不定积分的换元积分法
第四节 不定积分的换元积分法
不定积分时若凑微分法、分部法均解决不了问题,且被积函数中含有复杂的量(如:、
n、等),则可以考虑使用换元积分法. axb,arcsinx,,
一、换元积分法
1dx例6(4(1 求不定积分. ,1,x
2xt,解 这里主要障碍是 “”,不妨令 此时 xt,
x这样把复杂的量“”换元成最简单的变量“” t1dx则 ,1,x
12 ,dt,1,t
2t ,dt,1,t
t,,11 ,2dt,1,t
1 ,,2(1)dt,1,t
,,,,22ln1ttC
. ,,,,22ln1xxC,,
1dx例6(4(2 求不定积分. x,1,e
x解 同样令主要障碍e,此时 xt,ln,t
1dx则 x,1,e
1 ,dtln,1,t
1,dt ,1,tt,,
11 ,,()dt,tt1,
,,,,lnln1ttC
x,,,,xeCln(1).
例6(4(3 求不定积分. arcsinxdx,解 令,此时,则 arcsinxt,xt,sin
arcsinxdx,
,tdtsin ,
,,tttdtsinsin ,
,,,tttCsincos
2. ,,,,xxxCarcsin1
2xdx例6(4(4 求不定积分. 10,x,1,,解 令,此时,则 xt,,1xt,,1,,
2xdx 10,x,1,,
2t,1,, ,,dt1,,,10t
2tt,,21,dt 10,t
,,,8910,,,(2)tttdt ,
111 ,,,,,C789749ttt
111. ,,,,,C789714191xxx,,,,,,,,,
从以上例题可见,换元可使复杂积分变得简单,可关键是怎么换.
二、换元积分举例
例6(4(5 用换元法求下列不定积分:
xx,,11x(1); (2)edx; (3); dxdx,,,1,xx,,11
11dx(4); (5); (6). xxdx25,dx,,,x3xx1,e,1,,解
x(1) dx,1,x
t2,tdt x,1,t
22t,dt ,1,t
2t,,11,2dt ,1,t
1,,,,,21tdt ,,,1,t,,
2= tttC,,,,22ln1
=; xxxC,,,,22ln1,,
xedx(2) ,
t2xtedt, ,
t,2tedt ,
tt,,22teedt ,
t ,,,21etC,,
x=21exC,,; ,,
x,,11(3) dx,x,,11
t,12 xtdt,,,11,,,t,1
2tt,,2dt ,t,1
2tt,,,22,2dt ,t,1
2,,,,,22tdt ,,,t,1,,
2 ,,,,,tttC44ln1,,=; xxxC,,,,,,,1414ln11,,
(4) xxdx25,,
2222,,tt25(),,xttd ,554224tt,,dt ,25
2453 ,,,ttC12575
5324=; 2525,,,,xxC,,,,12575
1(5) dx,3xx1,,,
166 xtdt,,32tt1,,,26t66,,,dt 2,1,t
,,,66arctanttC
6666arctanxxC,,=;
1(6) dx,xe,1
1x2 etdt,,,1ln(1),t
1 ,2dt2,t,1
11,,,,dt ,,,tt,,11,,
t,1,,lnC t,1
2,,e,,11,,lnC. ,,2,,e,,11,,
可见前边例子中直接令“”或其它复杂的量“”也就行了.可若“”下含有,t
2“”项,问题就不是那么简单了. x
,,,,22t,,,1,xdx例6(4(6 利用三角公式()换元,求积分. 1sincos,,tt,,,22,,
21,xdx解 ,
xttdt,sincossin ,
2,costdt ,
1cos2,t ,dt,2
11 ,,dttdtcos22,,24
11 ,,,ttCsin224
11 ,,,tttCsincos22
11. ,,,arcsincosarcsinxxxC,,22
,,1,,2t,,,dx例6(4(7 利用三角公式()换元,求积分. 1tansec,,tt,,,222,,4,x
1解 dx ,24,x
1 xtdt,2tan2tan,2sect
,sectdt,
,,,lnsectanttC
xx,,. ,,,lnsecarctanC,,22,,
,dx,,2t,0,例6(4(8 利用三角公式()换元,求积分. sec1tantt,,,,,322,,xx,9
dx解 ,32xx,9
dt3sec xt,3sec3,27sec3tantt,
11 ,dt2,27sect
12 ,costdt,27
1 ,,(1cos2)tdt,54
11 ,,,ttCsin254108
11 ,,,tttCsincos5454
1313,,,,,arccossinarccosC. ,,5418xxx,,例6(4(9 求下列不定积分:
sinx1,x2xxdx,3(1);(2);(3). dxdx,,,sincosxx,1,x
解
sinx(1) dx,sincosxx,
1 ,dx,1cot,x
1cotxt, darctcot,1,t11 ,,dt2,11,,tt
111t,,,,,,dt ,,2,211,,tt,,
1111t, ,,,dtdt2,,2121,,tt
11111t ,,,,dtdtdt22,,,212121,,,ttt1112 ,,,,,,,ln1ln1cotttarctC,,242
1112=; ,,,,,,ln1cotln1cotxxxC,,242
1,x(2) dx,1,x
1,t2xtdt, ,21,t
2tt, ,2dt,21,t
2tt,,11 ,,22dtdt,,2211,,tt
122,,,,,,21212ttdtdt 查《积分表》(见文献文献×) ,,21,t
1t,,22,,,,,,,,212arcsin12arcsinttttC ,,22,,
22 ,,,,,,,211arcsinttttC
=; ,,,,2+1arcsinxxxC,,
2xxdx,3(3) ,
3sectx,3sec3tan3secttdt ,
22,33sectanttdt ,
2 ,33tantantdt,
3,,3tantC
,,33; ,,3tanarcscoC,,,,x,,
此题还可以用另一个很简单的解法:
2xxdx,3 ,
122 ,,xdx3,2
11222,,,xdx33 ,,,,,2
3122,,,xC3; ,,3
可见换元积分法不是一个很好的方法,凑微分法、分部法均解决不了,再考虑用它. 思考题6.4
1(本节介绍的换元积分法中,换元的根本目的是什么,应注意什么问题, 2(总结一下利用三角公式换元积分法(三角代换法)的三种类型. 3(思考凑微分法、分部法及换元法三种积分方法的优先次序,如何选用, 练习题6.4
1. 用换元法求下列不定积分:
1xxdxdx(1); (2); (3). dx3,,,31,x12,,x1,x,,2. 利用三角代换求下列不定积分:
1122axdx,dxdx(1); (2); (3)( a,0a,0,,,,,,,2222xxa,xx,4
练习题6.4答案
1.解
xdx(1) ,1,x
2t,1211,,,xtdt ,,,t2 ,,21tdt,,,
23 ,,,ttC23
32=; 121,,,,xxC3
1dx(2) ,312,,x
133 xtdt,,,22,,,1,t
23t,dt ,1,t
2t,,11,3dt ,1,t
1,,,,,31tdt ,,,1,t,,
32= tttC,,,,33ln12
23333=; xxxC,,,,,,,2323ln12,,2
x(3) dx3,1,x,,
1,t-x=t 11dt,,,3,t
t,1 ,dt3,t
11,,,,dt ,,23,tt,,
12 ,,,,C2tt
12=. ,,C2x,11,x,,
22axdx,2. 解(1) a,0,,,
xatatdat,sincos(sin),
22 ,atdtcos,
2a,,1cos2tdt ,,,2
22aa,,,ttCsin2 24
2axx22arcsin,,,axC=; 22a
1dx(2) ,22xx,4
1 x2tan2tan,tdt2,4tan2sectt
1cost,dt 2,4sint
11 ,dtsin2,4sint
1 ,,,C4sint
1x,,,,,cscarctanC; ,,42,,
1dx(3) a,0,,,22xxa,
1 xsecsec,atdat,atatsectan
1 ,dt,a
1 ,,tCa
1a( ,,arccosCax