计算下列定积分
习题5,3
1. 计算下列定积分:
,, (1); sin(x,)dx,,32
,,,,,,4211 解 . sin(x,dx),,cos(x,),,cos,cos,,,0,,,33332222
1dx (2); ,3,2(11,5x)
11dx111151,2,2,2 解 ,,,x,,,,,,. (115)161,3,2,2,521010512,x(115)
,3 (3)2sin,cos,d,; ,0
,,,,1111333332 解 22. d,,sd,,,,,,,,,,,,sincoscossincoscoscos0,,00044244
,3 (4)(1,sin,)d,; ,0
,,,,,322 解 (1,,sin,)d,,d,sin,dcos,,,,(1,cos,)dcos,,,,,00000
,143 . (coscos),,,,,,,,,033
,22 (5)cosudu; ,,6
,,,,111222 解 22cosudu,(1,cos2u)du,u,sin2u ,,,,,,2246666
,,,,113 . ,,(,),(sin,sin),,2264368
22 (6)2,xdx; ,0
,,x,2sint令2222 解 2,xdx2cost,2costdt,(1,cos2t)dt,,,000
,,12 . ,t,t,(sin2)022
22 (7)8,2ydy; ,,2
,22令y,2sinx224 解 8,2ydy,24,ydy22cosx,2cosxdx ,,,,,2,2,4
,,144,22(1,cos2xdx),22x(,sin2y),2(,,2). ,,,,,244
211,x (8)dx; 1,2x2
2,,,1,令x,t,xtsin1cos12 解 22. dx,tdt,,dt,,t,t,,cos(1)(cot)11,,,,,,2224xttsinsin4442
a222 (9)xa,xdx; ,0
4,,a令x,asinta22222222 解 xa,xdxasin,tacos,tacostdt,sin2tdt,,,0004
4444,,,aaaa,222 ,,tdt,t,t,. (1cos4)sin4,0008832163dx (10); ,122x1,x
,3令x,tantdx12 解 3,sectdt ,,,2122tant,sect4x1,x
,,cost1233 3,,,,,. dt2,,,2sint3sint44
1xdx (11); ,1,5,4x
111令,x,uxdx54111123 解 ,udu,,u,u,. (5)(5),,,1338836,x54
4dx (12) ; ,11,x
2422令x,udx112,2udu,2(1,du),2u(,ln|1,u|),2(1,ln) 解 . ,,,11111,u1,u31,x
1dx (13); 3,41,x,1
1110令1,x,udx1122 解 ,(,2udu),2(1,du),2u(,lnu|,1|),1,2ln2. 31,,,00u,1u,11,x,142
2axdx; ,0223a,x
(14)2a2a2axdx112222 解 . ,,d(3a,x),,3a,x,a(3,1),,000222223a,x3a,x
2t,12 (15)tedt; ,0
222ttt21,,,,111t2222 解 tedt,,ed(,),,e,1,e. ,,0002
2edx (16); ,1x1,lnx
222eeedx1 解 . ,dlnx,21,lnx,2(3,1),,111x1,lnx1,lnx
0dx (17); ,22,x,2x,2 000,dx1 解 ,dx,x,,,,,. arctan(1)arctan1arctan(1),,2222,,,22x,x,,x,221(1)
,2 (18)cosxcos2xdx; ,,,2
,,,22232 解 22xxdx,,xdx,x,x,. coscos2(12sin)sin(sinsin),,,,,,,,33222
,3 (19)2cosx,cosxdx; ,,,2
,,32 解 22cosx,cosxdx,cosx1,cosxdx ,,,,,,22
33,,00224222 2,x,xdx,xxdx,x,x, cos(sin)cossincoscos,,,,,0,033322
, (20)1,cos2xdx. ,0
,,, 解 1,cos2xdx,2sinxdx,,2cosx,22. ,,000
2. 利用函数的奇偶性计算下列积分:
,4xsinxdx; ,,,
, 44 解 因为x (1)sin x在区间[,, ]上是奇函数, 所以xsinxdx,0. ,,,,,,42 (2)4cos,d,; ,,,2
,,,,1cos2x442 解 222,,4cos,,d24cos,,d8()d, ,,,,00,22
,,312 22 ,2(1,2cos2x,cos2x)d,2(,2cos2x,cos4x)d,,,,0022
,,132 . ,,x,x,,(32sin2sin4)042
21(arcsinx)2 (3); dx1,2,21,x
22111(arcsinx)(arcsinx)2 解 222dx,2dx,2(arcsinx)d(arcsinx) ,,,100,222,1x,1x
31,232 . ,x,(arcsin)03324
325xsinx (4). dx,42,5x,2x,1
32325xsinxxsinx 解 因为函数是奇函数, 所以dx,0. ,4242,5x,2x,1x,2x,1
aa22 3. 证明: ,(x)dx,2,(x)dx, 其中,(u)为连续函数. ,,a,0
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. ,,a,0
bb 4. 设f(x)在[,b, b]上连续, 证明f(x)dx,f(,x)dx. ,,bb,,
证明 令x,,t, 则dx,,dt, 当x,,b时t,b, 当x,b时t,,b, 于是
bbb, f(x)dx,f(,t)(,1)dt,f(,t)dt, ,,,bbb,,
bb而 f(,t)dt,f(,x)dx, ,,bb,,
bbf(x)dx,f(,x)dx. ,,bb,,
bb 5. 设f(x)在[a, b]上连续., 证明所以 f(x)dx,f(a,b,x)dx. ,,aa
证明 令x,a,b,t, 则dx,d t, 当x,a时t,b, 当x,b时t,a, 于是
bab f(x)dx,f(a,b,t)(,1)dt,f(a,b,t)dt, ,,,aba
bb而 f(a,b,t)dt,f(a,b,x)dx, ,,aa
bb所以 f(x)dx,f(a,b,x)dx. ,,aa
11dxdxx 6. 证明: . ,(x,0),,221x1,x1,x
111 证明 令,,, 则, 当x,x时, 当x,1时t,1, 于是 dxdtx,t,2txt
111dx111x dtdt, ,,(,),1,,,222x11xtt1,1,x1,2t
1111而 xx, dtdx,,,2211tx1,1,
11dxdxx所以 . ,,,221x1,x1,x
11mnnm 7. 证明: x(1,x)dx,x(1,x)dx. ,,00
1011mnmnmnnm 证明 令1,x,t , 则x(1,x)dx,,(1,t)tdt,(1,t)tdt,x(1,x)dx, ,,,,0100
11mnnm即x(1,x)dx,x(1,x)dx. ,,00
,,nn 8. 证明: 2sinxdx,2sinxdx. ,,00
,,,nnn2 证明 sinxdx,sinxdx,sinxdx, ,,,,002
,,,,令0x,,tnnnn22而 sinxdxsin,(,t)(,dt),sintdt,sinxdx, ,,,,,,0022
,,nn所以 2sinxdx,2sinxdx. ,,00
a,1 9. 设f(x)是以l为周期的连续函数, 证明f(x)dx的值与a无关. ,a
a,la,lla,la10f(x)dx,f(x)dx,f(x)dx,f(x)dx,f(x)dx,f(x)dx,f(x)dx, ,,,,,,,aall000 证明 已知f(x,l),f(x).
a,laaa令x,t,l 而 f(x)dxf(t,l)dt,f(x,l)dx,f(x)dx, ,,,,000l
a,l1所以 f(x)dx,f(x)dx. ,,a0
a,1因此f(x)dx的值与a无关. ,a
x 10. 若f(t)是连续函数且为奇函数, 证明f(t)dt是偶函数; 若f(t)是连续函数且为偶函数, ,0
x证明f(t)dt是奇函数. ,0
x 证明 设F(x),f(t)dt. ,0
若f(t)是连续函数且为奇函数, 则f(,t),,f(t), 从而
,xxxx令t,,u F(,x),f(t)dtf(,u)(,1)du,f(u)dx,f(x)dx,F(x), ,,,,0000
x即F(x),f(t)dt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
,xxxx令t,,u F(,x),f(t)dtf(,u)(,1)du,,f(u)dx,,f(x)dx,,F(x), ,,,,0000
x即F(x),f(t)dt是奇函数. ,0
11. 计算下列定积分:
1,x (1)xedx; ,0
11111,x,x,x,x,1,x,1 解 xedx,,xde,,xe,edx,,e,e,1,2e. ,,,00000e (2)xlnxdx; ,1
eeeee1111111222222 解 . xlnxdx,lnxdx,xlnx,x,dx,e,x,(e,1),,,11011222x244
,2, (3)tsin,tdt(,为常数); ,0
2,222,,,111,,,, 解 ,tsintdt,,tdcos,t,,tcos,t,cos,tdt ,,,0000,,,
,2,,212,,sin. ,,,t,,2220,,,
, x3 (4)dx; ,,2xsin4
,,,,,,,x133 解 333dx,,xdcotx,,xcotx,cotxdx,,,,,lnsinx ,,,,,,,,234sinx344444
1313 . ,,(,),ln4922
4lnx (5)dx; ,1x
4444lnx1 解 dx2lnxdx2xlnx2xdx ,,,,,,,1111xx
441 . ,8ln2,2dx,8ln2,4x,4(2ln2,1),11x
1 (6)xarctanxdx; ,0
11111111222 解 xxdxxdxxxxdxarctan,arctan,arctan,,,,,20000222x1,
11,,,,,11111 . ,,,dx,,x,x,,,,,(1)(arctan)(1),200828282442x,1
,x22 (7)ecosxdx; ,0
,,,,xxxx22222222 解 ecosxdx,edsinx,esinx,2esinxdx,,,0000
,,,,,,,xxxx22222222 ,e,2edcosx,e,2ecosx,4ecosxdx,e,2,4ecosxdx,,,0000,1x,22所以 , ecosxdx,(e,2),05
于是
2 (8)xlogxdx; 2,1
21111222222 解 xlogxdxlogxdxxlogxxdx ,,,,222,,,1111222xln2
21132 22. ,,,x,,12ln224ln2
,2 (9)(xsinx)dx; ,0
,,,,1112232 (xsinx)dx,x(1,cos2x)dx,x,xdsin2x,,,0000264
33,,,,,,,1113 解 ,,xsin2x,sin2x,2xdx,,xdcos2x,,0000064464
333,,,,,,,,111 . ,,xx,xdx,,,x,,cos2cos2sin2,00064464864
e (10)sin(lnx)dx; ,1
e1令lnx,tt 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
11t所以 . esintdt,(e,sin1,e,cos1,1),02
e1因此 . sin(lnx)dx,(e,sin1,e,cos1,1),12
eeee1 解法二 sin(lnx)dx,x,sin(lnx),x,cos(lnx),dx,e,sin1,cos(lnx)dx,,,1111x
ee1 ,e,,x,x,x,x,dxsin1cos(ln)sin(ln),11x
e ,e,sin1,e,cos1,1,sin(lnx)dx, ,0
e1故 . sin(lnx)dx,(e,sin1,e,cos1,1),12
e (11)|lnx|dx; 1,e
e1eee11 解 |ln|xdx,,lnxdx,lnxdx,,lnxx,lnxx,dx,dx 111,,,,,1111eeee111 . ,,,e,(1,),(e,1),2(1,)eee
m122 (12)(1,x)dx(m为自然数); ,0
m,令,xsint1m,2122 解 ,(1x)dxcostdt. ,,00
,,,n1nn,222,, cosxdxcosxdx,,00n
,,,mm2m4531, 根据递推公式,,,,,,,, m为奇数m1,2,,,m1m1m364222 . ,,(1x)dx,,0,,mm2m4642,,,,,,,, m为偶数,,,m1m1m3753,
,m (13)J,xsinxdx(m为自然数). m,0
解 因为
0,,,令x,,,tmmmm xsinxdx,(,t)sin,(,t)(,1)dt,,sintdt,tsintdt, ,,,,000,
,,,,,,mmmm22所以 (用第8题结果). J,xsinxdx,sinxdx,,2sinxdx,sinxdx,m,,,,000022
,,,n1nn,222 根据递推公式,, sinxdxsinxdx,,00n
2,,m,m,m,135531,,,,,,,,m为偶数 ,mm,m,246422 . J,,mm,m,m,135642,,,,,,,,m为奇数 , mm,m,24753,