techniques of integration file4 integrals of rational functions 5 integrals involving quadratic...

Integration by PartsTrigonometric Integrals

Trigonometric SubstitutionsIntegrals of Rational Functions

Integrals Involving Quadratic ExpressionsIntegrals of Rational Functions of sin(x) and cos(x)

Miscellaneous SubstitutionsImproper Integrals

Techniques of Integration

Department of MathematicsKing Saud University

2016-2017






Table of contents

1 Integration by Parts

2 Trigonometric Integrals

3 Trigonometric Substitutions

4 Integrals of Rational Functions

5 Integrals Involving Quadratic Expressions and MiscellaneousSubstitutions

6 Integrals of Rational Functions of sin(x) and cos(x)

7 Miscellaneous Substitutions

8 Improper Integrals







The product rule of differentiation yields an integration techniqueknown as integration by parts. Let us begin with the product rule:

d

dx(u(x)v(x)) = u(x)

dv(x)

dx+

du(x)

dxv(x).

On integrating each term with respect to x from x = a to x = b,we get

∫ b

a

d

dx(u(x)v(x))dx =

∫ b

au(x)

(dv(x)

dx

)dx+

∫ b

av(x)

(du(x)

dx

)dx .

By using the differential notation and the fundamental theorem ofcalculus, we get

[u(x)v(x)]ba =

∫ b

au(x)v ′(x)dx +

∫ b

av(x)u′(x)dx .






The standard form of this integration by parts formula is written as∫ b

au(x)v ′(x)dx = [u(x)v(x)]ba −

∫ b

av(x)u′(x)dx .

and ∫udv = uv −

∫vdu.

We state this result in the following theorem






Theorem

(Integration by Parts)If u and v are two continuously differentiable functions on theinterval [a, b], then we have∫ b

au(x)v ′(x)dx = [u(x)v(x)]ba −

∫ b

av(x)u′(x)dx

and for the indefinite integrals∫udv = uv −

∫vdu.






Example 1 :Evaluate the following integrals:

1

∫sinh−1(x)dx ,

2

∫cosh−1(x)dx ,

3

∫tanh−1(x)dx ,

4

∫sin−1(x)dx ,

5

∫tan−1(x)dx .






Solution

1 By parts

∫sinh−1(x)dx = x sinh−1(x)−

∫x√

1 + x2dx =

x sinh−1(x)−√

1 + x2 + c ,

2

∫cosh−1(x)dx = x cosh−1(x)−

∫x√

x2 − 1dx =

x cosh−1(x)−√x2 − 1,

3

∫tanh−1(x)dx = x tanh−1(x)−

∫x

1− x2dx =

x tanh−1(x) +1

2ln(1− x2) + c,

4

∫sin−1(x)dx = x sin−1(x)−

∫x√

1− x2dx =

x sin−1(x) +√

1− x2 + c ,

5

∫tan−1(x)dx = x tan−1(x)−

∫x

1 + x2dx =

x tanh−1(x)− 1

2ln(1 + x2) + c .






Example 2 :Evaluate the following integrals:

1

∫ln(x)dx ,

2

∫ln2(x)dx ,

3

∫x cos(x)dx ,

4

∫xexdx ,

5

∫ex sin(x)dx ,

6

∫cosh(x) cos(x)dx .






Solution

1 By parts

∫ln(x)dx = x ln(x)− x + c ,

2

∫ln2(x)dx = x ln2(x)− 2(x ln(x)− x) + c ,

3

∫x cos(x)dx = x sin(x) + cos(x) + c,

4

∫xexdx = xex − ex + c ,

5

∫ex sin(x)dx =

ex

2(sin(x)− cos(x)) + c ,

6

∫cosh(x) cos(x)dx =

1

2(sin(x) cosh(x) + cos(x) sinh(x)) + c.






Trigonometric Integrals

Some Important Trigometric Formulas:

cos2(x) + sin2(x) = 1,

cos(2x) = cos2(x)− sin2(x),

cos2(x) =1 + cos(2x)

2,

sin2(x) =1− cos(2x)

2,

sec2(x) = 1 + tan2(x),

csc2(x) = 1 + cot2(x),






d

dxsin(x) = cos(x),

d

dxcos(x) = − sin(x),

d

dxtan(x) = sec2(x),

d

dxsec(x) = sec(x) tan(x),

d

dxcot(x) = − csc2(x),

d

dxcsc(x) = − csc(x) cot(x).






sin(x + y) = sin(x) cos(y) + cos(x) sin(y),

sin(x − y) = sin(x) cos(y)− cos(x) sin(y),

cos(x + y) = cos(x) cos(y)− sin(x) sin(y),

cos(x − y) = cos(x) cos(y) + sin(x) sin(y),

sin(x) cos(y) =1

2(sin(x + y) + sin(x − y)), (1)

sin(x) sin(y) =1

2(cos(x − y)− cos(x + y)), (2)

cos(x) cos(y) =1

2(cos(x + y) + cos(x − y)), (3)






Integrals of Type Kn,m =

∫cosn(x) sinm(x)dx , m, n ∈ N.

• If m = 2q + 1, we set u = cos x , then du = − sin(x)dx and

Kn,m =

∫cosn(x) sin2q+1(x)dx

=

∫cosn(x) sin2q(x). sin(x)dx

=

∫cosn(x)

(sin2(x)

)q. sin(x)dx

= −∫

cosn(x)(

1− cos2(x))q.(− sin(x))dx

= −∫

un(

1− u2)qdu.






• If n = 2p + 1, we set u = sin(x), then du = cos(x)dx and

Kn,m =

∫cos2p+1(x)(x) sinm(x)dx

=

∫cos2p(x) sinm(x). cos(x)dx

=

∫ (cos2(x)

)psinm(x). cos(x)dx

=

∫ (1− sin2(x)

)psinm(x). cos(x))dx

=

∫ (1− u2

)pumdu.






• If n = 2p and m = 2q,

Kn,m =

∫cos2p(x)(x) sin2q(x)dx

=

∫cos2p(x)(1− cos2(x))qdx .

We compute the integral Jn =

∫cos2n(x)dx by induction and by

parts: We set u = cos2n−1(x) and v ′ = cos x , then

Jn = sin(x) cos2n−1(x) + (2n − 1)

∫cos2n−2(x) sin2(x)dx

= sin(x) cos2n−1(x) + (2n − 1)Jn−1 − (2n − 1)Jn.






Thus Jn = 12n sin(x) cos2n−1(x) + 2n−1

2n Jn−1.

In particular

J1 =

∫cos2(x)dx =

sin(x) cos(x)

2+

x

2+ c =

sin(2x)

4+

x

2+ c .

J2 =

∫cos4(x)dx =

sin(x) cos3(x)

4+

3 sin(x) cos(x)

8+

3x

8+ c.






Examples 1 :

1 ∫sin2(x) cos2(x)dx =

∫(1− cos2(x)) cos2(x)dx

=sin(x) cos(x)

8+

x

8− sin(x) cos3(x)

4+ c .

We can also∫sin2(x) cos2(x)dx =

1

4

∫sin2(2x)dx =

1

8

∫(1− cos(4x))dx

=x

8− sin(4x)

32+ c .

2

∫sin2(x)dx =

∫1− cos2(x)dx =

x

2− sin(2x)

4+ c ,






3 ∫sin3(x)dx

u=cos(x)= −

∫(1− u2)du

= − cos(x) +1

3cos3(x) + c ,

4 ∫sin4(x)dx =

∫(1− cos2(x))2dx

= −5 sin(2x)

16+

3

4x +

sin(x) cos3(x)

4+ c ,






5

∫cos3(x)dx

u=sin(x)=

∫(1− u2)du = sin(x)− 1

3sin3(x) + c .,

6

∫sin5(x) cos4(x)dx

u=cos(x)= −

∫u4(1− u2)2du =

−cos5(x)

5− cos9(x)

9+

2 cos7(x)

7+ c ,

7

∫sin4(x) cos3(x)dx

u=sin(x)=

∫u4(1− u2)du =

sin5(x)

5− sin7(x)

7+ c .






Integrals of Type Ln,m =

∫secm(x) tann(x)dx , m, n ∈ N.

• If m = 2q and q 6= 0, we set u = tan x , then du = sec2(x)dx and

Ln,2q =

∫sec2q(x) tann(x)dx

=

∫un(1 + u2)q−1du.






Examples 2 :

1 ∫sec4(x) tan7(x)dx =

∫u7(1+u2)du =

tan8(x)

8+

tan10(x)

10+c .

2

∫sec2(x)dx = tan(x) + c .

3 ∫sec4(x)dx =

∫sec2(x) sec2(x)dx

=

∫(1 + tan2(x)) sec2(x)dx

u=tan(x)=

∫(1 + u2)du

= tan(x) +tan3(x)

3+ c .






4 ∫sec2p+2(x)dx =

∫sec2p(x) sec2(x)dx

=

∫(1 + tan2(x))p sec2(x)dx

u=tan(x)=

∫(1 + u2)pdu.






• If m = 0, ∫tan(x)dx = ln | sec(x)|+ c .

∫tan2(x)dx =

∫(sec2(x)− 1)dx = tan(x)− x + c .






For n ≥ 3,

Ln =

∫tann(x)dx =

∫tann−2(x) tan2(x)dx

=

∫tann−2(x) sec2(x)dx − Ln−2

=tann−1(x)

n − 1− Ln−2.






Example 3 :

L5 =

∫tan5(x)dx .

L5 =tan4(x)

4− L3 =

tan4(x)

4− tan2(x)

2+ L1

=tan4(x)

4− tan2(x)

2+ ln | sec(x)|+ c .






• If m = 2q + 1 and n = 2p + 1, we set u = sec(x), thendu = sec(x) tan(x).

Lm,n =

∫sec2q+1(x) tan2p+1(x)dx =

∫u2q(1− u2)pdu.

Example 4 :∫sec5(x) tan3(x)dx =

sec5(x)

5− sec7(x)

7+ c .






• If m = 2q + 1 and n = 2p. The result is obtained by integrationby parts and induction.

Example 5 :

1

∫sec(x)dx = ln | sec(x) + tan(x)|+ c .

2

∫sec3(x)dx =

∫sec(x) sec2(x)dx . Using integration by

parts, with u(x) = sec(x) and v ′(x) = sec2(x), we have∫sec3(x)dx =

∫sec(x) sec2(x)dx

= sec(x) tan(x)−∫

sec(x) tan2(x)dx

= sec(x) tan(x)−∫

sec3(x)dx +

∫sec(x)dx .Techniques of Integration





Therefore∫sec3(x)dx =

1

2sec(x) tan(x) +

1

2ln | sec(x) + tan(x)|+ c.

∫sec5(x)dx =

∫sec3(x) sec2(x)dx

= sec3(x) tan(x)− 3

∫sec3(x) tan2(x)dx

= sec3(x) tan(x)− 3

∫sec5(x)dx + 3

∫sec3(x)dx .






Then ∫sec5(x)dx =

1

4sec3(x) tan(x) +

3

8sec(x) tan(x)

+3

8ln | sec(x) + tan(x)|+ c.






Integrals of Type

∫sin(ax) sin(bx)dx,

∫sin(ax) cos(bx)dx and∫

cos(ax) cos(bx)dx

In use the formulas (1), we have

∫sin(ax) sin(bx)dx =

1

2

∫cos((a− b)x)− cos((a + b)x)dx ,

∫sin(ax) cos(bx)dx =

1

2

∫sin((a + b)x) + sin((a− b)x)dx ,

∫cos(ax) cos(bx)dx =

1

2

∫cos((a + b)x) + cos((a− b)x)dx ,






Examples 3 :• ∫

sin(5x) sin(3x)dx =1

2

∫cos(2x)− cos(8x)dx

=1

2

(sin(2x)

2− sin(8x)

8

)+ c

=sin(2x)

4− sin(8x)

16+ c .

• ∫sin(4x) cos(3x)dx =

1

2

∫sin(7x) + sin(x)dx

=1

2

(− cos(x)− cos(7x)

7

)+ c

= −cos(x)

2− cos(7x)

14+ c .






Trigonometric Substitutions

Let a > 0.• If in an integral we have

√a2 − x2, (−a ≤ x ≤ a), we set

x = a sin(θ), θ ∈ [−π2

;π

2].

Thendx = a cos(θ)dθ√a2 − x2 = a cos(θ).






Also, if we putx = a cos(θ), θ ∈ [0;π].

Thendx = −a sin(θ)dθ√a2 − x2 = a sin(θ).






• If in an integral we have√a2 + x2, (x ∈ R), we set

x = a tan(θ), θ ∈ (−π2

;π

2).

Thendx = a sec2(θ)dθ√a2 + x2 = a sec(θ).






• If in an integral we have√x2 − a2, (x > a), we set

x = a sec(θ), θ ∈ [0;π

2).

Thendx = a sec(θ) tan(θ)dθ√

x2 − a2 = a tan(θ).






Examples 4 :

1 ∫dx

x√

4− x2x=2 sin(θ)

=

∫dθ

2 sin(θ)

= −1

2ln(csc(θ) + cot(θ)) + c

= −1

2ln(

2 +√

4− x2

x) + c,

We know that∫dx

x√

4− x2= −1

2sech−1(

x

2) + c = −1

2ln(

2 +√4− x2

x) + c.






2 ∫dx

x2√x2 + 9

x=3 tan(θ)=

∫sec(θ)

9 tan2(θ)dθ

=1

9

∫cos(θ)

sin2(θ)dθ

= −csc(θ)

9+ c

= −√

9 + x2

9X+ c ,






3 ∫dx

x2√x2 − 25

x=5 sec(θ)=

∫dθ

25 sec(θ)

=1

25

∫cos(θ)dθ

=1

25sin(θ) + c

=

√x2 − 25

25x+ c .






Integrals of Rational Functions

In this section, we study the integrals of the form∫F (x)dx ,

where

F (x) =P(x)

Q(x), P, Q ∈ R[X ].

We shall describe a method for computing this type of integrals.The method is to decompose a given rational function into a sumof simpler fractions (called partial fractions) which is easier tointegrate.






Definition

The irreducible polynomials in R[X ] areThe irreducible linear polynomials are the polynomials of the form

R(x) = x − α, α ∈ R.

The irreducible quadratic polynomials are the polynomials of theform

R(x) = ax2 + bx + c , a, b, c ∈ R : b2 − 4ac < 0.






If the rational functionsP

Qin given which the degree P is not less

than that of Q, we can expressP

Qas the sum of a polynomial and

a proper rational function, that is

P

Q= R +

S

Q,

where R and S are polynomials and the degree S is less than thatof Q. For example,

x4 + 5x2 + 3

x3 − x= x +

6x2 + 3

x3 − x.






Example 1 :

Evaluation of the following integral

∫x4 + 5x2 + 3

x3 − xdx .

A general theorem in algebra states that there is three realnumbers a, b, c such that

6x2 + 3

x(x − 1)(x + 1)=

a

x+

b

x − 1+

c

x + 1.

Then

6x2 + 3

x(x − 1)(x + 1)=

a(x − 1)(x + 1) + bx(x + 1) + cx(x − 1)

x(x − 1)(x + 1)

=a(x2 − 1) + b(x2 + x) + c(x2 − x)

x(x − 1)(x + 1)

=ax2 − a + bx2 + bx + cx2 − cx

x(x − 1)(x + 1)

=(a + b + c)x2 + (b − c)x − a

x(x − 1)(x + 1).






By identification we have a = −3, b = 92 = c. So

6x2 + 3

x(x − 2)(x + 2)= −3.

1

x+

9

2.

1

x − 1+

9

2.

1

x + 1.

Then∫x4 + 5x2 + 3

x3 − xdx =

∫ (x − 3

x+

9

2(x − 1)+

9

2(x + 1)

)dx

=x2

2− 3 ln |x |+ 9

2ln |x − 1|+ 9

2ln |x + 1|+ c .






Example 2 :Evaluation of the following integral∫

x + 16

x2 + 2x − 8dx .

We have x2 + 2x − 8 = (x − 2)(x + 4) and

x + 16

(x − 2)(x + 4)=

3

x − 2− 2

x + 4.

Then ∫3

x − 2− 2

x + 4dx = 3 ln |x − 2| − 2 ln |x + 2|+ c .






Case of repeated linear factors of denominator:

Example 3 : Evaluation of the following integral∫2x2 − 25x − 33

(x + 1)2(x − 5)dx .

A general theorem in algebra states that there is three realnumbers a, b, c such that

2x2 − 25x − 33

(x + 1)2(x − 5)=

a

x + 1+

b

(x + 1)2+

c

x − 5.

We find

2x2 − 25x − 33

(x + 1)2(x − 5)=

5

x + 1+

1

(x + 1)2− 3

x − 5.

Then

I = 5 ln |x + 1| − 1

x + 1− 3 ln |x − 5|+ c .






Case of irreducible quadratic factor of denominator:

Example 4 :

Evaluate

∫x2 + 3x + 1

x4 + 5x2 + 4dx .

We havex4 + 5x2 + 4 = (x2 + 1)(x2 + 4).

The polynomials x2 + 1 and x2 + 4 are irreducible.A general theorem in algebra states that there is three realnumbers a, b, c, d such that

x2 + 3x + 1

(x2 + 1)(x2 + 4)=

ax + b

x2 + 1+

cx + d

x2 + 4.






We have

x2 + 3x + 1

(x2 + 1)(x2 + 4)=

x

x2 + 1+−x + 1

x2 + 4

and∫x2 + 3x + 1

x4 + 5x2 + 4dx =

∫x

x2 + 1+−x + 1

x2 + 4dx

=

∫1

2

2x

x2 + 1− 1

2

2x

x2 + 4+

1

x2 + 22dx

=1

2ln(x2 + 1)− 12 ln(x2 + 4) +

1

2tan−1(

x

2) + c .






Examples 5 :

1 ∫dx

x2 + 9=

1

3tan−1(

x

3) + c .

2

1

2

∫2x

x2 + 9dx =

1

2ln(x2 + 9) + c .

3 ∫x2

x2 + 9dx =

∫(x2 + 9)− 9

x2 + 9dx =

∫1− 9

x2 + 32dx

= x − 3 tan−1(x/3) + c .






4 ∫x + 1

x2 + 2x − 3dx =

1

2

∫dx

x − 1+

1

2

∫dx

x + 3

=1

2ln |x − 1|+ 1

2ln |x + 3|+ c .

5 ∫x2 + 3x − 1

x3 + x2 − 2xdx =

1

2

∫dx

x+

∫dx

x − 1− 1

2

∫dx

x + 2

=1

2ln |x |+ ln |x − 1| − 1

2ln |x + 2|+ c .






6 ∫x2 + 2x + 3

(x − 1)(x + 1)2dx =

3

2

∫dx

x − 1− 1

2

∫dx

x + 1−∫

dx

(x + 1)2

=3

2ln |x − 1| − 1

2ln |x + 1|+ 1

x + 1+ c .

7 ∫2x + 5

x2 + x + 1dx =

∫2x + 1

x2 + x + 1dx +

∫4

x2 + x + 1dx

= ln(x2 + x + 1) + 2

∫dx

(x + 12)2 + 3

4

= ln(x2 + x + 1) +8√3

tan−1(2x + 1√

3) + c.






8 ∫2x2 − x + 2

x(x2 + 1)2dx =

∫2dx

x−∫

2xdx

x2 + 1−∫

dx

(x2 + 1)2

= ln x2 − ln(x2 + 1) +1

2tan−1(x) +

x

2(x2 + 1)+ c .






Integrals Involving Quadratic Expressions

Example 1 : Evaluation of the following integral∫2x − 5

x2 − 6x + 13dx .

Using the method of completing square, we havex2 − 6x + 13 = (x − 3)2 + 4.






Then∫2x − 5

x2 − 6x + 13dx =

∫2x − 5

(x − 3)2 + 4dx

2u=x−3=

1

2

∫2(2u + 3)− 5

u2 + 1du

=1

2

∫4u

u2 + 1+

1

u2 + 1du

= ln(u2 + 1) +1

2tan−1(u) + c

= ln((x − 3)2 + 4) +1

2tan−1

(x − 3

2

)+ c .






Example 2 : Complete the square in the following cases:

x2 + 2x + 5, x2 + x + 1, x2 − x + 1, x2 + 5x .

x2 + 2x + 5 = (x2 + 2x + 1)− 1 + 5 = (x + 1)2 + 4.

x2 + x + 1 = (x2 + 2.1

2.x + (

1

2)2)− (

1

2)2 + 1,= (x +

1

2)2 +

3

4.

x2 − x + 1 = (x2 − 2.1

2.x + (−1

2)2)− (−1

2)2 + 1 = (x − 1

2)2 +

3

4.

x2 + 5x = (x2 + 2.5

2.x + (

5

2)2)− (

5

2)2 = (x +

5

2)2 − 25

4.






Example 3 :


∫dx

x2 − 2x + 2.

∫dx

x2 − 2x + 2u=x−1

=

∫du

u2 + 1= tan−1(u)+c = tan−1(x−1)+c .






Example 4 :


∫1√

7 + 6x − x2dx .

Using the method of completing square, we get∫1√

7 + 6x − x2dx =

∫1√

16− (x − 3)2dx

u=x−3=

∫1√

42 − u2du

= sin−1(u

4) + c = sin−1(

x − 3

4) + c .






Example 5 :


∫1

(x2 + 6x + 13)32

dx .

Using the method of completing square , we get∫1

(x2 + 6x + 13)32

dx =

∫1

((x + 3)2 + 4)32

dx

u=x+3=

∫dx

((x + 3)2 + 4)32

=

∫du

(u2 + 22)32

.

Put u = 2 tan(θ), θ ∈ (−π2 ; π2 ), then






∫1

(x2 + 6x + 13)32

dx =

∫2 sec2(θ)

8 sec3(θ)dθ =

1

4

∫dθ

sec(θ)

=1

4

∫cos(θ)dθ =

1

4sin(θ) + c.

We have√u2 + 22 = 2 sec(θ) and sin(θ) = u√

u2+22. Therefore∫

1

(x2 + 6x + 13)32

dx =1

4

u√u2 + 4

+ c =1

4

x + 3√(x + 3)2 + 4

+ c .






Example 6 :


∫ √x2 + 10x dx , x > 0.

Using the method of completing square, we get

∫ √x2 + 10xdx =

∫ √(x + 5)2 − 52dx

u=x+5=

∫ √u2 − 52du

u=5 sec(θ)=

∫5 tan(θ)5 sec(θ) tan(θ)dθ,

where θ ∈ [0; π2 ). Then






As

sec(θ) =u

5and tan(θ) =

√u2 − 25

5

then∫ √x2 + 10xdx =

1

2u√u2 − 25− 25

2ln∣∣∣u5

+

√u2 − 25

5

∣∣∣+ c .

Using the fact u = x + 5, we get

I =1

2(x + 5)

√x2 + 10x − 25

2ln∣∣∣x + 5

5+

√x2 + 10x

5

∣∣∣+ c .






Example 7 :

∫x + 3√4− x2

dx =

∫x√

4− x2+

3√4− x2

dx

=

∫−(

1

2)(−2x)(4− x2)−1/2 + 3

1√22 − x2

dx

= −(4− x2)1/2 + 3 sin−1(x/2) + c .






Integrals of Rational Functions of sin(x) and cos(x)

In this section, we treat the integrals of the following form∫P(

cos(x); sin(x))

Q(

cos(x); sin(x))dx

where P(X ;Y ) and Q(X ;Y ) are two polynomial functions inX , Y .






Method: Generally we use the following substitution

u = tan(x

2)⇒ du =

1

2sec2(

x

2)dx =

1

2

(1+tan2(

x

2))dx ⇒ dx =

2du

1 + u2.

We have

sin(x) =2u

1 + u2, cos(x) =

1− u2

1 + u2.






Indeed:

sin(x) = sin(2.x

2) = 2 sin(

x

2) cos(

x

2)

=tan( x2 )

sec2( x2 )=

2u

1 + u2.

and

cos(x) = cos2(x

2)− sin2(

x

2) = cos2(

x

2)(1− tan2(

x

2))

=1− tan2( x2 )

1 + tan2( x2 )=

1− u2

1 + u2.






Example 1 : Evaluation of the following integral

∫dx

2 + sin(x).

We set u = tan(x

2), then∫

dx

2 + sin(x)=

∫1

2 +2u

1 + u2

.2

1 + u2du

=

∫du

u2 + u + 1

=

∫du

(u + 12)2 + 3

4

=2√3

tan−1(2 tan( x2 ) + 1

√3

)+ c.






Miscellaneous Substitutions

Example 1 :

Evaluation of the integral

∫ √x

1 + 3√xdx .

We set x = u6,∫ √x

1 + 3√xdx =

∫u3

1 + u26u5du = 6

∫u8

1 + u2du.

Therefore∫ √x

1 + 3√xdx = 6

∫u6 − u4 + u2 − 1 +

1

1 + u2du

=6

7x

76 − 6

5x

56 + 2x

12 − 6x

16 + 6 tan−1(x

16 ) + c .






Example 2 :

Evaluation of the integral

∫ 4

0

2x + 3√1 + 2x

dx .

First compute the indefinite integral

∫2x + 3√1 + 2x

dx . For this set

u = 1 + 2x then∫2x + 3√1 + 2x

dxx= u−1

2=1

2

∫ 2(u − 1

2) + 3

u12

du

=1

2

∫u + 2

u12

du

=1

3(1 + 2x)

32 + (1 + 2x)

12 + c .

Therefore∫ 4

0

2x + 3√1 + 2x

dx =[1

3(1+2x)

32 +(1+2x)

12

]40

= (9+3)−(1

3+1) =

32

3.






Improper Integrals

Definition

Let f be a piecewise continuous function on the interval [a, b[,where a ∈ R, b ∈ R ∪ {+∞}.We say that the integral of f on the interval [a, b[ is convergent if

the function F (x) =

∫ x

af (t)dt defined on [a, b[ has a finite limit

when x tends to b (x < b). This limit is called the improper

integral of f on [a, b[ and will be denoted by:

∫ b

af (x)dx .






Example 1 :

1

∫ 1

0

esin−1(x)

√1− x2

dxu=sin−1(x)

=

∫ π2

0eudu = e

π2 − 1. This integral is

convergent.

2

∫ +∞

0xne−xdx = n!. This integral is convergent.

3

∫ +∞

0

x

1 + x2dx =

1

2

[ln(1 + x2)

]+∞0

= +∞. This integral is

divergent.






Definition

Let f a piecewise continuous function on the interval ]a, b], wherea ∈ R ∪ {−∞}, b ∈ R.We say that the integral of f on the interval ]a, b] is convergent if

the function G (x) =

∫ b

xf (t)dt defined on ]a, b] has a finite limit

when x tends to a (x > a). This limit is called the improper

integral of f on ]a, b] and will be denoted by:

∫ b

af (x)dx .






Example 2 :

1

∫ 1

0ln(x)dx = [x ln(x)− x ]10 = −1. This integral is

convergent.

2

∫ 0

−∞

1

(x − 3)2dx =

[−1

x − 3

]0−∞

=1

3. This integral is

convergent.

3

∫ 1

0

dx

x ln(x)= [ln(ln(x))]10 = −∞. This integral is divergent.






Definition

Let f be a piecewise continuous function on the interval ]a, b[,where a ∈ R ∪ {−∞}, b ∈ R ∪ {+∞}.We say that the integral of f on the interval ]a, b[ is convergent ifthe integral of f is convergent on ]a, c] and on [c , b[ for any c in]a, b[.






Example 3 :

1

∫ +∞

−∞

etan−1(x)

1 + x2dx

u=tan−1(x)=

∫ π2

−π2

eudu = 2 sinh(π

2). This

integral is convergent.

2

∫ 1

0

ln x

(1− x)32

dxx=1−t2

= 2

∫ 1

0

ln(1− t2)

t2dt = 2− 2 ln 2. This

integral is convergent.

3

∫ +∞

−∞exdx = [ex ]+∞−∞ = +∞. This integral is divergent.






Definition

Let f be a piecewise continuous function on an interval I . Thefunction is called integrable on I (or the integral is absolutelyconvergent) if the integral of |f | on the interval I is convergent.






Examples 6 :

1

∫ +∞

0

dx

1 + x= +∞. This integral is divergent.

2

∫ +∞

0

dx

1 + x2=π

2. This integral is divergent.

3

∫ 1

0

dx√x

= 2. This integral is divergent.

4 Let α ∈ R and a ∈ R∗+. The integral∫ +∞a

dxxα is convergent if

and only if α > 1 and the integral

∫ a

0

dx

xαis convergent if and

only if α < 1.






5 The integral

∫ +∞

0sin(x) is divergent since∫ x

0sin(t)dt = 1− cos(x) dont have a limit when x tends to

+∞.

6 The integral

∫ 1

0

sin t

tis convergent since the function sin t

t can

be considered continuous on the interval [0, 1].

7 The integral

∫ 1

0sin(

1

t) is convergent since the function sin(1t )

is continuous on the interval ]0, 1] and bounded.






8

∫ 1

0

x ln x

(1− x2)3/2dx , by integration by parts

∫ 1

0

x ln x

(1− x2)3/2dx =

x2=1−t2=

1

2

∫ 1

0

ln(1− t2)

t2dt = − ln 2.

9

∫ π2

0sin 2x ln(tan x)dx , by integration by parts

∫ π2

0sin 2x ln(tan x)dx = sin2 ln(tan x) + ln(cos x)

]π20

= 0.






10

∫ π2

0cos x ln(tan x)dx =∫ π

4

0cos x ln(tan x)dx −

∫ π4

0sin x ln(tan x)dx .

By integration by parts, we have∫ π4

0cos x ln(tan x)dx = −

∫ π4

0

dx

cos x= − ln(1 +

√2).

−∫ π

4

0sin x ln(tan x)dx = [(cos x − 1) ln(sin x) + ln(1 + cos x)]

π40

= −√

2

4ln 2− ln 2 + ln(1 +

√2).

Then

∫ π2

0cos x ln(tan x)dx = −

√2

4ln 2− ln 2.






11

∫ +∞

1

x4 + 1

x3(x + 1)(1 + x2)dx =∫ +∞

1(

1

x− 1

x2+

1

x3− 1

x + 1+ 2

x + 1

1 + x2)dx =

π

2− 1

2.

12

∫ +∞

1

dx

x 4√

1 + x2t4=1+x2

=

∫ +∞

214

2t2dt

(t4 − 1)=

∫ +∞

214

(1

2(t − 1)−

1

2(t + 1)+

1

1 + t2)dt =

1

2ln(

214 + 1

214 − 1

) +π

2− tan−1(2

14 ).






13

∫ 0

−1

dx√4− x2

=[sin−1(

x

2)]0−1

=π

2.

14

∫ 0

−2

dx3√x + 1

=

∫ −1−2

dx3√x + 1

+

∫ 0

−1

dx3√x + 1

=

3

2

[(x + 1)

23

]0−2

= 0.

15

∫ 1

−3

dx

x2=

∫ 0

−3

dx

x2+

∫ 1

0

dx

x2= +∞.


techniques of integration file4 integrals of rational functions 5 integrals involving quadratic...

Documents