Integration by Substitution Department of Mathematics and Statistics January 11, 2012 Calculus II (James Madison University) Math 236 January 11, 2012 1/8
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