高数上b1讲稿3第三章积分的计算及应用

b1p144-15..b1p144-16..b1p144-23..b1p144-24..b1p144-25..b1p144-26..b1p144-27..b1p144-28..b1p144-29..b1p144-30..b1p144-31..b1p144-32..b1p144-34..b1pP127P129,1-10,12-P134,4-11,13-P143-144,1-10,13,17-P123-128,例1-79,例12P130-134,例1-5P137-43§1Fu是fuFufuuux

那么复合函数F(ux))的导数dF(u(x))

f(u(x))uxf(u(

x

F(u(x))

C.f(u(x))dux

F(u(

C.ddux的一个函数，并且作为u的函Fu的积分，形式上变成关于u

f如果f(u)Fu

F(ux))fx

x

t

fttftt

FtfxdxF1x其中1x

x

t F1x

F1x1x

ft ttFt ftttfxdxF

x

C元x

t

数由fxftt

求I解I

xdx

cscxcscxcotx cscxcotcscxcot dcsccscxcot

csc

cotxcsc

cotx

1 sinx

cosxsin1cosxsinx

2sin2 2sinx

tan.2这与P125例8,答案一样..2a2x2求a2x2

a0解令xatan t dxa

a2x2asec2 aa2x2asec2 dt

csc

t

cot

t22sint 22

sec

tanta2x2a a2x2aa2x2a2x2

C1I

3a0a2x22令xasint

t a3cos3a3cos3

acostdt

1tanta2x2a2x2

Cx2x2

a0 1tan2tsec2

xa

x xa

,第一种情况,xa,令x则

asect,

0tx2a22x2a22dx

asecttantdt

xtanx2x2a2

tantdt,aaa

sec2

1dtatantatx2a2x aarccosx2a2x第二种情况,xa同样,令xasect,x2a20tx2a2atant,dx

asecttantdt

xtanx2x2a2

tantdt,aaa

sec2

1dtatantatxx2a2x2x2x2a2

aarccosaxx

aarccos

C,x.x2x2a2xxx

aarccos

C,

a1e31e3e3x1e3x

I

31e1e3

2

311 3x22323223

e3

C1e31e3

t23x1lnt3

,tdx

2t dt,3t2dt,3I3

2t t2

1 t

t1e3x1e3x1ln1e3x1e3x t

e3

tan2t,3e3xdx

2tan

tdt,dx

2tantsec2tdt23tan2

2sec23tan

dt,I1 I1

2sec22sec2

3sin33 csctdt33

2

csc

cott2

1e3x

e3

e3 3求Ix6x61

xtan dx

I

sec2tan6tsec

tan cos5 sin

usin

1 uu

12 u

2 1 1u42u21 求I

a2

x2a0解令

asin t Ia2

a2 1cos2ta21a2t1sin2t a2x2a2x2a2arcsin a2x2a2x2求I

7x

x27xx2

x

12 令t

x1

292I2

a2

t2a2ta2t2a2arcsina2ta2t229

arcsin2x1

2 7x7xx41xu1x

5 2I2

1

x2

2

1a1x2a

31x2 5,tx解法二 1,txdt

12x2Ix61x61

t2t2

t2t2t2t2

dt

t4

2 t2tt2t2 t21u11

u2

121

2

u t2t2

1111x21x11I

1x22 21x225 3

C.求I

x1.x1.xx解法一令u

x1,

则du

,

dx

2I

x12udu

u11du2u

2ln1

uxx

2ln1

C

u1

x1, du

,x xdx

x1du

2u

1I

2u

2lnuxx

2ln

xxuv

uv

uv

uxvx

uxvx

uxvxdx u

xdvx

u

xvx

vxdux uxvx u

xv

x

ln

cos

C;

ln

sin

C;1xCx2 1xCa0; x2a2 2a

xaxaaaaa2a2x2ax2x2a2

0; 0;

arcsinxxx2xx2a2C

ex2dx,

x2dxxx

1k2sin211k2sin21 1k20k1,求In

2(ta22n1,a0In

(t2a2

t(t2a2 (t2a2

t2a2a2(t2a2 (t2a2

2na2

(t2a2)n1

2na2(t2a2

(2n1)

. 2na2

2na2(t2a2I1

2dt1arctant(ta I2

12 I1

13arctant2a

(t2

2a 求I

x211Ix

x2

2xarccosxx

x2

2I11x2其中I1x2

1111

dx.

d

,11x1x2I1

arccos11x2

arccosx

dx1x211x221x211x22将I1

Ixarccosx

arccos

2x11x2求I

xarctanx31x223 d

x212I

arctanxd 11111x21x2arctan1xtan11x21x21

dx

sint1 1§3所谓 PxQx

a0xna1xn1 an1xanb0xmb1xm1 bm1,Px,Qx,a0 b0m ,A,xa

xan

,n

1; , ,n

1,x2

x2pxqn其中

B,C,

p

x2

px P(xQ(x

Qx

有 ,,

Qx

有 1i1,2

i2 ,l

il,ml,mlQ(

(x

(xa)nk(x2px

1

(x2x(x2 i,l.i其中p2q2i,l.i p24qi1, 式P(x式Q(x(xa1

P(x

Q(xxakxak2(xak

22Bm11xCmBm11xCm1(x2p1xq1x2p1B1lxB1lxC1x2plxqlBmllxCmll(x2plxql其中Anj

Cmi

规律：(x

a)k

(x

a)k

(x

a)kxx

,,

;xa;(x2

px

q)k

p24qM1x(x2pxq

M2xN2MkxNkx2pxqMkxNkx2pxq其中MiNi

,k);(i,k);k=1

MxN x2pxq积分，只出现下面四类情况：

Axa

=

d(xaxa=Aln

xa

(xaA(xa)n(= (=

a)n1 (1n)(xa

BxC x2pxqx2

px

(x

p

q2 2；2xp2

BxC x2pxqx2

px

(x

p

q设p2

24q0（2p24q

24记a2q 。2422txp，则x p，22dx BxC x2pxqdxp2

C

Bp 2 p2

x2q4

x2

ln(x2

px

q)

C

arctanxq qqq4

(

BxC2pxq BxC

dxB

dxp2

C(x2pxq

p2n2x2q424 4 221n2

xp2

q

1n

C

x,

Rxy变元x及yRsinxcosxR

x,

x2令ttan2则有

2arctant,dx

2t1t

dt,cos2

,2sec2 1t2sinx

2sinxcosx2tanxcos2x cosx

cos2

sin2

x1 R

x,

2t 1t

1t 2R 1tR RR 2t 1t 2t1t 1t 1t是关于 2通过变换ttanx，就把三角22ttan2

Rxy是关于变元xy的有理式,我们不定积naxnax

dxa0,其中n为自然数．对于这种类型naxnax时，x

tnb

dx

ntn1aaRx,

dx

Rtn

,t

ntn1nax nax上式右端的被积函数是t有理函naxbcxnaxbcxd

dxad

0,naxnaxbcxdaxbcxd

tnxdtnbactx

,dx

adbcntn1actn2I

x4x224I14

4x32x2x4x22414

x4x2

1x4x21d2x2

4x,2x2

12

4

4x2

x4x22

d2x2 4x44x2 d2x2 2x212

arctan

7722x2772

C.上式代入(14I1lnx4x2472x27272x272I

x4x22x2x22x2设 设x2x22x2

axb x22x2

x2

x x22x2

axbx2x22x2.c以x.c

将c

到

axb x22x2

1

axbx22x2 x22x2

ttan2 2

2t2 2 21tt

lnt

12lntanx12

cot2cot2cot2

cot2

cot2cot2

cotcsc2

1d33

cot3

cotsec4dsec2sec2

13 1tan33I

cos3 cosxsin1x1cos2x1sin2x 81ln2sin2x28J

sin3 cosxsin1x1cos2x1sin2x 81ln2sin2x282IJ11sin2x24x4

1cos2x

I

cos3xsin3xcosxsin41sin2x441ln2sin2x4

2

cos3xsin3xdx

2cos2x2sin2xcosxsin

241t1t1lnt

1 x dxcscx

xcsc2

xdcotx3cotx3

1cot3

C求I1x求I1x1解法一由1x11x 故设

1x11x1

11 t, 1sindx1sin

1I1

1x1x2

dx

cos21sin

1sin2t1

sint

arcsin

C

t, 1111x1t

,

1t22I

t 1t22

t2111t2224 22

1t24arctan

11t21

1t2

dt, ttan

dt

1t22

2 21u1sin2u 1arctant1 t I2arctant

2t1t

C.1x1x

2t 1x21x21t

1

1

x12t

1 x11x1x1

1t 2t 1t

.

1x1x22

t, 1111x1t

,

1t22I

t 1t22

1t2t1t

1t2arctant1x11x1t

2t1t11x,

则C.则

1x1xx

,

tt2It2

t12

2

td

t t t t

1x1x1x1x11

t t21x, 则2t

1x,dx I 2tdt2t2t2t22t22t22arcsint2

xcost,1cost1cost1cos

sint sin

2sin

22cos 2 t t

t

1

cost1x22tsin1x22arccosx 11xI

11

1 11111x2

dx

x1x1

C. x113 x113x11解令tx16 16tx16t

dx

6t5dt,I51tI1t

t51t

dt

t81t

I1

I2 1t1t 1t1t

t5

t42tut

u2111u3lnu

u2u

3lnt

1

3t

t2t

1t

dt

dt 1t

t

1t ,1t81t

1t

t4

t6I2

61t

t4

t6

dt6arctan6t76t52t

6arctantC2II1I26t7

6t5

3t4

2t

3t

6arctant3lnt

C.