Ponce 2011 Discussion Substitution Method

7/31/2019 Ponce 2011 Discussion Substitution Method

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A DISCUSSION ON THE SUBSTITUTION METHOD FOR

TRIGONOMETRIC RATIONAL FUNCTIONS

Juan Carlos Ponce-Campuzano and Antonio Rivera-FigueroaMathematics Education Department

Center for Research and Advanced StudiesCinvestav, IPN, DF, Mexico

[email protected]; [email protected]

SUBJECT: CALCULUS

INTRODUCTION

It is very common to see, in the books of Calculus, primitives of functions. However thegreat majority of authors pay scant attention to the domains over which the primitives arevalid, which could lead to errors in the evaluation of definite integrals. For example, the

substitution

tan

leads to

15 3 c o s 12 arctan12 tan 2 and 15 3 s i n 12 arctan54 tan 2 34 .However

15 3 cos

12 arctan12 tan 2

12 arctan

12 tan

22

12 arctan

12 tan

02

= 0and

15 3 s i n

12 arctan54 tan 2 34

12 arctan 54 tan 22 34 12 arctan 54 tan 02 34.= 0

both of which are clearly false as the integrands are continuous and positive on0,2.These situations can be found, for instance, in: [1], [3], [5], [6] and [7]. In the teaching ofcalculus, in particular, in the teaching of the substitution method it is necessary to payattention in the domains of functions, especially when we are trying to evaluate definiteintegrals by finding primitives. It turns out that the evaluation of definite integrals mayrequire partitioning the interval of integration. This paper attempts to do this for rationalfunctions involving sin and cos.


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THE SUBSTITUTION METHOD

The substitution method for the evaluation of primitives and integrals is a consequence ofthe chain rule. Recall that a differentiable function : is a primitive function of:

, if

for every

.

TheoremLet: be a continuous function. Let : be a differentiable function.Suppose that . Then for every and , we have

.Proof: Let : be a primitive of on . The function has a derivativeon and for

. Therefore,

is a primitive of

on

; whence

= .

because of the fundamental theorem of calculus.

So that

.This method is frequently taught in courses of Calculus. Now let us discuss this methodfor trigonometric rational functions.

INTEGRAL OF THE FORM , A function : is a rational function if there are two polynomials and , on , such that

, , , for , . Where is the domain of and it is equal to the set, : , 0We shall show how to integrate functions having the form sin,cos whereis arational function.


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Before proceeding further, let us recall thatsin 2 sin 2 cos 2cos cos 2 sin 2for

.

Hence if and cos 0, we havesin

2sin 2cos 21cos 2 2tan 21 t a n 2 (1)

and

cos 1 sin 2

cos

21cos 2 1 t a n

21 t a n 2(2)

We recall that cos 0 if and only if for ; hence cos 0 if and only if 2 1 for .It follows from (1) that if and , thensin2 2 arctan sin2arctan

2tanarctan

1 t a narctan 21 and from (2) that cos2 2 arctan cos2arctan

1 t a narctan1 t a narctan 1 1

Therefore, we have

sin2 2arctan 21 (3)cos2 2arctan 1 1 (4)Now let and : be a rational function. Let , be an interval andsuppose that if ,, then sin,cos . Then sin,cos is a well-defined continuous function on ,; hence we may consider the integral


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sin,cos

(5)We shall now show that the evaluation of integrals of the form (5) can be reduced to the

evaluation of integrals of rational functions.

We shall divide the discussion into three cases:

Case I: Suppose , 2 ,2 for some .Let tan and tan , and let : , be defined by 2 2arctanfor . Then and . Note thatarctantan2 2 for

2 ,2 .

It follows from the substitution method that if we put sin,cos for , ,then

From (3) and (4), we obtain

sin ,cos 21 , 1

1

for , tan ,tan . Since 21 for , we deduce that

sin,cos

21 , 1 1 21

(6)

Since is a rational function we can conclude that the mapping 21 , 1 1 21 of, into is a rational function. Hence the evaluation of (5) is reduced to that of theintegral of a rational function.


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Example 1. Let : 0, be a function defined as 1sinDetermine a primitive of on 0, .For 0, , we define 1sin

is a primitive on 0, ; it remains to determine explicitly. To do this take , : 0,, for , and 0, . Then sin,cos

.Using the method described above, we get

1 2 21 1

log| logtan 2 l o g tan

22 Since tan 1 and log1 0, it follows that

log tan2 .

Case II: Suppose , 2 1, 2 1 for some . Notice that if 2 1, then there is no such that ; similarly there is no suchthat $ . Nevertheless, we can reduce Case II to Case I by a process of limit. Let'sshow an example.

Example 2. Evaluate

sinsin 2 .

Notice that the mapping : of0, into is continuous; hence this functiondoes have a primitive : 0, . Moreover, since is continuous on 0, we have sinsin 2

0 lim 0


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lim sinsin 2

Using (6), we obtain, for 0,

sinsin 2

4 1 1

2 1 2 1

2arctantan 2 433 arctan33 2tan 2 1 39Since

lim arctantan

, we have

sinsin 2 lim sinsin 2 1 439 .

Case III: Suppose , is an interval to which Case II does not apply. In this case thereare , , such that 2 1, 2 1 and 2 1, 2 1.In this case, the integral can be written as a sum of integral over the intervals

, 2 1, 2 1, 2 3, 2 1, .Each case can be treated as in Case II.Example 3. Evaluate

15 3 c o s

.We have

15 3 c o s

15 3 cos

15 3 cos

By (6), we obtain 12 arctan12 tan 2for 0, and 12 arctan12 tan 2for ,2. Therefore


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15 3 cos

15 3 c o s

15 3 c o s

lim 0 lim 0 4 0 0 4 2since lim for 0, and lim for ,2.

FINAL REMARKS

It is possible to find out explicitly primitives of certain kind of trigonometric functions. For

instance, on the one hand, let , , such that ; then we have 1

cos 1

2arctan 2sin

2 cos

1 sin 1 2arctan 2cos2 sin In the other hand, let , , such that 0 and 0; then

1 cos 1 arctan sin2 cos 2 1 sin 1 arctan sin2 cos 2

The above results can be found by the conventional substitution method

tan, but

using appropriate trigonometric identities. Note that the integrands belong to the family ofrational functions sin,cos of into where is a well-defined continuousfunction on . The primitives showed above are well-defined for every and they canbe used in order to evaluate the integral in any interval .It still remains an opened question: Is there a general method for determining explicitprimitives of rational functions sin,cos of into where is a well-definedcontinuous function on ?REFERENCES

1. M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions withFormulas, Graphs, and Mathematical Tables, New York: Dover Publications, (1965).

2. T. M. Apostol, Calculus. Volume 1. One-Variable Calculus with an Introduction toLinear Algebra, 2nd Edition, Waltham. MA: Blaisdell, ISBN: 0471000051, (1967).

3. S. Banach, Differential and Integral Calculus, Publisher: Niriega Limusa, ISBN: 968-18-3949-8, (1996).


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Ponce 2011 Discussion Substitution Method

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