第三讲 换元积分法
第三讲 换元积分法
不定积分 一基本积分公式 (8) sinxdx =-cosx +c (1) kdx=kx+c (9) cosxdx sinx +c xu+1 (2) xudx +c(u≠-1) u+1 (10) sec2xdx tanx +c (3) dx =Inx+c (11) csc2xdx =-cotx +c (4) exdx =ex+c (12) secxtanxdx secx +c ax (5) axdx= +c(a>0,a≠1) (13) cscxcotxdx =-cscx +c 1 (6) 1 x2 dx=-+c dx arcsinx +c X (14) V1-x2 1元 dx =2vx+c 15) 1 1+x2 dx arctanx +c
(2) න 𝑥 𝑢𝑑𝑥 = 𝑥 𝑢+1 𝑢 + 1 + 𝑐 (𝑢 ≠ −1) (1) න 𝑘𝑑𝑥 = 𝑘𝑥 + 𝑐 (3) න 1 𝑥 𝑑𝑥 = 𝑙𝑛 |𝑥| + 𝑐 (4) න 𝑒 𝑥𝑑𝑥 = 𝑒 𝑥 + 𝑐 (5) න 𝑎 𝑥𝑑𝑥 = 𝑎 𝑥 𝑙𝑛𝑎 + 𝑐 (𝑎 > 0, 𝑎 ≠ 1) (6) න 1 𝑥 2 𝑑𝑥 = − 1 𝑥 + 𝑐 (9) න 𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑠𝑖𝑛𝑥 + 𝑐 (8) න 𝑠𝑖𝑛𝑥𝑑𝑥 = −𝑐𝑜𝑠𝑥 + 𝑐 (10) න 𝑠𝑒𝑐2𝑥𝑑𝑥 = 𝑡𝑎𝑛𝑥 + 𝑐 (11) න 𝑐𝑠𝑐 2𝑥𝑑𝑥 = −𝑐𝑜𝑡𝑥 + 𝑐 (12) න 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥𝑑𝑥 = sec𝑥 + 𝑐 (13) න 𝑐𝑠𝑐𝑥𝑐𝑜𝑡𝑥𝑑𝑥 = −𝑐𝑠𝑐𝑥 + 𝑐 (14) න 1 1 − 𝑥 2 𝑑𝑥 = 𝑎𝑟𝑐𝑠𝑖𝑛𝑥 + 𝑐 (15) න 1 1 + 𝑥 2 𝑑𝑥 = 𝑎𝑟𝑐𝑡𝑎𝑛𝑥 + 𝑐 (7) න 1 𝑥 𝑑𝑥 = 2 𝑥 + 𝑐 一.基本积分公式
不定积分 练习 )/=+c t> 7+C 2四jrx=-i+c-号 7+C (3)∫2*=篇+c (④)∫e3xdr=e3x+C (5)∫cos5xdx=号sin5x+C
�� (1( 𝟔𝒅𝒙 �� (2( −𝟖𝒅𝒙 �� (3( 𝒙𝒅𝒙 = 𝒙 𝟔+𝟏 𝟔+𝟏 + 𝑪 = 𝒙 𝟕 𝟕 + 𝑪 = 𝒙 −𝟖+𝟏 −𝟖+𝟏 + 𝑪 = 𝒙 −𝟕 −𝟕 + 𝑪 = 𝟐 𝒙 𝒍𝒏𝟐 + 𝑪 �� (4( 𝟑𝒙𝒅𝒙 练习 �𝒅𝒙𝟓𝒔𝒐𝒄� (5( = 𝟏 𝟑 𝒆 𝟑𝒙 + 𝑪 = 𝟏 𝟓 𝒔𝒊𝒏𝟓𝒙 + 𝑪
不定积分 二、微分公式 例1.求下列函数的微分： 1 d(ex)exdx d(nx)-dx 1 d(cosx)=sinxdx d(2x)= 1 d(sinx)cosxdx d(-1)=zdx
例1.求下列函数的微分： 𝒅( ) = 𝒆 𝒙𝒅𝒙 𝒅( ) = 𝒔𝒊𝒏𝒙𝒅𝒙 𝒅( ) = 𝒄𝒐𝒔𝒙𝒅𝒙 𝒅( ) = 𝟏 𝒙 𝒅𝒙 𝒔𝒊𝒏𝒙 𝒍𝒏𝒙 𝒆 𝒙 −𝒄𝒐𝒔𝒙 𝒅( ) = 𝟏 𝒙 𝟐 𝒙 𝒅𝒙 𝒅( ) = 𝟏 𝒙 𝟐 − 𝒅𝒙 𝟏 𝒙 二、.微分公式
不定积分 例2：求下列不定积分。 (1)∫e*cos exdx 2∫gdr 解：=∫cos ex.erdx 解：=∫lnx·dx =∫cos exdex =∫Inx dlnx =∫cos udu u=ex =∫udu u Inx sinex +C
例2：求下列不定积分。 �� (1( 𝒙 𝐜𝐨𝐬 𝒆 (2𝒙𝒅𝒙 ( 𝒍𝒏𝒙 𝒙 𝒅𝒙 �� �𝐨𝐜� = :解 𝒙 ∙ 𝒆 𝒙 𝒅𝒙 �� �𝐨𝐜� = 𝒙𝒅𝒆 𝒙 �� �𝒅𝒖� �𝐨𝐜� = �� = �� = 𝒔𝒊𝒏𝒆 𝒙 + 𝑪 ∙ �𝒏𝒍� = :解 𝟏 𝒙 𝒅𝒙 �𝒏𝒍𝒅� �𝒏𝒍� = �𝒏𝒍� = �� �𝒅𝒖� = = (𝒍𝒏𝒙) 𝟐 𝟐 + 𝑪
不定积分 练习 (1)∫exsin exdx =-cosex+C -+c 3 ax 4∫ dx arcsinex C
�� (1( 𝒙𝒔𝒊𝒏𝒆 (2𝒙𝒅𝒙 ( (𝒍𝒏𝒙) 𝟐 𝒙 𝒅𝒙 练习 = −𝒄𝒐𝒔𝒆 𝒙 + 𝑪 = (𝒍𝒏𝒙) 𝟑 𝟑 + 𝑪 (3( 𝒆 𝒙 𝟏−𝒆 𝟐𝒙 (4𝒅𝒙 ( (𝒍𝒏𝒙) 𝟑 𝒙 𝒅𝒙 = 𝒂𝒓𝒄𝒔𝒊𝒏𝒆 𝒙 + 𝑪 = (𝒍𝒏𝒙) 𝟒 𝟒 + 𝑪
不定积分 (3)∫sinx(cosx)2dx 4∫acdr=∫cotxdx 解：=∫(cosx)2·sinxdx 解：=∫品cosxdx =∫(cosx)2·d(-cosx) dsinx =-∫u2du u cosx -Sdu u=sinx 、 3+C In ul C In sinx+C
(�� �𝒐𝒄�) �𝒏𝒊𝒔� (3( (4𝟐𝒅𝒙 ( 𝒄𝒐𝒔𝒙 �𝒅𝒙𝒕𝒐𝒄� = �𝒅� �𝒏𝒊𝒔� (�� �𝒐𝒄�) = :解 𝟐 ∙ 𝒔𝒊𝒏𝒙𝒅𝒙 �𝒅𝟐� �𝒔𝒐𝒄� = �� �� − = = − 𝒖 𝟑 𝟑 + 𝑪 = :解 𝟏 𝒔𝒊𝒏𝒙 𝒄𝒐𝒔𝒙𝒅𝒙 = 𝟏 𝒔𝒊𝒏𝒙 𝒅𝒔𝒊𝒏𝒙 �� �𝒏𝒊𝒔� = �� = 𝒖 𝒅𝒖 = 𝒍𝒏 |𝒖| + 𝑪 (�� �𝒐𝒄�) = 𝟐 ∙ 𝒅(−𝒄𝒐𝒔𝒙) = − (𝒄𝒐𝒔𝒙) 𝟑 𝟑 + 𝑪 = 𝒍𝒏 |𝒔𝒊𝒏𝒙| + 𝑪
不定积分 练习 (5)∫cosx(sinx)5dx (6∫adx=∫tanxdx 解：=∫(sinx)5·cosxdx 解：=∫sinxdx =∫(sinx)5.d(sinx) =∫d(-cosx) =∫udu u=sinx -Sidu u=coSx u6 =+C -Inlul C (sinx 6+C =-In cosx|+C
(�� �𝒊𝒔�) �𝒔𝒐𝒄� (5( (6𝟓𝒅𝒙 ( 𝒔𝒊𝒏𝒙 𝒄𝒐𝒔𝒙 �𝒅𝒙𝒏𝒂𝒕� = �𝒅� (�� �𝒊𝒔�) =:解 𝟓 ∙ 𝒄𝒐𝒔𝒙𝒅𝒙 �𝒅𝟓� �𝒏𝒊𝒔� = �� �� = = 𝒖 𝟔 𝟔 + 𝑪 = :解 𝟏 𝒄𝒐𝒔𝒙 𝒔𝒊𝒏𝒙𝒅𝒙 = 𝟏 𝒄𝒐𝒔𝒙 𝒅(−𝒄𝒐𝒔𝒙) �� �𝒔𝒐𝒄� = �� −= 𝒖 𝒅𝒖 = −𝒍𝒏 |𝒖| + 𝑪 (�� �𝒊𝒔�) = 𝟓 ∙ 𝒅(𝒔𝒊𝒏𝒙) = (𝒔𝒊𝒏𝒙) 𝟔 𝟔 + 𝑪 = −𝒍𝒏 |𝒄𝒐𝒔𝒙| + 𝑪 练习
不定积分 5)了dx 解：=∫sin:dr 解品2x =∫sinv元·d(2vx) =2∫sinudu u=Vx =2∫du u=V屁 =-2cosu+C 2arcsinu +C =-2cosvx+C =2 arcsiny√x+C
(5( 𝒔𝒊𝒏 𝒙 𝒙 (6𝒅𝒙 ( 𝟏 𝒙−𝒙 𝟐 𝒅𝒙 ∙ �� �𝒊𝐬� = :解 𝟏 𝒙 𝒅𝒙 �� = �� �𝒅𝒖𝒏𝒊𝒔� �� = = −𝟐𝒄𝒐𝒔𝒖 + 𝑪 = :解 𝟏 𝟏−𝒙 ∙ 𝟏 𝒙 𝒅𝒙 = 𝟏 𝟏−( 𝒙) 𝟐 𝒅𝟐 𝒙 �� = �� �� = 𝟏 𝟏−𝒖𝟐 𝒅𝒖 = 𝟐𝒂𝒓𝒄𝒔𝒊𝒏𝒖 + 𝑪 (�� ��)�� ∙ �� �𝒊𝐬� = = −𝟐𝒄𝒐𝒔 𝒙 + 𝑪 = 𝟐𝒂𝒓𝒄𝒔𝒊𝒏 𝒙 + 𝑪
不定积分 sin- (7) dx dx x√x2- 解=∫sind =∫sin·d(-9 ∫ 1 d(- =-∫sinudu u= 1- X cosu+C =-arcsinu+C = cos=+C -arcsin+C
(7( 𝒔𝒊𝒏𝟏 𝒙 𝒙 (8𝟐 𝒅𝒙 ( 𝟏 𝒙 𝒙 𝟐−𝟏 𝒅𝒙 �� �𝒊𝐬� = :解 𝒙 ∙ 𝟏 𝒙 𝟐 𝒅𝒙 𝒖 = 𝟏 𝒙 �𝒅𝒖𝒏𝒊𝒔� − = = 𝒄𝒐𝒔𝒖 + 𝑪 = :解 𝟏 𝟏− 𝟏 𝒙 𝟐 ∙ 𝟏 𝒙 𝟐 𝒅𝒙 = −𝒂𝒓𝒄𝒔𝒊𝒏𝒖 + 𝑪 �� �𝒊𝐬� = 𝒙 ∙ 𝒅(− 𝟏 𝒙 ) = 𝒄𝒐𝒔 𝟏 𝒙 + 𝑪 = −𝒂𝒓𝒄𝒔𝒊𝒏 𝟏 𝒙 + 𝑪 = 𝟏 𝟏− 𝟏 𝒙 𝟐 ∙ 𝒅(− 𝟏 𝒙 ) − = 𝟏 𝟏− 𝒖𝟐 ∙ 𝒅𝒖