定积分换元积分法与分部

四、同步练习一、主要(一)定积分的换元

y a2a引 x2

y2

a b2 解(方法1)S

令xasinπa4b a2x2da

t( 2a4ba2arcsinx

a2x2

S4S4

a2x2d令xasinπ令xasinπ

atdπ

(形式上4ab1(tsin2t) 2abπ2

π 问题定积分是否有换元积分法 有定理5.4设fxC[ab单值函数x(t满足()a,()2)在[α, 或[β,α]上(t)具有 a(t)b baf(x)dxf[(t)(td——换元注1o不一定小 当时,换元公式仍成立2º

下限a下限上限b上限xxφ(t f

φ(t

φ(t)dt

f(x)db b或配元f[(t(tdt

f[φ(t

4,变量不代回(二)定积分的分部定理

设ux),vx)在[ab]abu(x)v(x)dxu(x)v(aababbvdaabudv

a bv(x)u'(x)dxaa二、典型例π例1

20

xsinxd解(方法

令tcosx, dtsinxdx,x :t0,2x

:t0

t6

xsinxd

t5dt 6 换元 20

xsinxd cos5xd(cos01cos6

π21 不明显写出新变量t,积分限不动1例 01

d11x2解2

x2d

令xsin

6costcostd x0:t x1:t 26(1cos2t)d0π1tsin2t

π

3π例3计算03π

sin3xsin5xd解Qfx

xsin5

cosxsinxπ sin3xsin5xdxπ0

πcosxsinx2d303π 2cos0π

3

dxπ2π

πcosxsinx2dπ3 2sinx2dsinx 0

sinx2dsin52sinx2

5

40 02例 设f(x)在[-a,a]上连续,

a

f(x)dx0[f(x)f(x)]da(2)a

f(x)dx

当fx)afx)dxfx)a0a 0a0a证0

a

f(x)dxa

f(x)dx0

f(x)d在afx)dxxt0a0

f(x)da

xa0a0

f(t)(dt)0

f(x)d af(x)d

0[f(x)f(x)]d fx)为偶函数，则f(xfa f(x)dxa

[f(

f(x)]d

f(x)d 若f(x)为奇函数， f(x)f a

f(x)dx0[f(x)f(x)]d

yyOyyO1x11

2x2111121 1

d11 d原式1

d 1

1x2(1 1x20 1

dx0 1(1x2 d11

1x2)d

4

101101x2d

42xx22xx2解

1(x1)2d101t

1

(t 1t2d 1 1t

dt1

d1偶函11t1t

π12π

cos24π1exd4

4[f(x)0

f(x)]dπ 40

cos21e

cos2(x)]d1eπ

x)[

]dxxx 1 1xxx4 4

cos

xdx

1(x2

1sin2x)

1(π

1

例 若f(x)在[0,1]上连续，证明πππ0

f(sinx)dx 0

f(cosx)dππ0 xf(sinx)dxππ

f(sinx)d0x0:tπx22:tx0:tπx22:t 证(1设xπt,dxdt,0π02f(sinx)dx

fsin(πt)dt

2f(cosx)d20 202π0(sinx)dxπ2π0f(sinx)d证(2) xπt,x0xππ

dxdt,tt00xf(sinx)dxπ(πt)f[sin(πt)]d 0(πt)f(sint)dπ

π0f(sint)d

0

xf(sinx)d 0

xf(sinx)dx

20

f(sinx)d(3)(3)π0f(sinx)dxπ20f(sinx)d

f(sinx)dx 0

f(sinx)dxππ2π

f(sinx)d Qππ

f(sinx)dt π

πf[sin(πt)](dt2 2f(sint)d0π

2f(sinx)d0π2 f(sinx)dx202

f(sinx)d例7计算0

xsinπ1cos2π

d解(方法10

xsinπ1cos2π

dx0

x sin dπ2sin2π

sin

dx

20

1cos2

20

1cos2 2(44)4.πxππ

π(πt)cos

xsin d01cos2

tx

dsin2 tcos tcos 1sin2

dt d1sin1sin2偶函 奇函 2

d01sin2 x) 040例 设f(x)是以T为周期的连续函数， f(x)dx0 f(x)d其中a证(方法 af(x)d

aTaf(x)dx

f(x)d

T f(x)da

xTt T

f(x)d

f(Tt)dt000

f(x)daaTTaf(x)dT0a0

f(x)dx

f(x)d

f(x)daa

f(x)dx

f(x)d0

a令F(a) f(x)dQfx)连续，F(a)F(a)f(aT)(aT)f(a)f(a)f(a)TF(a)C(常数 令aTT由F(00fx)dT

得C0fx)d命题得证.100 例 0 1cos2xdx. f(x) f(xπ)f( T 0 1cos2xdx0 1cos2xdxπ 1cos2xd L99 1cos2xd 1cos2xdx100 2sinxdx 2. 2例102

f(x)

1

求fx1)dx 11ex1解2fx1)dxtx0

111

f(t)d0 f(t)dt0

f(t)dt

dt d 0

111

011(11et)d

ln(1t)1ln(1et) ln2ln(1例11已知fx)连续π

tf(xt)dt1cosx0 2fxdx的值0解令uxt, 则有txu,dtd且当t0 ux;当tx u 0tf(xt)dtxxuf(u)dxx0xuf(u)dxxxx

f(u)du0uf(u)d

f(u)du0uf(u)du1cosx0f(u)duxf(x)xf(x)sinx 0x

f(u)dusin在上式中令xπ2π

2fudu0 2fxdx0例

xd 01cos2 Q1cos2x2cos2π

dx

dx

4xsec2xdπ01cos2ππ

20 22 1

xdtanx xtanx4

2 lnsecx4 例13

1ln(1x)d0(2x)21ln(1x)dx

ln(1 1解0(2x)2 01

ln(1

dln(12

021 d2x11 d2x1 0

( 1x 2xln2ln(1x)ln(2x)1 5ln2ln3例 证明定积分公式(Wallis公式In

20

xdx

20

xdn1n3L31π n 4 n1n

L42

n 5 证当n0 I0

2dx II12sinxdxcos 0 当n2时

20

xdx

20

xsinxd

20

xd(cos

[sinπ

xcosx00

2cosxd(sinn1

x)] 2cosx(n0

xcosxdIn

2cosx(n0π

n

xcosxd(n 0

n

xd1sin2π(n

2sinn0π

x(1sin2x)dπ(n

2sinn0

xdx(n

0

xd(n1)In2(nInn1IInn1I，nIInn1I(n nn3 12m12m32m5L5312m 2m6402m

nLL

(2m1)!!(2m (m1,2,L)

(2m1)!!I0π(2mI0πIInn1I(n

2m2IL422m

2m

2m

5 (2m (2m π

(2m(2mπ

(m1,2,L)

20

xdx

20

xd(n1)!!π

n为正偶数 (n 例

求

xsinnxd0π

(n In0

xsinnxdx(n20π

xdπ0π0xf(sinx)dxπ2π0f(sinx)dπ0f(sinx)dxπ20f(sinx)d 2sinnxd n1n3

31

n为偶数(n

n n

n

π

n为奇数(n n 例16

xxxxxxxx

d解令t

,则sin2t

xcos2t1sin2t1

tan2tx xxxxx dx0

3tdtan2t0πttan2t3π0π

30

2tdt

π33

3(sec2t1)d0πtant3

43

例

若fx)在[ab]bf(x)dxa

ba[f(a)f(b)]1b1 f(x)(xa)(xb)d2 bafxxaxb)dxaxa)(xb)db

(0(xa)(xb)0

(x)

a(2xab)

(x)daba(2xab)df(ab(2bf(x)(xa)(xb)dx(2

ab)df(xa[(2xab)f(x)b

b2f(x)dx] ab[(ba)f(b)(ab)f(a)2af(x)db(ba)[f(b)f(a)]2af(x)db f(x)dxa

ba[f(a)f(b)]1b1 f(x)(xa)(xb)d2例 设f(ex)1x,f(1)3

求fx)2f(x1)d解 uex 即xln f(u)1ln f(x)1ln f(x)f(1)

1

(t)dtx

(1lnt)d1 f(1)

f(x)1 (1lnt)dxf(x)1 (1lnt)dxt(1lnt)x

t d 1 [x(1lnx)1](x1)xln32f(x1)d3

ux1ux11

112 lnudu2122

1(u2lnu

ud1

2ln23.2(4ln u 2 例

xfx1xsin

d

xf(x)d0分析因

无法直接求出fx)，所以采用分部积分法.11解0xfx)dx11

f(x)d(x210112

f(x)

1x2df(201f(1)0

1x20

f(x)dQf(x)

x2sin

d

f(1)

1sin

dt f(x)

sinx22x

2sinx2x111 xf(x)dx

f(1)2

1x20

f(x)d2

12xsinx2dx

1sinx2dx20

1(cos1 例20(综合题

Fx)

xx2t2f(t)dt,fx)0且f(00若fx)Fx)为奇函数若fx0x0),则Fx)在[0，)当x0Fx)与x3是等价无穷小，求f证 f(x)f(QF(x)

x[(x)2t2]f(t)d0x(x2t2)f(t)d0F(x) x(x2t2)f(t)d0令u x[x2(u)2]f(u)(du)0

x(x2u)2f(u)duF(0 Fx)是奇函数xF(x)0( t)f(t)dx xx2

f(t)dt

xt

f(t)d xF(x)2x

f(t)dtx2

f(x)x

f(x)xF(x)2x0f(t)dx f(x)0(xx 当x0时，0f(t)dt从而当x0时，Fx)Fx)在)上单调增加x1x

F(

2

f(t)dx0 x

x0 x32lim0

f(t)dt

(002limf(x)limf(x)f

(f(0) f

x f(0)4三、同步4计算

x2x

dfx)在abπ2f(cosx)dx

12

f(cosx)d 4计算I2 d计算I2 d0sinxcos 1x2,x 设fx

,x

f(x2)d设fx)可导，且f(0F(x) x0

f(xn

tn)d

(nN求limFx).x0

x2n计算

ln4x4

dx

arctanxd1计算 1(1x0

1 d计 2(x3sin2x)cos2xd2100

1002

tan2xsin22xdx

计算1arcsinx2d0设fx在[0,1连续且f(01,f(23f(2)5, 1x0

(2x)dx 设f(x) e dt101

f(1)x1求x1

f(x)d4四、同步练习4计算

x2x

d解 2x1t,则x

t22

,dxtd

且当

于t21432x432 2x

dx

td t3dt

16 2

1t32

fx)在abπ2f(cosx)dx

12

f(cosx)d 4 f(cosx)d0

f(f(costππtπ

πfcos(πtπ

(dt)

dπf(cost)dtππf(cost)d2 2

2f 2

(du2 ππ π2π

偶函f(f(sinuπ402即0

f(sinu)du 0πf(cosx)dx40

f(cosu)df(cosx)dπ00

f(cosx)dx4

20f(cosx)d0计算I

dπ0sinxcosπππ dx ππ0sinxcos 1cos2

1sin2 dsinxsinx 1sin2 dxcosxcosx 1cos2

20sinxcos

d 2I

2sinxcosxdx2sinxcosxdx2dx2 0 0 I

2sinxcos

dx 4设fx)

1x2,x3ex,x3

求

f(x2)d分析先用换元法公式把被积函数化成f(t再把f(tf(t)t0解令x2t,则dxdt,且当x时当x3时 t1,于是

t31 f(x2)dx31

f(t)d00

t2t

f(x)f(x)1x2,x,x031 3

et

71.3

设fx)可导，且f(0F(x) x0

f(xn

tn)d

(nN求limFx).x1x0x1

x2nFx

xtn1

f(xn

tn)dt

n

f(xn

tn)d(tnxxn

f(xn

tn)d(xntn01uxntn01

nx

f(u)du

x1nx1

f(u)d xn F(x)n f(u)dF(x) 1f(xn)nxn1f(xn)n

F(x)

F(x)

f(xn)

x2n

x0

2n1

f(xnn

1

f(xn)fn

(f(0)2n

2n f(0).计算

ln4x4

dx解根据定积分的分部积 公式，4lnxdx 4

xln

42

xdln

1lnx

4ln4214ln421x4ln4x

x1dx1dx144ln41

arctanxd解根据定积分的分部积分公式，3arctanxdxxarctan 3 3xdarctan 3 3 d 1 3π1ln(1x2 3πln311

0(1x 1xd解令xsint,则dxcostdt,且当x0

t当x1时t