计算下列定积分【精品-doc】
习题5,3
1. 计算下列定积分:
,,sin(x,)dx (1); ,,32
,4211,,,,, 解 . sin(x,)dx,,cos(x,),,cos,cos,,,0,,,33332222
dx1 (2); ,3,2(11,5x)
11111511dx,2,2,2(115)161 解 . ,,,x,,,,,,,3,2,2521010512,(115),x
,32sin,cos,d, (3); ,0
,,,1111,33333222 解 . sincoscossincoscoscos0d,,sd,,,,,,,,,,,,,,00044244
,3(1,sin,)d, (4); ,0
,,,,,322 解 (1,sin,)d,,d,,sin,dcos,,,,(1,cos,)dcos,,,,,00000
,143(coscos),,,,,,,,, . 033
,22cosudu (5); ,,6
,,,,11122222cosudu,(1,cos2u)du,u,sin2u 解 ,,,,,,2246666113,,,,,(,),(sin,sin),, . ,2264368
222,xdx (6); ,0
,,2x,2sint令2222,xdx2cost,2costdt,(1,cos2t)dt 解 ,,,000
,1,2 . (sin2),t,t,022
228,2ydy (7); ,,2
,22令y,2sinx2248,2ydy,24,ydy22cosx,2cosxdx 解 ,,,,,,22,4
,,144,22(1,cos2x)dx,22(x,sin2y),2(,2), . ,,,,,244
211,x (8); dx1,2x2
2,,,sin1cos11令x,t,xt,222 解 . cos(1)(cot)1dx,tdt,,dt,,t,t,,1,,,,,,2224sinsinxtt4442
a222xa,xdx (9); ,0
4,,x,asintaa令22222222 解 xa,xdxasint,acost,acostdt,sin2tdt,,,0004
4444,,,aaaa,222 . (1cos4)sin4,,tdt,t,t,,0008832163dx (10); ,122x1,x
,3令x,tantdx123 解 ,sectdt ,,,2122tant,sect4x1,x
,,cost12333,dt,,,2, . ,,,2sint3sint44
xdx1 (11); ,,15,4x
15411令,x,u1111xdx23(5)(5),udu,,u,u, 解 . ,,,133883654,x
dx4 (12) ; ,11,x
2422令x,udx112,2udu,2(1,)du,2(u,ln|1,u|),2(1,ln) 解 . ,,,11111,u1,u31,x
dx1 (13); 3,1,x,14
111令1,x,u0dx1122,(,2u)du,2(1,)du,2(u,ln|u,1|),1,2ln2 解 . 31,,,00u,1u,141,x,12
xdx2a (14); ,0223a,x
2a2a2axdx112222 解 . ,,d(3a,x),,3a,x,a(3,1),,000222223a,x3a,x
2t,12 (15); tedt,0
222ttt21,,,,111t2222tedt,,ed(,),,e,1,e解 . ,,0002
2dxe (16); ,1x1,lnx
222eedxe1 解 . ,dlnx,21,lnx,2(3,1),,111x1,lnx1,lnx
0dx (17); ,2,2x,2x,2 010dx0,arctan(1)arctan1arctan(1) 解 ,dx,x,,,,,. ,,22,,22,22221(1)x,x,,x,
,2cosxcos2xdx (18); ,,,2
,,,2223222coscos2(12sin)sin(sinsin)xxdx,,xdx,x,x, 解 . ,,,,,,,,33222
,32cosx,cosxdx (19); ,,,2
,,3222cosx,cosxdx,cosx1,cosxdx 解 ,,,,,,22
33,,022402222cos(sin)cossincoscos,x,xdx,xxdx,x,x, ,,,,,0,033322
,1,cos2xdx (20). ,0
,,, 解 . 1,cos2xdx,2sinxdx,,2cosx,22,,000
2. 利用函数的奇偶性计算下列积分:
,4xsinxdx (1); ,,,
, 44xsinxdx,0解 因为xsin x在区间[,, ]上是奇函数, 所以. ,,,,,,42 (2); 4cos,d,,,,2
,,,1cos2x,442222 解 4cos,d,24cos,d,8()d, ,,,,,,00,22
,,31222,2(1,2cos2x,cos2x),d,2(,2cos2x,cos4x)d, ,,0022
,13,2 . (32sin2sin4),,x,x,,042
21(arcsinx)2 (3)dx; ,1,221,x
22111(arcsinx)(arcsinx)2222dx,2dx,2(arcsinx)d(arcsinx) 解 ,,,100,2221,x1,x
312,32 . (arcsin),x,03324
325xsinx (4)dx. ,42,5x,2x,1
3232xsinx5xsinx 解 因为函数是奇函数, 所以. dx,0,4242,5x,2x,1x,2x,1
aa22,(x)dx,2,(x)dx 3. 证明: , 其中,(u)为连续函数. ,,,0a
2 证明 因为被积函数,(x)是x的偶函数, 且积分区间[,a, a]关于原点对称, 所以有
aa22,(x)dx,2,(x)dx . ,,,0a
bbf(x)dx,f(,x)dx 4. 设f(x)在[,b, b]上连续, 证明. ,,,,bb
证明 令x,,t, 则dx,,dt, 当x,,b时t,b, 当x,b时t,,b, 于是
,bbbf(x)dx,f(,t)(,1)dt,f(,t)dt , ,,,,,bbb
bbf(,t)dt,f(,x)dx而 , ,,,,bb
bbf(x)dx,f(,x)dx所以 . ,,,,bb
bbf(x)dx,f(a,b,x)dx5. 设f(x)在[a, b]上连续., 证明. ,,aa
证明 令x,ab,t, 则dx,d t, 当x,a时t,b, 当x,b时t,a, 于是
babf(x)dx,f(a,b,t)(,1)dt,f(a,b,t)dt , ,,,ababbf(a,b,t)dt,f(a,b,x)dx而 , ,,aa
bbf(x)dx,f(a,b,x)dx所以 . ,,aa
11dxdxx 6. 证明: . ,(x,0),,22x11,x1,x
111dx,,dt 证明 令, 则, 当x,x时, 当x,1时t,1, 于是 x,t,2txt
111111dxx ,,(,),, dtdt1,,,222x111,1,xttx1,2t
1111xx而 ,, dtdx,,22111,1,tx
1dxdx1x所以 ,. ,,22x11,x1,x
11mnnmx(1,x)dx,x(1,x)dx 7. 证明: . ,,00
1011mnmnmnnmx(1,x)dx,,(1,t)tdt,(1,t)tdt,x(1,x)dx 证明 令1,x,t , 则, ,,,,0100
11mnnmx(1,x)dx,x(1,x)dx即. ,,00
,,nn2sinxdx,2sinxdx 8. 证明: . ,,00
,,,nnn2 证明 , sinxdx,sinxdx,sinxdx,,,,002
,,,0x,,t令,nnnn22而 , sinxdxsin(,,t)(,dt),sintdt,sinxdx,,,,,,0022
,,nn2sinxdx,2sinxdx所以 . ,,00
a,1f(x)dx 9. 设f(x)是以l为周期的连续函数, 证明的值与a无关. ,a
证明 已知f(xl),f(x).
10a,la,lla,laf(x)dx,f(x)dx,f(x)dx,f(x)dx,f(x)dx,f(x)dx,f(x)dx , ,,,,,,,000aall
令x,t,la,laaa而 , f(x)dxf(t,l)dt,f(x,l)dx,f(x)dx,,,,000l
1a,lf(x)dx,f(x)dx所以 . ,,0a
a,1f(x)dx因此的值与a无关. ,a
xf(t)dt 10. 若f(t)是连续函数且为奇函数, 证明是偶函数; 若f(t)是连续函数且为偶函数, ,0
xf(t)dt是奇函数. 证明,0
xF(x),f(t)dt 证明 设. ,0
若f(t)是连续函数且为奇函数, 则f(,t),,f(t), 从而
t,,u,xxxx令 F(,x),f(t)dtf(,u)(,1)du,f(u)dx,f(x)dx,F(x), ,,,,0000
xF(x),f(t)dt即是偶函数. ,0
若f(t)是连续函数且为偶函数, 则f(,t),f(t), 从而
t,,u,xxxx令 F(,x),f(t)dtf(,u)(,1)du,,f(u)dx,,f(x)dx,,F(x), ,,,,0000
xF(x),f(t)dt即是奇函数. ,0
11. 计算下列定积分:
1,xxedx (1); ,0
11111,x,x,x,x,1,x,1 解 . xedx,,xde,,xe,edx,,e,e,1,2e,,,00000
exlnxdx (2); ,1
eeeee1111111222222xlnxdx,lnxdx,xlnx,x,dx,e,x,(e,1) 解 . ,,,11011222x244
,2,tsin,tdt (3)(,为常数); ,0
,2,,,222111,,,, 解 tsin,tdt,,tdcos,t,,tcos,t,cos,tdt,,,0000,,,
,2,,212,, . ,,,t,,sin2220,,,
,x3 (4); dx,,2sinx4
,,,,,x1,,33333dx,,xdcotx,,xcotx,cotxdx,,,,,lnsinx 解 ,,,,,,,,234sinx344444
1313 . ,(,),ln,4922
4lnx (5); dx,1x
4444lnx1 解 ,,,,dx2lnxdx2xlnx2xdx,,,1111xx
441 . ,8ln2,2dx,8ln2,4x,4(2ln2,1),11x
1xarctanxdx (6); ,0
11111111222arctan,arctan,arctan,,xxdxxdxxxxdx 解 ,,,200002221,x
1111111,,,,,(1)(arctan)(1),,,dx,,x,x,,,,, . ,2008282824421,x
,2x2 (7); ecosxdx,0
,,,,xxxx22222222ecosxdx,edsinx,esinx,2esinxdx 解 ,,,0000
,,,,xxxx2222,,,2222,e,2edcosx,e,2ecosx,4ecosxdx,e,2,4ecosxdx ,,,0000,12x,2ecosxdx,(e,2)所以 , ,05
于是
2xlogxdx (8); 2,1
22221111222xlogxdxlogxdxxlogxxdx,,,, 解 222,,,1111222xln2
2113222,,,x,, . 12ln224ln2
,2(xsinx)dx (9); ,0
,,,,1112232 解 (xsinx)dx,x(1,cos2x)dx,x,xdsin2x,,,000026433,,,111,,,,3 ,,xsin2x,sin2x,2xdx,,xdcos2x,,0000064464333,,111,,,,,, . cos2cos2sin2,,xx,xdx,,,x,,,00064464864
esin(lnx)dx (10); ,1
lnx,te1令t 解法一 . sin(lnx)dxsint,edt,,10
1111tttt因为 sint,edt,sintde,esint,ecostdt,,,0000
111ttt ,e,sin1,costde,e,sin1,ecost,esintdt,,000
1t,e,sin1,e,cos1,1,esintdt , ,0
11tesintdt,(e,sin1,e,cos1,1) . 所以,02
e1sin(lnx)dx,(e,sin1,e,cos1,1)因此 . ,12
eeee1sin(lnx)dx,x,sin(lnx),x,cos(lnx),dx,e,sin1,cos(lnx)dx 解法二 ,,,1111x
ee1,e,,x,x,x,x,dx sin1cos(ln)sin(ln) ,11x
e,e,sin1,e,cos1,1,sin(lnx)dx , ,0
e1sin(lnx)dx,(e,sin1,e,cos1,1)故 . ,12
e|lnx|dx (11); 1,e
1e11eee|lnx|dx,,lnxdx,lnxdx,,xlnx,xlnx,dx,dx 解 111,,,,,1111eeee111,,,e,(1,),(e,1),2(1,) . eee
m122(1,x)dx (12)(m为自然数); ,0
m,1xsint令,2m,122(1x)dxcostdt, 解 . ,,00
,,n,1nn,222cosxdxcosxdx, 根据递推公式, ,,00n
mm,2m,4531,,,, , , , ,,, m为奇数m,1m,1m,1m,3642222 . (1,x)dx,,,0mm,2m,4642,,, , , , ,, m为偶数m1m1m3753,,,,
,mJ,xsinxdx (13)(m为自然数). m,0
解 因为
0,x,,t,,,令mmmm , xsinxdx(,,t)sin(,,t)(,1)dt,,sintdt,tsintdt,,,,000,
,,,,,,mmmm22J,xsinxdx,sinxdx,,2sinxdx,,sinxdx所以 (用第8题结果). m,,,,000022
,,n,1nn,222sinxdxsinxdx, 根据递推公式, ,,00n
2,,13553m,m,m,1 ,,,,,,,,m为偶数,mm,2m,46422J, . ,m1356m,m,m,42, , ,,,,,,,m为奇数mm,2m,4753,