx”，所以只需直接分部积分即可。即： arcsinx dx DBP======== d(arcsin(x) � x)� x d arcsinx = d(arcsin(x) � x)� xp 1� x2dx = d(arcsin(x) � x) + 1 2 1p 1� x2d(1� x 2) = d(arcsin(x) � x) + d(1 2 � 2 � p 1� x2) = d(arcsin(x) � x+ p 1� x2); 从而 原式 = arcsin(x) � x+ p 1� x2 + C: (4)此题与 (2)题类似。 x2exdx ex右移========= 幂 �指 x2dex DBP======== d(x2 � ex)� exd(x2) = d(x2 � ex)� ex � 2xdx ex右移========= 指 �幂 d(x2 � ex)� 2(xdex) DBP======== d(x2 � ex)� 2�d(x � ex)� exdx� = d(x2 � e2)� 2�d(x � ex)� dex� = d�x2 � ex � 2xex + 2ex� = d � (x2 � 2x+ 2) � ex�; School of Mathematics and Statistics, Yunnan University Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Av ec Yu anh on g Z hi 3.1 不定积分和定积分的计算 31 从而 原式 = (x2 � 2x+ 2) � ex + C: (5) x2 lnx dx x 2右移========= 幂 �对 lnx d � x3 3 � DBP======== d � ln(x) � � x3 3 �� � x 3 3 d lnx = d � ln(x) � x 3 3 � � x 2 3 dx = d � ln(x) � x 3 3 � + d � �x 3 9 � = d � ln(x) � x 3 3 � x 3 9 � ; 从而 原式 = ln(x) � x 3 3 � x 3 9 + C: (6)只要将微分形式改写为 x�3 lnx dx，则即为与 (5)类似的问题，于是 lnx x3 dx = x�3 lnx dx x �3右移=========== 幂 �对 lnx d � x�2 �2 � DBP======== d � ln(x) � x �2 �2 � � x �2 �2 d lnx = d � ln(x) � x �2 �2 � + x�3 2 dx = d � ln(x) � x �2 �2 + x�2 �4 � = d � 2 lnx+ 1 �4x2 � 从而 原式 = 2 lnx+ 1�4x2 + C: (7)此题被积函数为两个对数的乘积，优先级相同，此时只需直接对微分形式进行DBP 即可（以达到降次目的）。 ln2 x dx DBP======== d(x ln2 x)� xd ln2 x = d(x ln2 x)� x � 2 � ln(x) � 1 x dx = d(x ln2 x)� 2 ln(x) dx DBP======== d(x ln2 x)� d(2x lnx) + 2xd lnx = d(x ln2 x)� d(2x lnx) + 2dx = d(x ln2 x)� d(2x lnx) + d(2x) = d(x ln2 x� 2x lnx+ 2x); 从而 原式 = x ln2 x� 2x lnx+ 2x+ C: (8)此题被积函数是“幂 � 三”形式，其中的“三”指 sec2 x，故 x sec2 x dx sec 2 x右移============= xd tanx DBP======== d(x tanx)� tanxdx = d(x tanx)� sinxcosxdx Avec Yuanhong Zhi Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd Sta tist ics , Y un nan Un ive rst iy Sc ho ol of Ma the ma tic s a nd St