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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
Techniques of Integration
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
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
Techniques of Integration
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
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 .
Techniques of Integration
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
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
Techniques of Integration
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
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.
Techniques of Integration
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
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 .
Techniques of Integration
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
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 .
Techniques of Integration
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
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 .
Techniques of Integration
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
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.
Techniques of Integration
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
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),
Techniques of Integration
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
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).
Techniques of Integration
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
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)
Techniques of Integration
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
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.
Techniques of Integration
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
• 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.
Techniques of Integration
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
• 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.
Techniques of Integration
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
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.
Techniques of Integration
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
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 ,
Techniques of Integration
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
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 ,
Techniques of Integration
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
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 .
Techniques of Integration
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
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.
Techniques of Integration
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
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 .
Techniques of Integration
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
4 ∫sec2p+2(x)dx =
∫sec2p(x) sec2(x)dx
=
∫(1 + tan2(x))p sec2(x)dx
u=tan(x)=
∫(1 + u2)pdu.
Techniques of Integration
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
• If m = 0, ∫tan(x)dx = ln | sec(x)|+ c .
∫tan2(x)dx =
∫(sec2(x)− 1)dx = tan(x)− x + c .
Techniques of Integration
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
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.
Techniques of Integration
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
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 .
Techniques of Integration
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
• 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 .
Techniques of Integration
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
• 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
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
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 .
Techniques of Integration
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
Then ∫sec5(x)dx =
1
4sec3(x) tan(x) +
3
8sec(x) tan(x)
+3
8ln | sec(x) + tan(x)|+ c.
Techniques of Integration
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
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 ,
Techniques of Integration
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
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 .
Techniques of Integration
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
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).
Techniques of Integration
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
Also, if we putx = a cos(θ), θ ∈ [0;π].
Thendx = −a sin(θ)dθ√a2 − x2 = a sin(θ).
Techniques of Integration
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
• 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).
Techniques of Integration
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
• 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).
Techniques of Integration
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
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.
Techniques of Integration
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
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 ,
Techniques of Integration
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
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 .
Techniques of Integration
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
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.
Techniques of Integration
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
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.
Techniques of Integration
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
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.
Techniques of Integration
Integration by PartsTrigonometric Integrals
Trigonometric SubstitutionsIntegrals of Rational Functions
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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).
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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 .
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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 .
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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 .
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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.
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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 .
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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 .
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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 .
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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.
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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 .
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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.
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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 .
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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.
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Example 3 :
Evaluation of the following integral
∫dx
x2 − 2x + 2.
∫dx
x2 − 2x + 2u=x−1
=
∫du
u2 + 1= tan−1(u)+c = tan−1(x−1)+c .
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Example 4 :
Evaluation of the following integral
∫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 .
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Example 5 :
Evaluation of the following integral
∫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
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∫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 .
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Example 6 :
Evaluation of the following integral
∫ √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
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I = 25
∫sec(θ) tan2(θ)dθ
= 25
∫sec(θ)(sec2(θ)− 1)dθ
= 25
∫sec3(θ)− sec(θ)dθ
= 25
∫sec3(θ)dθ − 25
∫sec(θ)dθ
= 25(1
2sec(θ) tan(θ) +
1
2ln | sec(θ) + tan(θ)|
)−25 ln | sec(θ) + tan(θ)|+ c
=25
2sec(θ) tan(θ)− 25
2ln | sec(θ) + tan(θ)|+ c .
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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 .
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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 .
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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 .
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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.
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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.
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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.
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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 .
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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.
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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 .
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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.
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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 .
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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.
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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[.
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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.
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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.
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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.
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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.
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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.
Techniques of Integration
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
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.
Techniques of Integration
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
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 ).
Techniques of Integration
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
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