Integration by Substitution - James Madison University

Integration by Substitution

Department of Mathematics and Statistics

January 11, 2012

Calculus II (James Madison University) Math 236 January 11, 2012 1 / 8

An Overview of Integration Techniques

We will study several methods for finding antiderivatives.With each method, the goal is to transform a given integral into one ofthe “basic forms”.The methods we will discuss are:

Integration by Substitution

Integration by Parts

Partial Fractions

Trigonometric Integrals

Trigonometric Substitution

The first two methods, above, are the most general and useful. The finalthree are more specialized.As we study these methods, keep in mind which method might work on aparticular integral and the reason(s) the method might apply to aparticular integral.

Calculus II (James Madison University) Math 236 January 11, 2012 2 / 8

An Overview of Integration Techniques

We will study several methods for finding antiderivatives.With each method, the goal is to transform a given integral into one ofthe “basic forms”.The methods we will discuss are:

Integration by Substitution

Integration by Parts

Partial Fractions

Trigonometric Integrals

Trigonometric Substitution

The first two methods, above, are the most general and useful. The finalthree are more specialized.As we study these methods, keep in mind which method might work on aparticular integral and the reason(s) the method might apply to aparticular integral.

Calculus II (James Madison University) Math 236 January 11, 2012 2 / 8

An Overview of Integration Techniques

We will study several methods for finding antiderivatives.With each method, the goal is to transform a given integral into one ofthe “basic forms”.The methods we will discuss are:

Integration by Substitution

Integration by Parts

Partial Fractions

Trigonometric Integrals

Trigonometric Substitution

The first two methods, above, are the most general and useful. The finalthree are more specialized.As we study these methods, keep in mind which method might work on aparticular integral and the reason(s) the method might apply to aparticular integral.

Calculus II (James Madison University) Math 236 January 11, 2012 2 / 8

An Overview of Integration Techniques

We will study several methods for finding antiderivatives.With each method, the goal is to transform a given integral into one ofthe “basic forms”.The methods we will discuss are:

Integration by Substitution

Integration by Parts

Partial Fractions

Trigonometric Integrals

Trigonometric Substitution

The first two methods, above, are the most general and useful. The finalthree are more specialized.As we study these methods, keep in mind which method might work on aparticular integral and the reason(s) the method might apply to aparticular integral.

Calculus II (James Madison University) Math 236 January 11, 2012 2 / 8

An Overview of Integration Techniques

We will study several methods for finding antiderivatives.With each method, the goal is to transform a given integral into one ofthe “basic forms”.The methods we will discuss are:

Integration by Substitution

Integration by Parts

Partial Fractions

Trigonometric Integrals

Trigonometric Substitution

The first two methods, above, are the most general and useful. The finalthree are more specialized.As we study these methods, keep in mind which method might work on aparticular integral and the reason(s) the method might apply to aparticular integral.

Calculus II (James Madison University) Math 236 January 11, 2012 2 / 8

An Overview of Integration Techniques

We will study several methods for finding antiderivatives.With each method, the goal is to transform a given integral into one ofthe “basic forms”.The methods we will discuss are:

Integration by Substitution

Integration by Parts

Partial Fractions

Trigonometric Integrals

Trigonometric Substitution

The first two methods, above, are the most general and useful. The finalthree are more specialized.As we study these methods, keep in mind which method might work on aparticular integral and the reason(s) the method might apply to aparticular integral.

Calculus II (James Madison University) Math 236 January 11, 2012 2 / 8

An Overview of Integration Techniques

We will study several methods for finding antiderivatives.With each method, the goal is to transform a given integral into one ofthe “basic forms”.The methods we will discuss are:

Integration by Substitution

Integration by Parts

Partial Fractions

Trigonometric Integrals

Trigonometric Substitution

The first two methods, above, are the most general and useful. The finalthree are more specialized.As we study these methods, keep in mind which method might work on aparticular integral and the reason(s) the method might apply to aparticular integral.

Calculus II (James Madison University) Math 236 January 11, 2012 2 / 8

An Overview of Integration Techniques

We will study several methods for finding antiderivatives.With each method, the goal is to transform a given integral into one ofthe “basic forms”.The methods we will discuss are:

Integration by Substitution

Integration by Parts

Partial Fractions

Trigonometric Integrals

Trigonometric Substitution

The first two methods, above, are the most general and useful. The finalthree are more specialized.As we study these methods, keep in mind which method might work on aparticular integral and the reason(s) the method might apply to aparticular integral.

Calculus II (James Madison University) Math 236 January 11, 2012 2 / 8

An Overview of Integration Techniques

We will study several methods for finding antiderivatives.With each method, the goal is to transform a given integral into one ofthe “basic forms”.The methods we will discuss are:

Integration by Substitution

Integration by Parts

Partial Fractions

Trigonometric Integrals

Trigonometric Substitution

The first two methods, above, are the most general and useful. The finalthree are more specialized.As we study these methods, keep in mind which method might work on aparticular integral and the reason(s) the method might apply to aparticular integral.

Calculus II (James Madison University) Math 236 January 11, 2012 2 / 8

An Overview of Integration Techniques

We will study several methods for finding antiderivatives.With each method, the goal is to transform a given integral into one ofthe “basic forms”.The methods we will discuss are:

Integration by Substitution

Integration by Parts

Partial Fractions

Trigonometric Integrals

Trigonometric Substitution

The first two methods, above, are the most general and useful. The finalthree are more specialized.As we study these methods, keep in mind which method might work on aparticular integral and the reason(s) the method might apply to aparticular integral.

Calculus II (James Madison University) Math 236 January 11, 2012 2 / 8

An Overview of Integration Techniques

We will study several methods for finding antiderivatives.With each method, the goal is to transform a given integral into one ofthe “basic forms”.The methods we will discuss are:

Integration by Substitution

Integration by Parts

Partial Fractions

Trigonometric Integrals

Trigonometric Substitution

The first two methods, above, are the most general and useful. The finalthree are more specialized.As we study these methods, keep in mind which method might work on aparticular integral and the reason(s) the method might apply to aparticular integral.

Calculus II (James Madison University) Math 236 January 11, 2012 2 / 8

Undoing the Chain Rule

Theorem (The Chain Rule)

Given functions f and u, for all values of x at which u is differentiable at xand f is differentiable at u(x), we have:

ddx (f (u(x))) = f ′(u(x))u′(x).

Therefore, we have the corresponding antiderivative formula:

Integration by Substitution∫

f ′(u(x))u′(x) dx = f (u(x)) + C .

The key is to identify a factor of the integrand which is the derivative ofan “inside” function in a composition.

Calculus II (James Madison University) Math 236 January 11, 2012 3 / 8

Undoing the Chain Rule

Theorem (The Chain Rule)

Given functions f and u, for all values of x at which u is differentiable at xand f is differentiable at u(x), we have:

ddx (f (u(x))) = f ′(u(x))u′(x).

Therefore, we have the corresponding antiderivative formula:

Integration by Substitution∫

f ′(u(x))u′(x) dx = f (u(x)) + C .

The key is to identify a factor of the integrand which is the derivative ofan “inside” function in a composition.

Calculus II (James Madison University) Math 236 January 11, 2012 3 / 8

Undoing the Chain Rule

Theorem (The Chain Rule)

Given functions f and u, for all values of x at which u is differentiable at xand f is differentiable at u(x), we have:

ddx (f (u(x))) =

f ′(u(x))u′(x).

Therefore, we have the corresponding antiderivative formula:

Integration by Substitution∫

f ′(u(x))u′(x) dx = f (u(x)) + C .

The key is to identify a factor of the integrand which is the derivative ofan “inside” function in a composition.

Calculus II (James Madison University) Math 236 January 11, 2012 3 / 8

Undoing the Chain Rule

Theorem (The Chain Rule)

Given functions f and u, for all values of x at which u is differentiable at xand f is differentiable at u(x), we have:

ddx (f (u(x))) = f ′(u(x))u′(x).

Therefore, we have the corresponding antiderivative formula:

Integration by Substitution∫

f ′(u(x))u′(x) dx = f (u(x)) + C .

The key is to identify a factor of the integrand which is the derivative ofan “inside” function in a composition.

Calculus II (James Madison University) Math 236 January 11, 2012 3 / 8

Undoing the Chain Rule

Theorem (The Chain Rule)

Given functions f and u, for all values of x at which u is differentiable at xand f is differentiable at u(x), we have:

ddx (f (u(x))) = f ′(u(x))u′(x).

Therefore, we have the corresponding antiderivative formula:

Integration by Substitution∫

f ′(u(x))u′(x) dx = f (u(x)) + C .

The key is to identify a factor of the integrand which is the derivative ofan “inside” function in a composition.

Calculus II (James Madison University) Math 236 January 11, 2012 3 / 8

Undoing the Chain Rule

Theorem (The Chain Rule)

Given functions f and u, for all values of x at which u is differentiable at xand f is differentiable at u(x), we have:

ddx (f (u(x))) = f ′(u(x))u′(x).

Therefore, we have the corresponding antiderivative formula:

Integration by Substitution∫

f ′(u(x))u′(x) dx = f (u(x)) + C .

The key is to identify a factor of the integrand which is the derivative ofan “inside” function in a composition.

Calculus II (James Madison University) Math 236 January 11, 2012 3 / 8

Undoing the Chain Rule

Theorem (The Chain Rule)

Given functions f and u, for all values of x at which u is differentiable at xand f is differentiable at u(x), we have:

ddx (f (u(x))) = f ′(u(x))u′(x).

Therefore, we have the corresponding antiderivative formula:

Integration by Substitution∫

f ′(u(x))u′(x) dx = f (u(x)) + C .

The key is to identify a factor of the integrand which is the derivative ofan “inside” function in a composition.

Calculus II (James Madison University) Math 236 January 11, 2012 3 / 8

The Differential

Definition

If u(x) is a differentiable function of x , then the differential du is equal todu = u′(x) dx .

Calculus II (James Madison University) Math 236 January 11, 2012 4 / 8

The Differential

Definition

If u(x) is a differentiable function of x , then the differential du is equal todu = u′(x) dx .

Calculus II (James Madison University) Math 236 January 11, 2012 4 / 8

Main Ideas

Within the integrand locate a factor which is the derivative of an“inside” function. The “inside” function is your “u”.

Find dudx .

Write this in differential form, du = u′(x) dx .

Replace all occurrences of the original variable, x , with the newvariable u.

The new integral should only contain one variable.The new integral should be a basic form, whose antiderivative yourecognize.

Find the antiderivative.

If you are working with an indefinite integral, you must rewrite theanswer in terms of the original variable.

Don’t forget to include a constant of integration in every indefiniteintegral.

Calculus II (James Madison University) Math 236 January 11, 2012 5 / 8

Main Ideas

Within the integrand locate a factor which is the derivative of an“inside” function. The “inside” function is your “u”.

Find dudx .

Write this in differential form, du = u′(x) dx .

Replace all occurrences of the original variable, x , with the newvariable u.

The new integral should only contain one variable.The new integral should be a basic form, whose antiderivative yourecognize.

Find the antiderivative.

If you are working with an indefinite integral, you must rewrite theanswer in terms of the original variable.

Don’t forget to include a constant of integration in every indefiniteintegral.

Calculus II (James Madison University) Math 236 January 11, 2012 5 / 8

Main Ideas

Within the integrand locate a factor which is the derivative of an“inside” function. The “inside” function is your “u”.

Find dudx .

Write this in differential form, du = u′(x) dx .

Replace all occurrences of the original variable, x , with the newvariable u.

The new integral should only contain one variable.The new integral should be a basic form, whose antiderivative yourecognize.

Find the antiderivative.

If you are working with an indefinite integral, you must rewrite theanswer in terms of the original variable.

Don’t forget to include a constant of integration in every indefiniteintegral.

Calculus II (James Madison University) Math 236 January 11, 2012 5 / 8

Main Ideas

Within the integrand locate a factor which is the derivative of an“inside” function. The “inside” function is your “u”.

Find dudx .

Write this in differential form, du = u′(x) dx .

Replace all occurrences of the original variable, x , with the newvariable u.

The new integral should only contain one variable.The new integral should be a basic form, whose antiderivative yourecognize.

Find the antiderivative.

If you are working with an indefinite integral, you must rewrite theanswer in terms of the original variable.

Don’t forget to include a constant of integration in every indefiniteintegral.

Calculus II (James Madison University) Math 236 January 11, 2012 5 / 8

Main Ideas

Within the integrand locate a factor which is the derivative of an“inside” function. The “inside” function is your “u”.

Find dudx .

Write this in differential form, du = u′(x) dx .

Replace all occurrences of the original variable, x , with the newvariable u.

The new integral should only contain one variable.The new integral should be a basic form, whose antiderivative yourecognize.

Find the antiderivative.

If you are working with an indefinite integral, you must rewrite theanswer in terms of the original variable.

Don’t forget to include a constant of integration in every indefiniteintegral.

Calculus II (James Madison University) Math 236 January 11, 2012 5 / 8

Main Ideas

Within the integrand locate a factor which is the derivative of an“inside” function. The “inside” function is your “u”.

Find dudx .

Write this in differential form, du = u′(x) dx .

Replace all occurrences of the original variable, x , with the newvariable u.

The new integral should only contain one variable.The new integral should be a basic form, whose antiderivative yourecognize.

Find the antiderivative.

If you are working with an indefinite integral, you must rewrite theanswer in terms of the original variable.

Don’t forget to include a constant of integration in every indefiniteintegral.

Calculus II (James Madison University) Math 236 January 11, 2012 5 / 8

Main Ideas

Within the integrand locate a factor which is the derivative of an“inside” function. The “inside” function is your “u”.

Find dudx .

Write this in differential form, du = u′(x) dx .

Replace all occurrences of the original variable, x , with the newvariable u.

The new integral should only contain one variable.The new integral should be a basic form, whose antiderivative yourecognize.

Find the antiderivative.

If you are working with an indefinite integral, you must rewrite theanswer in terms of the original variable.

Don’t forget to include a constant of integration in every indefiniteintegral.

Calculus II (James Madison University) Math 236 January 11, 2012 5 / 8

Main Ideas

Within the integrand locate a factor which is the derivative of an“inside” function. The “inside” function is your “u”.

Find dudx .

Write this in differential form, du = u′(x) dx .

Replace all occurrences of the original variable, x , with the newvariable u.

The new integral should only contain one variable.The new integral should be a basic form, whose antiderivative yourecognize.

Find the antiderivative.

If you are working with an indefinite integral, you must rewrite theanswer in terms of the original variable.

Don’t forget to include a constant of integration in every indefiniteintegral.

Calculus II (James Madison University) Math 236 January 11, 2012 5 / 8

Main Ideas

Within the integrand locate a factor which is the derivative of an“inside” function. The “inside” function is your “u”.

Find dudx .

Write this in differential form, du = u′(x) dx .

Replace all occurrences of the original variable, x , with the newvariable u.

The new integral should only contain one variable.The new integral should be a basic form, whose antiderivative yourecognize.

Find the antiderivative.

If you are working with an indefinite integral, you must rewrite theanswer in terms of the original variable.

Don’t forget to include a constant of integration in every indefiniteintegral.

Calculus II (James Madison University) Math 236 January 11, 2012 5 / 8

Main Ideas

Within the integrand locate a factor which is the derivative of an“inside” function. The “inside” function is your “u”.

Find dudx .

Write this in differential form, du = u′(x) dx .

Replace all occurrences of the original variable, x , with the newvariable u.

The new integral should only contain one variable.The new integral should be a basic form, whose antiderivative yourecognize.

Find the antiderivative.

If you are working with an indefinite integral, you must rewrite theanswer in terms of the original variable.

Don’t forget to include a constant of integration in every indefiniteintegral.

Calculus II (James Madison University) Math 236 January 11, 2012 5 / 8

Definite Integrals Using Substitution

If f (x) = g(u(x)) and G (u) is an antiderivative of g(u), we have two

choices when we evaluate the definite integral

∫ b

af (x) dx :

Obvious∫ b

af (x) dx =

∫ x=b

x=ag(u) du = [G (u)]x=b

x=a = [G (u(x))]ba

= G (u(b))− G (u(a)).

Cooler (change the limits as you change the variable)

∫ b

af (x) dx =

∫ u(b)

u(a)g(u) du = [G (u)]u(b)

u(a) = G (u(b))− G (u(a)).

Calculus II (James Madison University) Math 236 January 11, 2012 6 / 8

Definite Integrals Using Substitution

If f (x) = g(u(x)) and G (u) is an antiderivative of g(u), we have two

choices when we evaluate the definite integral

∫ b

af (x) dx :

Obvious∫ b

af (x) dx =

∫ x=b

x=ag(u) du = [G (u)]x=b

x=a = [G (u(x))]ba

= G (u(b))− G (u(a)).

Cooler (change the limits as you change the variable)

∫ b

af (x) dx =

∫ u(b)

u(a)g(u) du = [G (u)]u(b)

u(a) = G (u(b))− G (u(a)).

Calculus II (James Madison University) Math 236 January 11, 2012 6 / 8

Definite Integrals Using Substitution

If f (x) = g(u(x)) and G (u) is an antiderivative of g(u), we have two

choices when we evaluate the definite integral

∫ b

af (x) dx :

Obvious∫ b

af (x) dx =

∫ x=b

x=ag(u) du = [G (u)]x=b

x=a = [G (u(x))]ba

= G (u(b))− G (u(a)).

Cooler (change the limits as you change the variable)

∫ b

af (x) dx =

∫ u(b)

u(a)g(u) du = [G (u)]u(b)

u(a) = G (u(b))− G (u(a)).

Calculus II (James Madison University) Math 236 January 11, 2012 6 / 8

Tangent and Cotangent

∫tan x dx =

∫sin xcos x dx = − ln | cos x | + C .

∫cot x dx =

∫cos xsin x dx = ln | sin x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 7 / 8

Tangent and Cotangent

∫tan x dx =

∫sin xcos x dx = − ln | cos x | + C .

∫cot x dx =

∫cos xsin x dx = ln | sin x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 7 / 8

Tangent and Cotangent

∫tan x dx =

∫sin xcos x dx =

− ln | cos x | + C .∫

cot x dx =

∫cos xsin x dx = ln | sin x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 7 / 8

Tangent and Cotangent

∫tan x dx =

∫sin xcos x dx = − ln | cos x | + C .

∫cot x dx =

∫cos xsin x dx = ln | sin x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 7 / 8

Tangent and Cotangent

∫tan x dx =

∫sin xcos x dx = − ln | cos x | + C .

∫cot x dx =

∫cos xsin x dx = ln | sin x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 7 / 8

Tangent and Cotangent

∫tan x dx =

∫sin xcos x dx = − ln | cos x | + C .

∫cot x dx =

∫cos xsin x dx =

ln | sin x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 7 / 8

Tangent and Cotangent

∫tan x dx =

∫sin xcos x dx = − ln | cos x | + C .

∫cot x dx =

∫cos xsin x dx = ln | sin x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 7 / 8

Secant and Cosecant

∫sec x dx =

∫sec x(sec x+tan x)

sec x+tan x dx = ln |sec x + tan x | + C .

∫csc x dx =

∫csc x(csc x+cot x)

csc x+cot x dx = − ln |csc x + cot x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 8 / 8

Secant and Cosecant

∫sec x dx =

∫sec x(sec x+tan x)

sec x+tan x dx = ln |sec x + tan x | + C .∫

csc x dx =

∫csc x(csc x+cot x)

csc x+cot x dx = − ln |csc x + cot x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 8 / 8

Secant and Cosecant

∫sec x dx =

∫sec x(sec x+tan x)

sec x+tan x dx =

ln |sec x + tan x | + C .∫

csc x dx =

∫csc x(csc x+cot x)

csc x+cot x dx = − ln |csc x + cot x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 8 / 8

Secant and Cosecant

∫sec x dx =

∫sec x(sec x+tan x)

sec x+tan x dx = ln |sec x + tan x | + C .

∫csc x dx =

∫csc x(csc x+cot x)

csc x+cot x dx = − ln |csc x + cot x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 8 / 8

Secant and Cosecant

∫sec x dx =

∫sec x(sec x+tan x)

sec x+tan x dx = ln |sec x + tan x | + C .∫

csc x dx =

∫csc x(csc x+cot x)

csc x+cot x dx = − ln |csc x + cot x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 8 / 8

Secant and Cosecant

∫sec x dx =

∫sec x(sec x+tan x)

sec x+tan x dx = ln |sec x + tan x | + C .∫

csc x dx =

∫csc x(csc x+cot x)

csc x+cot x dx =

− ln |csc x + cot x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 8 / 8

Secant and Cosecant

∫sec x dx =

∫sec x(sec x+tan x)

sec x+tan x dx = ln |sec x + tan x | + C .∫

csc x dx =

∫csc x(csc x+cot x)

csc x+cot x dx = − ln |csc x + cot x | + C .

Calculus II (James Madison University) Math 236 January 11, 2012 8 / 8