EULER-SU
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Expository article by Khristo N. Boyadzhiev of Ohio Northern University, dated August 2006, found in a folder on the 1/sqrt(R) integral. It explains the first, second and third Euler substitutions, which remove the radical and reduce the integral to a rational function. It notes that the first two cover all cases, and works examples comparing the first, second and third substitutions. The formulas are garbled in the extracted text.
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1Euler Substitutions
Khristo N. Boyadzhiev
Ohio Nor thern University
August 2006
Euler su bstituti ons a re us ed to ev aluate i ntegrals of the f orm
,
by “removing ” the radica l. The re are three specific subst itutions s uggested by E uler. In eac h one
of them the idea is t o eliminate the term with . We assume that and that the poly nomi al
is no t strictly neg ative. Note also that if the radical can be removed by a
trigonomet ric or hype rbolic subst itution.
1. First Eul er substitution .
When we introduce a new vari able by setting
. (1)
Then
,
so that the term cancels out an d sol ving for we find
.
Thus the i ntegral takes the f orm
, (2)
where is a ra tional func tion.
22. Second Euler s ubstitution .
Suppose the polynomial has two dif feren t real root s, i.e.
.
Then we s et
, (3)
which yields
,
,
,
and aga in the integral is red uced to the f orm ( 2).
3. Third Euler substitution
The third E uler subs titution ca n be us ed w hen . In such ca se we set
, (4)
and then ,
,
, etc,
leading to the form ( 2).
4. Remarks.
The first two Euler substitutions are sufficient to cover all p ossible c ases, because if
, then the roots of the polynomial are real and different (the graph of this
3polynomial is a parabola open ing down a nd w ith vertex above th e x-axis). Ne vertheless, the th ird
Euler substitution is usef ul because it often requires less computations.
It is good to keep in mind that the radical can be simplif ied by co mplet ing
the polynomial to a perfect square and then using a trigonometric or hyperbolic
substitution .
5.Examples.
1. W e shall evaluate
, (5)
by the fi rst Euler subs titution. Set
.
Then
,
,
,
and
. (6)
2. The sam e integral ca n be evaluated by the third Eul er substi tution. F or this we set
,
4and the refore, sol ving for ,
,
,
, (7)
which is (6) in a diffe rent form.
3. Now con sider
, (8)
for . We sh all use the second Eul er substi tution. Si nce we
set
,
which yields
, ,
.
Finally,
5.
Of course , in this integral we ca n use the second Eul er su bstituti on in the form
,
with the same s uccess.