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Module 2

Integral calculus

General characteristic of the module. This chapter is designed to give a brief review of basic methods of indefinite and definite integrals application.

Text organization

Topic 1. Indefinite integrals. Table of integrals. Evaluating techniques. Topic 2. Polynomial functions. Rational functions.

Topic 3. Integrating of rational functions by partial fractions. Topic 4. Integrals involving powers of trigonometric functions. Topic 5. Integrating of irrational functions.

Topic 6. Definite integrals. Newton-Leibniz fundamental theorem. Properties of definite integrals. Evaluating techniques.

Topic 7. Improper integrals. Integrand unbounded. Topic 8. Application of the definite integrals.

Basic concepts. 1. Antiderivative. 2. Indefinite integral. 3. Definite integral. 4. Improper integral. 5. Convergence of improper integrals.

Basic problems. 1. Finding indefinite integrals. 2. Definite integrals evaluation techniques. 3. Application of definite integrals. 4. Investigation to convergence of improper integrals.

Topic 1. Indefinite integral

Concepts antiderivative and the indefinite integral. The table of the integrals. The substitution technique. Integration by parts.

Literature: [1, section 6, ch. ch. 6.1 – 6.3], [2, section 2, ch. 2.1], [3, section 7, §1], [4, section 7, §22], [6, section 8], [7, section 10, §§1–6], [9, §§29–30].

T1. Main concepts

1.1.Some basic indefinite integrals

There are two main concepts in calculus: the derivative and integral. The derivative provides information at a point or at a particular instant. The integral, the other major concept of calculus, does just the opposite. It is a tool for obtaining the numerical value of some overall quantity from local information.This section describes the substitution technique that changes the form of an integral, preferably to that of an easier integral. Before describing

this technique, we collect some basic facts about integrals in order to have a supply of integrands.

Every formula for a derivative provides a corresponding formula for an antiderivative of integral. For instance, since (3x4+5) =12x3 and (3x4 7) =12x3 that does not imply that 3x4+5=3x4 7.

An antiderivative of a function f is a function F such that F (x) = f(x),

or equivalently, in differential notation, dF = f(x)dx.

For any constant C, (3x4+C) =12x3.

Any two antiderivatives of a function differ only by a constant.

Since 3x4 + C describes all antiderivatives of 12x3, we can refer to it as

being the most general antiderivative of 12 x3, denoted by 12x3dx, which is read “the indefinite integral of 12x3, with respect to x”. Thus we write

12x3dx = 3x4+ C.

The symbol is called the integral sign, 12x3 is the integrand, and C is the constant of integration. The dx is a part of the integral notation and indicates the variable involved. Here x is the variable of integration.

More generally, the indefinite integral of any function f with respect to x

is written f(x)dx and denotes the most general antiderivative of f. Since all antiderivatives of f differ only by a constant, if F is any antiderivative of f, then

f(x)dx = F(x) + C, where C is a constant. To integrate f means to find f(x)dx. In summary,

f(x)dx = F(x) + C if and only if F (x) = f(x).

1.2.Properties of integrals

1. If k is a constant, then kf (x)dx k f (x)dx and if k ≠ 0, then
f (x)dx	1	f (x)dx. The constant can be taken outside the integral sign.
k	k

2.For two functions f(x) and g(x):

( f (x)) g(x)dx f (x)dx g(x)dx.

An antiderivative of the sum of functions can be found by adding antiderivatives of these functions.

3. If k ≠ 0 and l are constants and f (x)dx F(x) C then

f (kx l)dx	1	F(kx l) C.	(2.1)
	k

4.d f (x)dx f x dx.

5.F (x)dx F(x) C .

The following miniature integral table list has a few formulas that should be memorized. Each can be checked by differentiation.

1.3. Table of integrals

Let a function u(x) be continuous on an open interval [a,b] and have the

continuous derivative u (x)

on [a,b]. Then on [a,b], the following formulas are

accurated (table. 2.1):

Table 2.1

1. 0du C .

2. k du ku C .

u n 1

4. ò

3. u

C , n 1 .

u =ln | u | +C .

n 1

C .

6. eu du eu C .

ln a

sin udu cos u C .

cos udu sin u C .

tg u C .

10.

ctg u C .

cos

sin

11.

arctg u

C .

12.

u a

C .

u a

13.

arcsin u

C .

14.

u2 a2

C .

a 2 u 2

15.

sh udu ch u C .

16.

ch udu sh u C .

17.

th u C .

18.

cth u C .

sh2 u

ch 2

19.

ò tg udu =-ln | cos u | +C .

20.

ò ctg udu =ln | sin u | +C .

21.

ln | tg u | C

22.

ln | tg(u

) | C

sin u

cos u

Methods of integration: 1. Integration by tables.

2. The substitution technique.

3. Integration by parts.

1.4. Integration by tables

Certain forms of integrals that occur frequently may be found in standard tables of integration formulas. A shot table appears in table 2.1 and its use will be illustrated in this section.

A given integral may have to be replaced by an equivalent form before it fits a formula in the table. The equivalent form must match the formula exactly. Consequently, the steps that you perform should not be done mentally. Failure to do this can easily lead to incorrect results.

1.5. The substitution technique

This method involves the introduction of a function that changes the form of the integrand, hopefully to that of a simpler integrand. Several examples will illustrate the mechanics of the method, known as substitution. The proof that it works will be given after these examples. Substitution is the most important tool in computing integrals.

There are two cases in order to apply the substitution technique to find

f (x)dx :

x φ(t)

;

ω (x) t

Here φ(t) and ω(x) are both continuous and differentiable functions. In the first case: dx φ (t)dt and

f (x)dx f (φ (t)) φ (t)dt .

The second substitution is used if a function f has the form


f (x)dx g(ω(x)) ω (x)dx ,
then
		g(t)dt . ( 2.2)
f (x)dx g(ω(x)) ω (x)dx g(ω (x))d ω(x)		g(t)dt . ( 2.2)
In this case a function (x) can be written:
	dt
ω (x)dx d (ω (x))	dt

and is called basic integration form.

It is important to keep in mind that there is no simple routine method for antidifferentiation of elementary functions. This is in contrast with the routine that exists for differentiation. Practice in integration pays off in the quick recognition of which technique is most promising.

1.6. Integration by parts

Integration by parts is used not only to obtain integrals, but also to establish properties of functions.

Just as the chain rule is the basis for integration by substitution, the formula for the derivative of a product is the basis for integration by parts. Let

u u(x) and v v(x) be differentiable functions on some interval. Then

udv uv vdu

This differential form is the most useful and is called the integration by parts formula.

The key to applying integration by parts is the labeling of u and dv. Usually three conditions should be met.

First: v can be found by integrating and should not be too messy. Second: du should not be messier than u.

Third: vdu should be easier than the original udv .

Some types of integrals where need that use of integration by parts are: 1) integrals in forms


	P(x)ekx dx , P(x)sin kxdx , P(x) cos kxdx ,
where	P(x) is polynomial. In this case we consider the factor P(x) as u
and dv	is the remainder;
2) integrals in forms
P(x) ln xdx ,		P(x) arcsin xdx , P(x) arccos xdx , P(x) arctg xdx ,

where P(x) is polynomial. In this case dv P(x)dx and u is remainder;

3)integrals in forms

e x sin xdx , e x cos xdx ,

where , are real numbers. In this case integration by parts is applied twice and provides an equation for the unknown integral.

T1. Typical problems

І. Using a table of integrals 1. Find 1dx.

Solution. By formula 2 with k=1, 1dx 1x C. Usually we write 1dx

as dx. Thus dx x C.
2. Find x3dx .
Solution. By formula 3 with n = 3: x3dx	x3 1	C	x4	C .
Solution. By formula 3 with n = 3: x3dx	3 1	C		C .
	3 1	4

3. Find 5xdx .

Solution. By property 1 with k = 5 , f(x) = x and by formula 3 with n = 1

we have: 5xdx 5 xdx 5	x1 1	C	5x2	C.
			2
1 1

4. Find 3x2 4x 5 dx .

Solution. By property 2 and by formulas 2 and 3:

3x2 4x 5 dx = 3x2dx – 4xdx + 5dx =

x2 1

x1 1

5x C x3 2x2 5x C .

2 1

1 1

One constant of integration is enough.

5. Find xdx .

Solution. Here x is variable of integration. We rewrite the integrand so that

a basic formula can be used .Since

x x 2 , application of formula 3 gives:

xdx x0,5dx

x0,5 1

x C .

0,5 1

3 x2

6: Find I

dx .

					2				1		2x		5 1		x	2	1		3x		1	1
					2				1				5 1			3	1				2	1
		5			3				2
Solution.	2x		x			3x				dx
Solution.	2x		x			3x				dx		5 1			2					1
												5 1			2	1				1	1
															3	1				2	1
				x 4				5		1									5			1
							3x3			1					1			3x	3			1
C							3x3			6x 2	C				1			3x	3	6x 2 C .
C				2			5			6x 2	C			2x4				5		6x 2 C .
				2			5							2x4				5
7. Find x		x3 2 x2dx .

Solution. Sometimes , in order to apply the basic integration formulas it is necessary to first perform algebraic manipulations on an integrand, as is shown:

2x3

2x 2

In Example 7 we first multiplied the factors in the integrand. We point out

that x

x3 2

x2dx x

x3 2

dx x2dx .

More generally, f x g x dx f x dx g x dx .

The integral of a product is not the product of the integrals.

8. Find x3 2x2 x 3dx.

Solution. We can break up the integrand into fractions by dividing each term in the numerator by the denominator:

				3								x	2						1					x	2	4	3
								2											2			1
					2			2											2			1
		2x				3x									4x					3x			C					C.
	x	2x				3x				dx		2			4x					3x			C	2		x	x	C.
			3x 5 2 dx									2												2		x	x
9. Find			3x 5 2 dx						.
9. Find									.
						x
Solution. By property 3 with k														= 7 , l = –5 and by formula 3 with n, = 5 we
get:	9x2 30x 25						dx				3					1					1
	9x2 30x 25										2					2					2
		1									2		30x			2			25x		2
	x	1								9x			30x						25x			dx
	x		2
											5						3				1
								=		18 x 2		20x						50x 2 C.
								=		18 x 2		20x					2	50x 2 C.
										5


10. Find sin		2x dx
10. Find sin	37	2x dx
	37

Solution. By property 3 with k = –2, l = π/37 and by formula 7:

sin

2x dx

cos

11. Find

dx .

2 58

Solution. By property 3 with k =

2 dx

ln 5

12. Find 3x exdx .

Solution. As is generally known a x b x

(ab) x , then

3x e x dx (3e) x dx

(3e) x

C .

ln(3e)

13. Find а)

;

2 2x2

9 4x 2

Solution.

а)

ln | x +

1+x2 | +C ;

2x2

1 arcsin

C .

9 4x

9 / 4 x

14. Find

1 cos

Solution.

1 tg x C .

cos 2x

cos

15. Find

Solution.

C .

( 5)

2 5

16. Find

sin

x cos

Solution.

(sin 2 x cos2 x)dx

sin

x cos

sin

x cos

cosdx2 x sindx2 x tg x ctg x C .

17.Find cos2 2x dx .

Solution. cos2

(1 cos x)dx

(x sin x) C .

18. Find

4 x

dx .

Solution.

4 x 4

(2 x

2)(2

dx (2 x

2)dx

2 x

2x C .

2 2

ln 2

19. Find

1 9x

Solution.

1 arctg 3x C .

1/ 9 x

(1/ 3)

20. Find

x 2 dx

Solution. As the power of the numerator is not less than that of the

denominator, we use long division:

x 2 dx

(x 2

1) 1

x arctg x C .

21. Find

5x 3

Solution.

1 ln | 5x 3 | C.

22. Find cos3xdx .

Solution. cos 3xdx

1 sin 3x C .

23. Find (2x 1)5 dx .

Solution.

(2x 1)

5 dx

1 (2x 1)6

(2x 1)6

C .

24. Find

xdx

x 2

Solution.

xdx

(x 2) 2

x 2

= x -2 ln

x +2

+C .

25. Find

5dx

5 3x 8 2

Solution. By formula 13 and (2.1):

5dx

5 arcsin

3x 8

5 3x 8 2

26. Find

4dx

6x 7

Solution. First perform algebraic manipulations on an integrand, as is

shown:

– 6x + 7

Now it is easy to find the integral:

															3
		4dx									1	2			2x			4		4x 3
		4dx									1	2			2x	2	C	4		4x 3	C.
										4	2			arctg					arctg
			3 2					19		4	2		19	arctg	19			19	arctg	19
	2x
								4							2

			2
ІІ. Using substitution
27. Find					2xdx
									.
				x2			7		.
Solution. Note that 2x is the derivative of x2 + 7. Introduce u = x2+7. Then
du = 2xdx, and						2xdx					du
											u .
						x2 7					u .

Now it’s easy to find duu .

Replacing u by x2 + 7 in ln|u| yields ln|x2 + 7| . Since x2 + 7 is always positive we can omit the absolute-value bars:

		2xdx		ln x2			7 C.
		2
		x	7	d x2 7
We can write this as	2xdx							ln x2 7	C.
	2			x	2	7
	x	7

28. Find x 2 xdx .

Solution. Using the substitution 2 x t , we get

2 x t 2 , x 2 t 2 and dx 2tdt . Then evaluate the new indefinite integral:

x 2 xdx

(2 t 2 ) t ( 2t)dt

2 (t 4

2t 2 )dt

t 5

t 3

C .

Replacing t by

2 x

we have

2 xdx

(2 x)5

(2 x)3

C .

29. Find

Solution. Using the substitution 1 e x

t . In this case it’s more useful to

find dt : dt e x dx

and just as

e x t 1 so

dt (t 1)dx , or

Since

t 1

t (t 1)

d(t 1)

1 ex

t(t

t(t 1)

t 1

ln | t 1| ln | t | C ln ex ln 1 ex C x ln1 e x C .

30. Find 4 x 2 dx .

Solution. Using the substitution x 2 sin t , dx 2 cos tdt . Thereby

4 x 2		4 4 sin 2 t 2	1 sin 2 t 2 cos t .
Therefore

4 x 2 dx 2 cos t 2 cos tdt 4 cos2 tdt

2 (1 cos 2t)dt 2t sin 2t C .

Replacing t, we have:

sin t		x	, t arcsin		x		,		sin 2t 2 sin t cos t

	2			2
2 sin t 1 sin 2 t				2		x				1		x 2		x	4 x 2 ,
				2							4		2
	4 x 2 dx 2 arcsin								x		x		4 x 2		C .
								2			2

Frequently the basic integration form (2.2) is useful. It is provided to perform the following algebraic manipulations of the differential.

dx d (x b) , b is a constant,

dx 1a d (ax b) , а, b are the constants,

xdx	1 d (x2 ) ,			dx d(ln x) ,
	2			x
cos xdx d (sin x) ,			sin xdx d (cos x) ,
dx		d (tg x) ,		dx		d (ctg x) .
				sin2
cos2 x					x

31. Find x 1 x2 dx .

Solution. As in our case xdx 12 d(1 x2 ) , for the original integral we can state

x 1 x2 dx					1 1 x2 d(1 x2 ) .
					2
					1	3
The table integral is provided u 2 dx							2
							2	u	2	C , we get
						3		u	2	C , we get
						3
x 1 x2 dx				1		1
x 1 x2 dx				1	(1 x2 ) 2 d(1 x2 )
				2
	3									3
1	2	(1 x2 )		C		1 (1 x2 )						C .
	2		2								2
	3		2								2
2	3					3

32. Find x3e x4 dx .

Solution. Finding d( x4 ) 4x3dx , then x3dx 14 d ( x4 ) . Hence

x3e x4 dx 14 e x4 d( x 4 ) 14 e x4 C .

33. Find		dx	.
	x	1 ln 2 x

Solution. As in our case d(ln x) 1x dx , for the original integral we can state

d ln x

x 1 ln 2 x

1 ln 2 x

We get the table integral

arcsin u C . Therefore,

1 u 2

arcsin(ln x) C .

1 ln2

34. Find

arctg x

dx .

1 x

Solution.

arctg x

arctg xd(arctg x)

arctg2

C .

1 x

35. Find

7dx .

Solution. We can write this as x2 x3

7 2 dx . Notice that the

integrand contains a power of the function x3 – 7 . If u = x3 – 7 then du = 3x2dx.

Since the constant factor is 3 in du, the constant 7 does not appear in the integrand, this integral does not have the form of un du . However we can put

the given integral in this form by first multiplying and dividing it by 3. This does not change the value. Thus:

x2								x3	1					u x		3	7
									1							3
		x3 7dx					1		7 2 3x2dx
							3							du 3x2dx
		1				3			3								3
		1		u 2					3								3
1	u 2 du		1							2				2	x3 7
1			1					C		2	u	2	C	2				2	C.
3			3					C			u	2	C	9				2	C.
3			3		3					3 3				9
				2

Or x2

x3 7dx 1

x3 7d x3 7 2 x3 7 2 C.

36. Find

4x3 3x

dx .

Solution. If

u = 2x4– 3x2 + 7 then du = (8x3 – 6x)dx which is two times

greater then

the numerator.

We multiply the integrand by 2 and compensate it by the factor

putting

it before the integral sign.

1 2 4x3 3x dx 1

8x3 6x dx

u 2x4 3x2 7

1 du

8x3 6x dx

2x4 3x2 7

1 ln

2x4 3x2 7

37. Find

sin xdx

2 cos2

sin x dx

Solution.

sin xdx

t cos x

2 cos x

dt sin xdx

2 cos2 x

2 t

C ln

cos x

2 cos

2 t2

2dx

38. Find

Solution.

x ln

2dx

t ln x

2 dt3 2 t

3dt

t 2 C

x ln

ІІІ. Using integration by parts

39. Find

x cos xdx .

Solution.

To use the formula (2.1), it is necessary to write x cos xdx in

the form udu . The integrand is so simple that there is not much choice. Try u = x and dv = cosxdx; that is break up the integrand this way:

x cosx dx u dv

Then find du and v. Since u = x, it follows that du = dx. Since dv = cos x dx, we choose v = sin x. Of course v could be sin x + C for any constant C, but choose the simplest v whose derivative is cos x. Applying integration by parts yields:

x cos x dx = x sin x sinx dx.

u dv

u v

v du

v du is simpler than the original integral u dv.

sin x dx = cosx C.

Hence x cos x dx = x sinx + cosx + C. Save the constant of integration till the end. The reader may check this by differentiation.

We denote: u = cos x; dv = x dx. Let us see what this method gives:

x cos xdx

u cos x du sin xdx

x2 cos x

sin xdx.

dv xdx v 2

The new integrand is harder than the one we started with. Though sinx is

not harder than cosx;

sin x is definitely harder than xcosx for the exponent

has increased from 1 to 2. 40. Find x ln xdx.

Solution. Setting dv = lnxdx is not a wise move since v = ln xdx is not immediately apparent. But setting u = lnx is promising since du = dxx is much easier than lnx . This second approach goes through smoothly:

		u ln x du				dx			x2				x2	dx
						dx
x ln xdx							x
x ln xdx		dv xdx v				x		2		2			2	x
						x				2			2	x
						2
						2
	x2 ln x		1	xdx	x2						x2
=								ln x				C.
=	2		2		2			ln x		4		C.
	2		2		2					4

The result may be checked by differentiation.

Remark 1. The key to applying integration by parts is the correct choice of u and dv. Usually three conditions should be met:

First: v can be found by integrating and should not be too messy. Second: du should not be messier than u.

Third: vdu should be easier than the original udv .

For instance x2e2 xdx , with u = x2 and dv = e2xdx, meets these criteria. In

this case v = e2x can be found and is not too messy; du = dx2 =

2xdx is

easier than u as example 25 shows. vdu

is indeed simpler than udv .

41. Find ln xdx .

Solution.

x ln x x dx x ln x x C.

ln xdx

u ln x du x

dv dx v x

42. Find ln2 xdx.

Solution. ln2 xdx

ln2 x du

2ln xdx

x ln2

dv dx v x

x ln xdx

x ln x2 2 ln xdx.

By example 41 ln xdx x ln x x C.

Thus: ln2 xdx x ln2

x 2x ln x x C.

43. Find. arctg 2xdx.

Solution. arctg 2xdx

u arctg 2xdx du

2dx

4x2

dv dx v x

x arctg 2x

2xdx

1 4x2 t

x arctg 2x

8xdx

x arctg 2x

1 ln

x arctg 2x 14 ln 1 4x2 C .

44.

Find x2e2 xdx.

Solution.

2 x

u x2

2xdx

dv e2 xdx v 1 e2x

2 x

2 e

2xdx 2 x

u x du dx

2 x

dv e2xdx v 1 e2x

2 x

1 xe2x

1 e2x dx

1 x2e2x

1 xe2x

1 e2x

Integration by parts with u = x3

could be used to express x3e2xdx in

terms of

x2e2 xdx. Another integration

by parts with u=x2, then expresses

x2exdx

terms

xexdx,

as was

done in Example 25. Each time

integration by parts lowers the exponent by 1.

45. Find (x2

2x 5)e x dx .

Solution. In this case

u x2 2x 5

and dv e x dx , therefore

v e x

and du (2x 2)dx .Hence

(x2 2x 5)e x dx (x2 2x 5)e x 2 (x 1)e x dx .

Using another integration by parts to the last integral:

u x 1 , dv e x dx , du dx , v e x ,

(x 1)e x dx (x 1)e x e x dx (x 1)e x e x C1

xe x C1 .

Thus

(x2 2x 5)e x dx (x2 2x 5)e x 2xe x C (x2 5)e x C Remark 2. If P(x) is a polynomial of the form

P(x)e x dx Q(x)e x C ,

where Q(x) is a polynomial which has the same degree of P(x) , then the

indeterminate coefficient method is used. The following example shows how to integrate in this case.

46. Find (x3 18)e2x dx .

Solution. (x3 18)e2x dx (Ax3 Bx 2 Cx D)e2x C1 , where A, B, C, D are unknown coefficients.

Differentiating both parts of the expression with respect to х:

(x3 18)e2x (3Ax2 2Bx C)e2x 2(Ax3 Bx2 Cx D)e2x ,

then

x3 18 (3Ax2 2Bx C) 2(Ax3 Bx2 Cx D) ,

x3 18 2Ax3 (3A 2B)x2 (2B 2C)x C 2D .

Since, the coefficients of the same degrees of х are equal and we have 1 2A, 0 3A 2B, 0 2B 2C, 18 C 2D,

therefore A 12 , B 34 , C 34 , D 698 . Hence

(x3 18)e2x dx 18 4x3 6x2 6x 69 e2x C1 .

Remark 3. Reduction or recursion formulas. Many formulas in a table of integrals express the integral of a function that involves the n-th power of some expression in terms of the integral of a function that involves the (n-1)-th or lower power of the same expression. These are reduction formulas. Usually they are obtainned by an integration by parts as examples 47 and 48 shows.

47. Find cosln xdx .

Solution. Letting u cos ln x and dv dx we have

du sin ln x dx , v x ,	cos ln xdx x cos ln x sin ln xdx .
x
Using other integration by parts to the last integral:
u sin ln x ,	dv dx , du cos ln x dx	, v x ,
	x

sin ln xdx x sin ln x cos ln xdx .

Since

cos ln xdx x cosln x x sin ln x cos ln xdx ,

therefore

cos ln xdx 1 (x cos ln x x sin ln x) C .

48. Find

Solution.

a x2 x2

x2 a n

a x2

x2dx

x xdx

x2 a n

x2 a n 1

x2 a n

u x du

xdx

2 1 n x2 a n 1

x2 a n 1

x2 a

2 1 n x2 a n 1

2 1 n

x2 a n 1

2a 1 n x2 a n 1

2a 1 n

x2 a

n 1

2 1 n 1

2a 1 n

x2 a

n 1

2a 1 n x2 a n 1

Hence

2n 3

(2.3)

x2 a n

2a n 1 x2

a n 1

2a n 1

n 1

This is a reduction formula.

For example, this formula is spelled in the following case.

arctg

C is the table integral;

x2 a2

J 2

(x2 a 2 )2

2a 2

x 2 a 2

a 2

arctg

C etc.

2a 2

2 a 2

2a3

49. Find

J n, m

sin n

dx .

cosm x

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