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Section 5.5
           Integration by Substitution

                    V63.0121.041, Calculus I

                        New York University


                       December 13, 2010



Announcements
   ”Wednesday”, December 15: Review, Movie
   Monday, December 20, 12:00pm–1:50pm: Final Exam
Announcements




        ”Wednesday”, December 15:
        Review, Movie
        Monday, December 20,
        12:00pm–1:50pm: Final
        Exam




V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   2 / 37
Resurrection Policy
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Image credit: Scott Beale / Laughing Squid
   V63.0121.041, Calculus I (NYU)          Section 5.5 Integration by Substitution   December 13, 2010   3 / 37
Objectives



         Given an integral and a
         substitution, transform the
         integral into an equivalent
         one using a substitution
         Evaluate indefinite integrals
         using the method of
         substitution.
         Evaluate definite integrals
         using the method of
         substitution.




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   4 / 37

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Lesson 25: Evaluating Definite Integrals (Section 021 handout)
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Outline



Last Time: The Fundamental Theorem(s) of Calculus


Substitution for Indefinite Integrals
   Theory
   Examples


Substitution for Definite Integrals
   Theory
   Examples




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   5 / 37
Differentiation and Integration as reverse processes


Theorem (The Fundamental Theorem of Calculus)

 1. Let f be continuous on [a, b]. Then
                                                    x
                                        d
                                                        f (t) dt = f (x)
                                        dx      a

 2. Let f be continuous on [a, b] and f = F for some other function F .
    Then
                                        b
                                            f (x) dx = F (b) − F (a).
                                    a




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   6 / 37
Techniques of antidifferentiation?



So far we know only a few rules for antidifferentiation. Some are general,
like
                [f (x) + g (x)] dx = f (x) dx + g (x) dx




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   7 / 37
Techniques of antidifferentiation?



So far we know only a few rules for antidifferentiation. Some are general,
like
                [f (x) + g (x)] dx = f (x) dx + g (x) dx

Some are pretty particular, like
                                     1
                                   √       dx = arcsec x + C .
                                  x x2 − 1




 V63.0121.041, Calculus I (NYU)    Section 5.5 Integration by Substitution   December 13, 2010   7 / 37

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Techniques of antidifferentiation?



So far we know only a few rules for antidifferentiation. Some are general,
like
                [f (x) + g (x)] dx = f (x) dx + g (x) dx

Some are pretty particular, like
                                     1
                                   √       dx = arcsec x + C .
                                  x x2 − 1
What are we supposed to do with that?




 V63.0121.041, Calculus I (NYU)    Section 5.5 Integration by Substitution   December 13, 2010   7 / 37
No straightforward system of antidifferentiation



So far we don’t have any way to find
                                                   2x
                                              √          dx
                                                  x2 + 1
or
                                                tan x dx.




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   8 / 37
No straightforward system of antidifferentiation



So far we don’t have any way to find
                                                   2x
                                              √          dx
                                                  x2 + 1
or
                                                tan x dx.

Luckily, we can be smart and use the “anti” version of one of the most
important rules of differentiation: the chain rule.




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   8 / 37
Outline



Last Time: The Fundamental Theorem(s) of Calculus


Substitution for Indefinite Integrals
   Theory
   Examples


Substitution for Definite Integrals
   Theory
   Examples




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   9 / 37

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Substitution for Indefinite Integrals



Example
Find
                                                   x
                                             √               dx.
                                                 x2    +1




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   10 / 37
Substitution for Indefinite Integrals



Example
Find
                                                   x
                                             √               dx.
                                                 x2    +1

Solution
Stare at this long enough and you notice the the integrand is the
derivative of the expression 1 + x 2 .




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   10 / 37
Say what?


Solution (More slowly, now)
Let g (x) = x 2 + 1.




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   11 / 37
Say what?


Solution (More slowly, now)
Let g (x) = x 2 + 1. Then g (x) = 2x and so

                          d                       1                            x
                                  g (x) =                  g (x) = √
                          dx                 2     g (x)                      x2   +1




 V63.0121.041, Calculus I (NYU)     Section 5.5 Integration by Substitution             December 13, 2010   11 / 37

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Say what?


Solution (More slowly, now)
Let g (x) = x 2 + 1. Then g (x) = 2x and so

                          d                         1                                x
                                   g (x) =                   g (x) = √
                          dx                   2     g (x)                       x2      +1

Thus
                               x                        d
                         √            dx =                      g (x)           dx
                             x2 + 1                     dx
                                           =       g (x) + C =                  1 + x2 + C .




 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution                 December 13, 2010   11 / 37
Leibnizian notation FTW


Solution (Same technique, new notation)
Let u = x 2 + 1.




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   12 / 37
Leibnizian notation FTW


Solution (Same technique, new notation)
                                                                            √
Let u = x 2 + 1. Then du = 2x dx and                        1 + x2 =         u.




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution         December 13, 2010   12 / 37
Leibnizian notation FTW


Solution (Same technique, new notation)
                                                                               √
Let u = x 2 + 1. Then du = 2x dx and                           1 + x2 =         u. So the integrand
becomes completely transformed into
                                                     1
                              √
                                  x dx               2 du
                                                     √
                                                                       1
                                                                       √ du
                                         =                   =
                                  x2 + 1                u             2 u




 V63.0121.041, Calculus I (NYU)      Section 5.5 Integration by Substitution        December 13, 2010   12 / 37

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Leibnizian notation FTW


Solution (Same technique, new notation)
                                                                               √
Let u = x 2 + 1. Then du = 2x dx and                           1 + x2 =         u. So the integrand
becomes completely transformed into
                                                     1
                              √
                                  x dx               2 du
                                                     √
                                                                       1
                                                                       √ du
                                         =                   =
                                  x2 + 1                u             2 u
                                                    1 −1/2
                                           =        2u     du




 V63.0121.041, Calculus I (NYU)      Section 5.5 Integration by Substitution        December 13, 2010   12 / 37
Leibnizian notation FTW


Solution (Same technique, new notation)
                                                                               √
Let u = x 2 + 1. Then du = 2x dx and                           1 + x2 =         u. So the integrand
becomes completely transformed into
                                                     1
                              √
                                  x dx               2 du
                                                     √
                                                                       1
                                                                       √ du
                                         =                   =
                                  x2 + 1                u             2 u
                                                    1 −1/2
                                           =        2u     du
                                               √
                                           =       u+C =              1 + x2 + C .




 V63.0121.041, Calculus I (NYU)      Section 5.5 Integration by Substitution        December 13, 2010   12 / 37
Useful but unsavory variation

Solution (Same technique, new notation, more idiot-proof)
                                              √
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:”
                                                        du
                                              dx =
                                                        2x




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   13 / 37
Useful but unsavory variation

Solution (Same technique, new notation, more idiot-proof)
                                              √
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:”
                                                              du
                                                    dx =
                                                              2x
So the integrand becomes completely transformed into
                                   x                       x du
                            √               dx =          √ ·
                                  x2   +1                   u 2x




 V63.0121.041, Calculus I (NYU)         Section 5.5 Integration by Substitution   December 13, 2010   13 / 37

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Useful but unsavory variation

Solution (Same technique, new notation, more idiot-proof)
                                              √
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:”
                                                              du
                                                    dx =
                                                              2x
So the integrand becomes completely transformed into
                                   x                       x du                    1
                            √               dx =          √ ·    =                 √ du
                                  x2   +1                   u 2x                  2 u




 V63.0121.041, Calculus I (NYU)         Section 5.5 Integration by Substitution      December 13, 2010   13 / 37
Useful but unsavory variation

Solution (Same technique, new notation, more idiot-proof)
                                              √
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:”
                                                              du
                                                    dx =
                                                              2x
So the integrand becomes completely transformed into
                                   x                       x du                    1
                            √               dx =          √ ·    =                 √ du
                                  x2   +1                   u 2x                  2 u
                                                          1 −1/2
                                                 =        2u     du




 V63.0121.041, Calculus I (NYU)         Section 5.5 Integration by Substitution      December 13, 2010   13 / 37
Useful but unsavory variation

Solution (Same technique, new notation, more idiot-proof)
                                              √
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:”
                                                              du
                                                    dx =
                                                              2x
So the integrand becomes completely transformed into
                                   x                       x du                    1
                            √               dx =          √ ·    =                 √ du
                                  x2   +1                   u 2x                  2 u
                                                          1 −1/2
                                                 =        2u     du
                                                     √
                                                 =       u+C =              1 + x2 + C .




 V63.0121.041, Calculus I (NYU)         Section 5.5 Integration by Substitution      December 13, 2010   13 / 37
Useful but unsavory variation

Solution (Same technique, new notation, more idiot-proof)
                                              √
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:”
                                                              du
                                                    dx =
                                                              2x
So the integrand becomes completely transformed into
                                   x                       x du                    1
                            √               dx =          √ ·    =                 √ du
                                  x2   +1                   u 2x                  2 u
                                                          1 −1/2
                                                 =        2u     du
                                                     √
                                                 =       u+C =              1 + x2 + C .


Mathematicians have serious issues with mixing the x and u like this.
However, I can’t deny that it works.
 V63.0121.041, Calculus I (NYU)         Section 5.5 Integration by Substitution      December 13, 2010   13 / 37

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Theorem of the Day

Theorem (The Substitution Rule)
If u = g (x) is a differentiable function whose range is an interval I and f
is continuous on I , then

                                  f (g (x))g (x) dx =                f (u) du

That is, if F is an antiderivative for f , then

                                   f (g (x))g (x) dx = F (g (x))

In Leibniz notation:
                                            du
                                    f (u)      dx =             f (u) du
                                            dx


 V63.0121.041, Calculus I (NYU)    Section 5.5 Integration by Substitution      December 13, 2010   14 / 37
A polynomial example


Example
Use the substitution u = x 2 + 3 to find                       (x 2 + 3)3 4x dx.




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution    December 13, 2010   15 / 37
A polynomial example


Example
Use the substitution u = x 2 + 3 to find                        (x 2 + 3)3 4x dx.


Solution
If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So

                           (x 2 + 3)3 4x dx




 V63.0121.041, Calculus I (NYU)    Section 5.5 Integration by Substitution    December 13, 2010   15 / 37
A polynomial example


Example
Use the substitution u = x 2 + 3 to find                       (x 2 + 3)3 4x dx.


Solution
If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So

                           (x 2 + 3)3 4x dx =           u 3 2du = 2         u 3 du




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution       December 13, 2010   15 / 37

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A polynomial example


Example
Use the substitution u = x 2 + 3 to find                       (x 2 + 3)3 4x dx.


Solution
If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So

                           (x 2 + 3)3 4x dx =           u 3 2du = 2         u 3 du
                                                1
                                               = u4
                                                2




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution       December 13, 2010   15 / 37
A polynomial example


Example
Use the substitution u = x 2 + 3 to find                       (x 2 + 3)3 4x dx.


Solution
If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So

                           (x 2 + 3)3 4x dx =           u 3 2du = 2         u 3 du
                                                1     1
                                               = u 4 = (x 2 + 3)4
                                                2     2




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution       December 13, 2010   15 / 37
A polynomial example, by brute force


Compare this to multiplying it out:

                 (x 2 + 3)3 4x dx




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   16 / 37
A polynomial example, by brute force


Compare this to multiplying it out:

                 (x 2 + 3)3 4x dx =            x 6 + 9x 4 + 27x 2 + 27 4x dx




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   16 / 37

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A polynomial example, by brute force


Compare this to multiplying it out:

                 (x 2 + 3)3 4x dx =            x 6 + 9x 4 + 27x 2 + 27 4x dx

                                    =          4x 7 + 36x 5 + 108x 3 + 108x dx




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   16 / 37
A polynomial example, by brute force


Compare this to multiplying it out:

                 (x 2 + 3)3 4x dx =            x 6 + 9x 4 + 27x 2 + 27 4x dx

                                    =          4x 7 + 36x 5 + 108x 3 + 108x dx
                                     1
                                    = x 8 + 6x 6 + 27x 4 + 54x 2
                                     2




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   16 / 37
A polynomial example, by brute force


Compare this to multiplying it out:

                 (x 2 + 3)3 4x dx =            x 6 + 9x 4 + 27x 2 + 27 4x dx

                                    =          4x 7 + 36x 5 + 108x 3 + 108x dx
                                     1
                                    = x 8 + 6x 6 + 27x 4 + 54x 2
                                     2
Which would you rather do?




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   16 / 37
A polynomial example, by brute force


Compare this to multiplying it out:

                 (x 2 + 3)3 4x dx =            x 6 + 9x 4 + 27x 2 + 27 4x dx

                                    =          4x 7 + 36x 5 + 108x 3 + 108x dx
                                     1
                                    = x 8 + 6x 6 + 27x 4 + 54x 2
                                     2
Which would you rather do?
      It’s a wash for low powers




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   16 / 37

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A polynomial example, by brute force


Compare this to multiplying it out:

                 (x 2 + 3)3 4x dx =            x 6 + 9x 4 + 27x 2 + 27 4x dx

                                    =          4x 7 + 36x 5 + 108x 3 + 108x dx
                                     1
                                    = x 8 + 6x 6 + 27x 4 + 54x 2
                                     2
Which would you rather do?
      It’s a wash for low powers
      But for higher powers, it’s much easier to do substitution.




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   16 / 37
Compare

We have the substitution method, which, when multiplied out, gives
                               1
            (x 2 + 3)3 4x dx = (x 2 + 3)4
                               2
                               1 8
                             =   x + 12x 6 + 54x 4 + 108x 2 + 81
                               2
                               1                            81
                             = x 8 + 6x 6 + 27x 4 + 54x 2 +
                               2                             2

and the brute force method

                              1
            (x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2
                              2
Is there a difference? Is this a problem?


 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   17 / 37
Compare

We have the substitution method, which, when multiplied out, gives
                               1
            (x 2 + 3)3 4x dx = (x 2 + 3)4 + C
                               2
                               1 8
                             =   x + 12x 6 + 54x 4 + 108x 2 + 81 + C
                               2
                               1                            81
                             = x 8 + 6x 6 + 27x 4 + 54x 2 +    +C
                               2                             2

and the brute force method

                              1
            (x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2 + C
                              2
Is there a difference? Is this a problem? No, that’s what +C means!


 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   17 / 37
A slick example


Example
Find       tan x dx.




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A slick example


Example
                                              sin x
Find       tan x dx. (Hint: tan x =                 )
                                              cos x




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   18 / 37
A slick example


Example
                                              sin x
Find       tan x dx. (Hint: tan x =                 )
                                              cos x

Solution
Let u = cos x . Then du = − sin x dx . So

                                          sin x
                         tan x dx =             dx
                                          cos x




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   18 / 37
A slick example


Example
                                              sin x
Find       tan x dx. (Hint: tan x =                 )
                                              cos x

Solution
Let u = cos x . Then du = − sin x dx . So

                                          sin x
                         tan x dx =             dx
                                          cos x




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   18 / 37
A slick example


Example
                                              sin x
Find       tan x dx. (Hint: tan x =                 )
                                              cos x

Solution
Let u = cos x . Then du = − sin x dx . So

                                          sin x
                         tan x dx =             dx
                                          cos x




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A slick example


Example
                                              sin x
Find       tan x dx. (Hint: tan x =                 )
                                              cos x

Solution
Let u = cos x . Then du = − sin x dx . So

                                          sin x                        1
                         tan x dx =             dx = −                   du
                                          cos x                        u




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution     December 13, 2010   18 / 37
A slick example


Example
                                              sin x
Find       tan x dx. (Hint: tan x =                 )
                                              cos x

Solution
Let u = cos x . Then du = − sin x dx . So

                                       sin x                           1
                         tan x dx =           dx = −                     du
                                       cos x                           u
                                  = − ln |u| + C




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution     December 13, 2010   18 / 37
A slick example


Example
                                              sin x
Find       tan x dx. (Hint: tan x =                 )
                                              cos x

Solution
Let u = cos x . Then du = − sin x dx . So

                                       sin x                           1
                         tan x dx =           dx = −                     du
                                       cos x                           u
                                  = − ln |u| + C
                                  = − ln | cos x| + C = ln | sec x| + C




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution     December 13, 2010   18 / 37
Can you do it another way?

Example
                                              sin x
Find       tan x dx. (Hint: tan x =                 )
                                              cos x




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Can you do it another way?

Example
                                              sin x
Find       tan x dx. (Hint: tan x =                 )
                                              cos x

Solution
                                                                         du
Let u = sin x. Then du = cos x dx and so dx =                                 .
                                                                        cos x




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution         December 13, 2010   19 / 37
Can you do it another way?

Example
                                                  sin x
Find       tan x dx. (Hint: tan x =                     )
                                                  cos x

Solution
                                                                             du
Let u = sin x. Then du = cos x dx and so dx =                                     .
                                                                            cos x
                                         sin x          u     du
                    tan x dx =                 dx =
                                         cos x        cos x cos x
                                          u du         u du                            u du
                                  =             =              =
                                         cos 2x     1 − sin2 x                        1 − u2
At this point, although it’s possible to proceed, we should probably back
up and see if the other way works quicker (it does).

 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution         December 13, 2010   19 / 37
For those who really must know all

Solution (Continued, with algebra help)
Let y = 1 − u 2 , so dy = −2u du. Then
                                 u du         u dy
            tan x dx =                2
                                         =
                                1−u           y −2u
                               1    dy       1                 1
                          =−             = − ln |y | + C = − ln 1 − u 2 + C
                               2     y       2                 2
                                   1                      1
                          = ln √          + C = ln                +C
                                 1 − u2                1 − sin2 x
                                  1
                          = ln          + C = ln |sec x| + C
                               |cos x|


There are other ways to do it, too.

 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   20 / 37
Outline



Last Time: The Fundamental Theorem(s) of Calculus


Substitution for Indefinite Integrals
   Theory
   Examples


Substitution for Definite Integrals
   Theory
   Examples




 V63.0121.041, Calculus I (NYU)   Section 5.5 Integration by Substitution   December 13, 2010   21 / 37

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Substitution for Definite Integrals


Theorem (The Substitution Rule for Definite Integrals)
If g is continuous and f is continuous on the range of u = g (x), then
                                  b                                    g (b)
                                      f (g (x))g (x) dx =                      f (u) du.
                              a                                      g (a)




 V63.0121.041, Calculus I (NYU)          Section 5.5 Integration by Substitution           December 13, 2010   22 / 37
Substitution for Definite Integrals


Theorem (The Substitution Rule for Definite Integrals)
If g is continuous and f is continuous on the range of u = g (x), then
                                  b                                    g (b)
                                      f (g (x))g (x) dx =                      f (u) du.
                              a                                      g (a)



Why the change in the limits?
      The integral on the left happens in “x-land”
      The integral on the right happens in “u-land”, so the limits need to
      be u-values
      To get from x to u, apply g



 V63.0121.041, Calculus I (NYU)          Section 5.5 Integration by Substitution           December 13, 2010   22 / 37
Example
                   π
Compute                cos2 x sin x dx.
               0




V63.0121.041, Calculus I (NYU)      Section 5.5 Integration by Substitution   December 13, 2010   23 / 37
Example
                    π
Compute                 cos2 x sin x dx.
                0


Solution (Slow Way)

First compute the indefinite integral                        cos2 x sin x dx and then evaluate.
Let u = cos x . Then du = − sin x dx and

                          cos2 x sin x dx




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Example
                    π
Compute                 cos2 x sin x dx.
                0


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First compute the indefinite integral                        cos2 x sin x dx and then evaluate.
Let u = cos x . Then du = − sin x dx and

                          cos2 x sin x dx




 V63.0121.041, Calculus I (NYU)      Section 5.5 Integration by Substitution   December 13, 2010   23 / 37
Example
                    π
Compute                 cos2 x sin x dx.
                0


Solution (Slow Way)

First compute the indefinite integral                        cos2 x sin x dx and then evaluate.
Let u = cos x . Then du = − sin x dx and

                          cos2 x sin x dx




 V63.0121.041, Calculus I (NYU)      Section 5.5 Integration by Substitution   December 13, 2010   23 / 37
Example
                    π
Compute                 cos2 x sin x dx.
                0


Solution (Slow Way)

First compute the indefinite integral                        cos2 x sin x dx and then evaluate.
Let u = cos x . Then du = − sin x dx and

                          cos2 x sin x dx = −            u 2 du




 V63.0121.041, Calculus I (NYU)      Section 5.5 Integration by Substitution   December 13, 2010   23 / 37
Example
                    π
Compute                 cos2 x sin x dx.
                0


Solution (Slow Way)

First compute the indefinite integral                        cos2 x sin x dx and then evaluate.
Let u = cos x . Then du = − sin x dx and

                          cos2 x sin x dx = −            u 2 du

                                            = − 3 u 3 + C = − 1 cos3 x + C .
                                                1
                                                              3




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Example
                     π
Compute                  cos2 x sin x dx.
                 0


Solution (Slow Way)

First compute the indefinite integral                         cos2 x sin x dx and then evaluate.
Let u = cos x . Then du = − sin x dx and

                           cos2 x sin x dx = −            u 2 du

                                             = − 3 u 3 + C = − 1 cos3 x + C .
                                                 1
                                                               3

Therefore
                 π                                           π
                                        1
                     cos2 x sin x dx = − cos3 x
             0                          3                    0



 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution   December 13, 2010   23 / 37
Example
                     π
Compute                  cos2 x sin x dx.
                 0


Solution (Slow Way)

First compute the indefinite integral                         cos2 x sin x dx and then evaluate.
Let u = cos x . Then du = − sin x dx and

                           cos2 x sin x dx = −            u 2 du

                                             = − 3 u 3 + C = − 1 cos3 x + C .
                                                 1
                                                               3

Therefore
                 π                                           π
                                        1                                1
                     cos2 x sin x dx = − cos3 x                  =−        (−1)3 − 13
             0                          3                    0           3


 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution     December 13, 2010   23 / 37
Example
                     π
Compute                  cos2 x sin x dx.
                 0


Solution (Slow Way)

First compute the indefinite integral                         cos2 x sin x dx and then evaluate.
Let u = cos x . Then du = − sin x dx and

                           cos2 x sin x dx = −            u 2 du

                                             = − 3 u 3 + C = − 1 cos3 x + C .
                                                 1
                                                               3

Therefore
                 π                                           π
                                        1                                1             2
                     cos2 x sin x dx = − cos3 x                  =−        (−1)3 − 13 = .
             0                          3                    0           3             3


 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution    December 13, 2010   23 / 37
Definite-ly Quicker

Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
                        π
                            cos2 x sin x dx
                    0




 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution   December 13, 2010   24 / 37

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This document contains notes from a calculus class. It provides the outline and key points about the Fundamental Theorem of Calculus. It discusses the first and second Fundamental Theorems of Calculus, including proofs and examples. It also provides brief biographies of several important mathematicians that contributed to the development of calculus, including the Fundamental Theorem of Calculus, such as Isaac Newton, Gottfried Leibniz, James Gregory, and Isaac Barrow.

v6301210342009fcalculusintegral
Lesson 28: The Fundamental Theorem of Calculus
Lesson 28: The Fundamental Theorem of CalculusLesson 28: The Fundamental Theorem of Calculus
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This document contains lecture notes on the Fundamental Theorem of Calculus. It begins with announcements about the final exam date and current grade distribution. The outline then reviews the Evaluation Theorem and introduces the First and Second Fundamental Theorems of Calculus. It provides examples of how the integral can represent total change in concepts like distance, cost, and mass. Biographies are also included of mathematicians like Gregory, Barrow, Newton, and Leibniz who contributed to the development of calculus.

calculusintegralv6301212009f
Lesson 5: Continuity (Section 21 slides)
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This document is from a Calculus I class at New York University and covers continuity. It provides announcements about office hours and homework deadlines. It then discusses the objectives of understanding the definition of continuity and applying it to piecewise functions. Examples are provided to demonstrate how to show a function is continuous at a point by evaluating the limit as x approaches the point and showing it equals the function value. The students are asked to determine at which other points the example function is continuous.

v6301210212010fv6301212010ffunction
Definite-ly Quicker

Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
                        π
                            cos2 x sin x dx
                    0




 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution   December 13, 2010   24 / 37
Definite-ly Quicker

Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
                        π
                            cos2 x sin x dx
                    0




 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution   December 13, 2010   24 / 37
Definite-ly Quicker

Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
                        π                               −1                     1
                            cos2 x sin x dx =                −u 2 du =              u 2 du
                    0                               1                          −1




 V63.0121.041, Calculus I (NYU)      Section 5.5 Integration by Substitution            December 13, 2010   24 / 37
Definite-ly Quicker

Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
                        π                               −1                     1
                            cos2 x sin x dx =                −u 2 du =              u 2 du
                    0                               1                          −1
                                                             1
                                                  1 3                 1            2
                                             =      u             =     1 − (−1) =
                                                  3          −1       3            3




 V63.0121.041, Calculus I (NYU)      Section 5.5 Integration by Substitution            December 13, 2010   24 / 37

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v6301210212010fv6301212010ffunction
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calculusinverse trigonometricfunction
Definite-ly Quicker

Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
                        π                               −1                     1
                            cos2 x sin x dx =                −u 2 du =              u 2 du
                    0                               1                          −1
                                                             1
                                                  1 3                 1            2
                                             =      u             =     1 − (−1) =
                                                  3          −1       3            3


      The advantage to the “fast way” is that you completely transform the
      integral into something simpler and don’t have to go back to the
      original variable (x).


 V63.0121.041, Calculus I (NYU)      Section 5.5 Integration by Substitution            December 13, 2010   24 / 37
Definite-ly Quicker

Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
                        π                               −1                     1
                            cos2 x sin x dx =                −u 2 du =              u 2 du
                    0                               1                          −1
                                                             1
                                                  1 3                 1            2
                                             =      u             =     1 − (−1) =
                                                  3          −1       3            3


      The advantage to the “fast way” is that you completely transform the
      integral into something simpler and don’t have to go back to the
      original variable (x).
      But the slow way is just as reliable.

 V63.0121.041, Calculus I (NYU)      Section 5.5 Integration by Substitution            December 13, 2010   24 / 37
An exponential example
Example
               √
          ln          8
Find          √           e 2x   e 2x + 1 dx
         ln       3




 V63.0121.041, Calculus I (NYU)          Section 5.5 Integration by Substitution   December 13, 2010   25 / 37
An exponential example
Example
               √
          ln          8
Find          √           e 2x   e 2x + 1 dx
         ln       3

Solution
Let u = e 2x , so du = 2e 2x dx. We have
                                       √
                                  ln          8                                           8√
                                                                                 1
                                      √           e 2x   e 2x + 1 dx =                         u + 1 du
                                 ln       3                                      2    3




 V63.0121.041, Calculus I (NYU)                      Section 5.5 Integration by Substitution         December 13, 2010   25 / 37

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About those limits




Since
                                                    √                 √   2
                                           e 2(ln       3)
                                                             = e ln       3
                                                                              = e ln 3 = 3

we have                         √
                           ln          8                                               8√
                                                                               1
                               √           e 2x     e 2x + 1 dx =                           u + 1 du
                          ln       3                                           2   3




 V63.0121.041, Calculus I (NYU)                Section 5.5 Integration by Substitution            December 13, 2010   26 / 37
An exponential example
Example
               √
          ln          8
Find          √           e 2x       e 2x + 1 dx
         ln       3

Solution
Let u = e 2x , so du = 2e 2x dx. We have
                                           √
                                      ln          8                                            8√
                                                                                      1
                                          √           e 2x   e 2x + 1 dx =                          u + 1 du
                                     ln       3                                       2    3

Now let y = u + 1, dy = du. So
                                     8√                                     9                           9
                          1                                        1            √              1
                                           u + 1 du =                               y dy =                  y 1/2 dy
                          2      3                                 2    4                      2    4
                                                                                      9
                                                               1 2                         1           19
                                                              = · y 3/2                   = (27 − 8) =
                                                               2 3                    4    3           3
 V63.0121.041, Calculus I (NYU)                          Section 5.5 Integration by Substitution                December 13, 2010   27 / 37
About those fractional powers



We have

                                      93/2 = (91/2 )3 = 33 = 27
                                      43/2 = (41/2 )3 = 23 = 8

so
                             9                                 9
                    1                         1 2 3/2                  1            19
                                 y 1/2 dy =    · y                 =     (27 − 8) =
                    2    4                    2 3              4       3            3




 V63.0121.041, Calculus I (NYU)        Section 5.5 Integration by Substitution    December 13, 2010   28 / 37
An exponential example
Example
               √
          ln          8
Find          √           e 2x       e 2x + 1 dx
         ln       3

Solution
Let u = e 2x , so du = 2e 2x dx. We have
                                           √
                                      ln          8                                            8√
                                                                                      1
                                          √           e 2x   e 2x + 1 dx =                          u + 1 du
                                     ln       3                                       2    3

Now let y = u + 1, dy = du. So
                                     8√                                     9                           9
                          1                                        1            √              1
                                           u + 1 du =                               y dy =                  y 1/2 dy
                          2      3                                 2    4                      2    4
                                                                                      9
                                                               1 2                         1           19
                                                              = · y 3/2                   = (27 − 8) =
                                                               2 3                    4    3           3
 V63.0121.041, Calculus I (NYU)                          Section 5.5 Integration by Substitution                December 13, 2010   29 / 37

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Another way to skin that cat

Example
               √
          ln          8
Find          √           e 2x   e 2x + 1 dx
         ln       3


Solution
Let u = e 2x + 1,




 V63.0121.041, Calculus I (NYU)          Section 5.5 Integration by Substitution   December 13, 2010   30 / 37
Another way to skin that cat

Example
               √
          ln          8
Find          √           e 2x   e 2x + 1 dx
         ln       3


Solution
Let u = e 2x + 1,so that du = 2e 2x dx.




 V63.0121.041, Calculus I (NYU)          Section 5.5 Integration by Substitution   December 13, 2010   30 / 37
Another way to skin that cat

Example
               √
          ln          8
Find          √           e 2x   e 2x + 1 dx
         ln       3


Solution
Let u = e 2x + 1,so that du = 2e 2x dx. Then
                                       √
                                  ln          8                                           9√
                                                                                 1
                                      √           e 2x   e 2x + 1 dx =                         u du
                                 ln       3                                      2    4




 V63.0121.041, Calculus I (NYU)                      Section 5.5 Integration by Substitution          December 13, 2010   30 / 37
Another way to skin that cat

Example
               √
          ln          8
Find          √           e 2x   e 2x + 1 dx
         ln       3


Solution
Let u = e 2x + 1,so that du = 2e 2x dx. Then
                                       √
                                  ln          8                                           9√
                                                                                 1
                                      √           e 2x   e 2x + 1 dx =                             u du
                                 ln       3                                      2    4
                                                                                               9
                                                                             1
                                                                            = u 3/2
                                                                             3                 4




 V63.0121.041, Calculus I (NYU)                      Section 5.5 Integration by Substitution              December 13, 2010   30 / 37

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v6301212010fv6301210412010f
Another way to skin that cat

Example
               √
          ln          8
Find          √           e 2x   e 2x + 1 dx
         ln       3


Solution
Let u = e 2x + 1,so that du = 2e 2x dx. Then
                                       √
                                  ln          8                                           9√
                                                                                 1
                                      √           e 2x   e 2x + 1 dx =                         u du
                                 ln       3                                      2    4
                                                                             1 3/2 9
                                                                            = u
                                                                             3     4
                                                                             1           19
                                                                            = (27 − 8) =
                                                                             3           3


 V63.0121.041, Calculus I (NYU)                      Section 5.5 Integration by Substitution          December 13, 2010   30 / 37
A third skinned cat


Example
               √
          ln          8
Find          √           e 2x   e 2x + 1 dx
         ln       3


Solution
Let u =               e 2x + 1, so that

                                  u 2 = e 2x + 1




 V63.0121.041, Calculus I (NYU)          Section 5.5 Integration by Substitution   December 13, 2010   31 / 37
A third skinned cat


Example
               √
          ln          8
Find          √           e 2x   e 2x + 1 dx
         ln       3


Solution
Let u =               e 2x + 1, so that

                                  u 2 = e 2x + 1 =⇒ 2u du = 2e 2x dx




 V63.0121.041, Calculus I (NYU)          Section 5.5 Integration by Substitution   December 13, 2010   31 / 37
A third skinned cat


Example
               √
          ln          8
Find          √           e 2x   e 2x + 1 dx
         ln       3


Solution
Let u =               e 2x + 1, so that

                                  u 2 = e 2x + 1 =⇒ 2u du = 2e 2x dx

Thus                                     √
                                    ln          8              3                          3
                                                                                  1 3             19
                                        √            =             u · u du =       u         =
                                   ln       3              2                      3       2       3



 V63.0121.041, Calculus I (NYU)                     Section 5.5 Integration by Substitution            December 13, 2010   31 / 37

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Lesson 1: Functions
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No, this is not a function because the same input of 2 is mapped to two different outputs of 4 and 5. For a relation to be a function, each input must map to a unique output. V63.0121.021/041, Calculus I (NYU) Section 1.1 Functions September 8, 2010 17 / 33

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v6301212010ffunctionv6301210412010f
A Trigonometric Example



Example
Find
                                  3π/2
                                                      θ                θ
                                          cot5              sec2                dθ.
                                  π                   6                6




 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution         December 13, 2010   32 / 37
A Trigonometric Example



Example
Find
                                  3π/2
                                                      θ                θ
                                          cot5              sec2                dθ.
                                  π                   6                6

Before we dive in, think about:
       What “easy” substitutions might help?
       Which of the trig functions suggests a substitution?




 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution         December 13, 2010   32 / 37
Solution
       θ           1
Let ϕ = . Then dϕ = dθ.
       6           6
                3π/2                                                    π/4
                                  θ              θ
                       cot5           sec2             dθ = 6                   cot5 ϕ sec2 ϕ dϕ
               π                  6              6                    π/6
                                                                       π/4
                                                                                sec2 ϕ dϕ
                                                             =6
                                                                      π/6         tan5 ϕ




 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution          December 13, 2010   33 / 37
Solution
       θ           1
Let ϕ = . Then dϕ = dθ.
       6           6
                3π/2                                                    π/4
                                  θ              θ
                          cot5        sec2             dθ = 6                   cot5 ϕ sec2 ϕ dϕ
                π                 6              6                     π/6
                                                                        π/4
                                                                                sec2 ϕ dϕ
                                                             =6
                                                                       π/6        tan5 ϕ

Now let u = tan ϕ. So du = sec2 ϕ dϕ, and
                     π/4                           1
                           sec2 ϕ dϕ                       −5
            6                        =6            √ u          du
                    π/6      tan5 ϕ              1/ 3
                                                                   1
                                                   1                             3
                                        =6        − u −4             √
                                                                             =     [9 − 1] = 12.
                                                   4               1/ 3          2



 V63.0121.041, Calculus I (NYU)       Section 5.5 Integration by Substitution          December 13, 2010   33 / 37

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v6301212010fv6301210412010f
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Lesson 6: Limits Involving ∞ (Section 41 slides)
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infinityv6301212010ffunction
Graphs
                       3π/2                                                            π/4
                                         θ                θ
                              cot   5
                                               sec   2
                                                                 dθ                          6 cot5 ϕ sec2 ϕ dϕ
                     π                   6                6                           π/6

                 y                                                                y




                                                                       θ                              ϕ
                                        3π       π                                    ππ
                                         2                                            64
The areas of these two regions are the same.


 V63.0121.041, Calculus I (NYU)         Section 5.5 Integration by Substitution              December 13, 2010   35 / 37
Graphs
                       π/4                                                       1
                                                                                          −5
                             6 cot5 ϕ sec2 ϕ dϕ                                    √ 6u        du
                     π/6                                                         1/ 3

                 y                                                           y




                                   ϕ                                                 u
                      ππ                                    1 1
                      64                                   √
                                                             3
                      The areas of these two regions are the same.


 V63.0121.041, Calculus I (NYU)    Section 5.5 Integration by Substitution           December 13, 2010   36 / 37
Summary


     If F is an antiderivative for f , then:

                                     f (g (x))g (x) dx = F (g (x))

     If F is an antiderivative for f , which is continuous on the range of g ,
     then:
                   b                               g (b)
                       f (g (x))g (x) dx =                 f (u) du = F (g (b)) − F (g (a))
               a                                 g (a)

     Antidifferentiation in general and substitution in particular is a
     “nonlinear” problem that needs practice, intuition, and perserverance
     The whole antidifferentiation story is in Chapter 6


V63.0121.041, Calculus I (NYU)     Section 5.5 Integration by Substitution   December 13, 2010   37 / 37

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Lesson 27: Integration by Substitution (Section 041 slides)

  • 1. Section 5.5 Integration by Substitution V63.0121.041, Calculus I New York University December 13, 2010 Announcements ”Wednesday”, December 15: Review, Movie Monday, December 20, 12:00pm–1:50pm: Final Exam
  • 2. Announcements ”Wednesday”, December 15: Review, Movie Monday, December 20, 12:00pm–1:50pm: Final Exam V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 2 / 37
  • 3. Resurrection Policy If your final score beats your midterm score, we will add 10% to its weight, and subtract 10% from the midterm weight. Image credit: Scott Beale / Laughing Squid V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 3 / 37
  • 4. Objectives Given an integral and a substitution, transform the integral into an equivalent one using a substitution Evaluate indefinite integrals using the method of substitution. Evaluate definite integrals using the method of substitution. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 4 / 37
  • 5. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 5 / 37
  • 6. Differentiation and Integration as reverse processes Theorem (The Fundamental Theorem of Calculus) 1. Let f be continuous on [a, b]. Then x d f (t) dt = f (x) dx a 2. Let f be continuous on [a, b] and f = F for some other function F . Then b f (x) dx = F (b) − F (a). a V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 6 / 37
  • 7. Techniques of antidifferentiation? So far we know only a few rules for antidifferentiation. Some are general, like [f (x) + g (x)] dx = f (x) dx + g (x) dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
  • 8. Techniques of antidifferentiation? So far we know only a few rules for antidifferentiation. Some are general, like [f (x) + g (x)] dx = f (x) dx + g (x) dx Some are pretty particular, like 1 √ dx = arcsec x + C . x x2 − 1 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
  • 9. Techniques of antidifferentiation? So far we know only a few rules for antidifferentiation. Some are general, like [f (x) + g (x)] dx = f (x) dx + g (x) dx Some are pretty particular, like 1 √ dx = arcsec x + C . x x2 − 1 What are we supposed to do with that? V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
  • 10. No straightforward system of antidifferentiation So far we don’t have any way to find 2x √ dx x2 + 1 or tan x dx. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 8 / 37
  • 11. No straightforward system of antidifferentiation So far we don’t have any way to find 2x √ dx x2 + 1 or tan x dx. Luckily, we can be smart and use the “anti” version of one of the most important rules of differentiation: the chain rule. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 8 / 37
  • 12. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 9 / 37
  • 13. Substitution for Indefinite Integrals Example Find x √ dx. x2 +1 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 10 / 37
  • 14. Substitution for Indefinite Integrals Example Find x √ dx. x2 +1 Solution Stare at this long enough and you notice the the integrand is the derivative of the expression 1 + x 2 . V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 10 / 37
  • 15. Say what? Solution (More slowly, now) Let g (x) = x 2 + 1. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
  • 16. Say what? Solution (More slowly, now) Let g (x) = x 2 + 1. Then g (x) = 2x and so d 1 x g (x) = g (x) = √ dx 2 g (x) x2 +1 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
  • 17. Say what? Solution (More slowly, now) Let g (x) = x 2 + 1. Then g (x) = 2x and so d 1 x g (x) = g (x) = √ dx 2 g (x) x2 +1 Thus x d √ dx = g (x) dx x2 + 1 dx = g (x) + C = 1 + x2 + C . V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
  • 18. Leibnizian notation FTW Solution (Same technique, new notation) Let u = x 2 + 1. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
  • 19. Leibnizian notation FTW Solution (Same technique, new notation) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
  • 20. Leibnizian notation FTW Solution (Same technique, new notation) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into 1 √ x dx 2 du √ 1 √ du = = x2 + 1 u 2 u V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
  • 21. Leibnizian notation FTW Solution (Same technique, new notation) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into 1 √ x dx 2 du √ 1 √ du = = x2 + 1 u 2 u 1 −1/2 = 2u du V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
  • 22. Leibnizian notation FTW Solution (Same technique, new notation) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into 1 √ x dx 2 du √ 1 √ du = = x2 + 1 u 2 u 1 −1/2 = 2u du √ = u+C = 1 + x2 + C . V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
  • 23. Useful but unsavory variation Solution (Same technique, new notation, more idiot-proof) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
  • 24. Useful but unsavory variation Solution (Same technique, new notation, more idiot-proof) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2x So the integrand becomes completely transformed into x x du √ dx = √ · x2 +1 u 2x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
  • 25. Useful but unsavory variation Solution (Same technique, new notation, more idiot-proof) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2x So the integrand becomes completely transformed into x x du 1 √ dx = √ · = √ du x2 +1 u 2x 2 u V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
  • 26. Useful but unsavory variation Solution (Same technique, new notation, more idiot-proof) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2x So the integrand becomes completely transformed into x x du 1 √ dx = √ · = √ du x2 +1 u 2x 2 u 1 −1/2 = 2u du V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
  • 27. Useful but unsavory variation Solution (Same technique, new notation, more idiot-proof) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2x So the integrand becomes completely transformed into x x du 1 √ dx = √ · = √ du x2 +1 u 2x 2 u 1 −1/2 = 2u du √ = u+C = 1 + x2 + C . V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
  • 28. Useful but unsavory variation Solution (Same technique, new notation, more idiot-proof) √ Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2x So the integrand becomes completely transformed into x x du 1 √ dx = √ · = √ du x2 +1 u 2x 2 u 1 −1/2 = 2u du √ = u+C = 1 + x2 + C . Mathematicians have serious issues with mixing the x and u like this. However, I can’t deny that it works. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
  • 29. Theorem of the Day Theorem (The Substitution Rule) If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then f (g (x))g (x) dx = f (u) du That is, if F is an antiderivative for f , then f (g (x))g (x) dx = F (g (x)) In Leibniz notation: du f (u) dx = f (u) du dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 14 / 37
  • 30. A polynomial example Example Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
  • 31. A polynomial example Example Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx. Solution If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So (x 2 + 3)3 4x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
  • 32. A polynomial example Example Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx. Solution If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So (x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
  • 33. A polynomial example Example Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx. Solution If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So (x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du 1 = u4 2 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
  • 34. A polynomial example Example Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx. Solution If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So (x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du 1 1 = u 4 = (x 2 + 3)4 2 2 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
  • 35. A polynomial example, by brute force Compare this to multiplying it out: (x 2 + 3)3 4x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
  • 36. A polynomial example, by brute force Compare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
  • 37. A polynomial example, by brute force Compare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx = 4x 7 + 36x 5 + 108x 3 + 108x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
  • 38. A polynomial example, by brute force Compare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx = 4x 7 + 36x 5 + 108x 3 + 108x dx 1 = x 8 + 6x 6 + 27x 4 + 54x 2 2 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
  • 39. A polynomial example, by brute force Compare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx = 4x 7 + 36x 5 + 108x 3 + 108x dx 1 = x 8 + 6x 6 + 27x 4 + 54x 2 2 Which would you rather do? V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
  • 40. A polynomial example, by brute force Compare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx = 4x 7 + 36x 5 + 108x 3 + 108x dx 1 = x 8 + 6x 6 + 27x 4 + 54x 2 2 Which would you rather do? It’s a wash for low powers V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
  • 41. A polynomial example, by brute force Compare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx = 4x 7 + 36x 5 + 108x 3 + 108x dx 1 = x 8 + 6x 6 + 27x 4 + 54x 2 2 Which would you rather do? It’s a wash for low powers But for higher powers, it’s much easier to do substitution. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
  • 42. Compare We have the substitution method, which, when multiplied out, gives 1 (x 2 + 3)3 4x dx = (x 2 + 3)4 2 1 8 = x + 12x 6 + 54x 4 + 108x 2 + 81 2 1 81 = x 8 + 6x 6 + 27x 4 + 54x 2 + 2 2 and the brute force method 1 (x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2 2 Is there a difference? Is this a problem? V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 17 / 37
  • 43. Compare We have the substitution method, which, when multiplied out, gives 1 (x 2 + 3)3 4x dx = (x 2 + 3)4 + C 2 1 8 = x + 12x 6 + 54x 4 + 108x 2 + 81 + C 2 1 81 = x 8 + 6x 6 + 27x 4 + 54x 2 + +C 2 2 and the brute force method 1 (x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2 + C 2 Is there a difference? Is this a problem? No, that’s what +C means! V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 17 / 37
  • 44. A slick example Example Find tan x dx. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
  • 45. A slick example Example sin x Find tan x dx. (Hint: tan x = ) cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
  • 46. A slick example Example sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x . Then du = − sin x dx . So sin x tan x dx = dx cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
  • 47. A slick example Example sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x . Then du = − sin x dx . So sin x tan x dx = dx cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
  • 48. A slick example Example sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x . Then du = − sin x dx . So sin x tan x dx = dx cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
  • 49. A slick example Example sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x . Then du = − sin x dx . So sin x 1 tan x dx = dx = − du cos x u V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
  • 50. A slick example Example sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x . Then du = − sin x dx . So sin x 1 tan x dx = dx = − du cos x u = − ln |u| + C V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
  • 51. A slick example Example sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x . Then du = − sin x dx . So sin x 1 tan x dx = dx = − du cos x u = − ln |u| + C = − ln | cos x| + C = ln | sec x| + C V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
  • 52. Can you do it another way? Example sin x Find tan x dx. (Hint: tan x = ) cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
  • 53. Can you do it another way? Example sin x Find tan x dx. (Hint: tan x = ) cos x Solution du Let u = sin x. Then du = cos x dx and so dx = . cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
  • 54. Can you do it another way? Example sin x Find tan x dx. (Hint: tan x = ) cos x Solution du Let u = sin x. Then du = cos x dx and so dx = . cos x sin x u du tan x dx = dx = cos x cos x cos x u du u du u du = = = cos 2x 1 − sin2 x 1 − u2 At this point, although it’s possible to proceed, we should probably back up and see if the other way works quicker (it does). V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
  • 55. For those who really must know all Solution (Continued, with algebra help) Let y = 1 − u 2 , so dy = −2u du. Then u du u dy tan x dx = 2 = 1−u y −2u 1 dy 1 1 =− = − ln |y | + C = − ln 1 − u 2 + C 2 y 2 2 1 1 = ln √ + C = ln +C 1 − u2 1 − sin2 x 1 = ln + C = ln |sec x| + C |cos x| There are other ways to do it, too. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 20 / 37
  • 56. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 21 / 37
  • 57. Substitution for Definite Integrals Theorem (The Substitution Rule for Definite Integrals) If g is continuous and f is continuous on the range of u = g (x), then b g (b) f (g (x))g (x) dx = f (u) du. a g (a) V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 22 / 37
  • 58. Substitution for Definite Integrals Theorem (The Substitution Rule for Definite Integrals) If g is continuous and f is continuous on the range of u = g (x), then b g (b) f (g (x))g (x) dx = f (u) du. a g (a) Why the change in the limits? The integral on the left happens in “x-land” The integral on the right happens in “u-land”, so the limits need to be u-values To get from x to u, apply g V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 22 / 37
  • 59. Example π Compute cos2 x sin x dx. 0 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
  • 60. Example π Compute cos2 x sin x dx. 0 Solution (Slow Way) First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and cos2 x sin x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
  • 61. Example π Compute cos2 x sin x dx. 0 Solution (Slow Way) First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and cos2 x sin x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
  • 62. Example π Compute cos2 x sin x dx. 0 Solution (Slow Way) First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and cos2 x sin x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
  • 63. Example π Compute cos2 x sin x dx. 0 Solution (Slow Way) First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and cos2 x sin x dx = − u 2 du V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
  • 64. Example π Compute cos2 x sin x dx. 0 Solution (Slow Way) First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and cos2 x sin x dx = − u 2 du = − 3 u 3 + C = − 1 cos3 x + C . 1 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
  • 65. Example π Compute cos2 x sin x dx. 0 Solution (Slow Way) First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and cos2 x sin x dx = − u 2 du = − 3 u 3 + C = − 1 cos3 x + C . 1 3 Therefore π π 1 cos2 x sin x dx = − cos3 x 0 3 0 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
  • 66. Example π Compute cos2 x sin x dx. 0 Solution (Slow Way) First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and cos2 x sin x dx = − u 2 du = − 3 u 3 + C = − 1 cos3 x + C . 1 3 Therefore π π 1 1 cos2 x sin x dx = − cos3 x =− (−1)3 − 13 0 3 0 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
  • 67. Example π Compute cos2 x sin x dx. 0 Solution (Slow Way) First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and cos2 x sin x dx = − u 2 du = − 3 u 3 + C = − 1 cos3 x + C . 1 3 Therefore π π 1 1 2 cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = . 0 3 0 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
  • 68. Definite-ly Quicker Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π cos2 x sin x dx 0 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
  • 69. Definite-ly Quicker Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π cos2 x sin x dx 0 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
  • 70. Definite-ly Quicker Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π cos2 x sin x dx 0 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
  • 71. Definite-ly Quicker Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π −1 1 cos2 x sin x dx = −u 2 du = u 2 du 0 1 −1 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
  • 72. Definite-ly Quicker Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π −1 1 cos2 x sin x dx = −u 2 du = u 2 du 0 1 −1 1 1 3 1 2 = u = 1 − (−1) = 3 −1 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
  • 73. Definite-ly Quicker Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π −1 1 cos2 x sin x dx = −u 2 du = u 2 du 0 1 −1 1 1 3 1 2 = u = 1 − (−1) = 3 −1 3 3 The advantage to the “fast way” is that you completely transform the integral into something simpler and don’t have to go back to the original variable (x). V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
  • 74. Definite-ly Quicker Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π −1 1 cos2 x sin x dx = −u 2 du = u 2 du 0 1 −1 1 1 3 1 2 = u = 1 − (−1) = 3 −1 3 3 The advantage to the “fast way” is that you completely transform the integral into something simpler and don’t have to go back to the original variable (x). But the slow way is just as reliable. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
  • 75. An exponential example Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 25 / 37
  • 76. An exponential example Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution Let u = e 2x , so du = 2e 2x dx. We have √ ln 8 8√ 1 √ e 2x e 2x + 1 dx = u + 1 du ln 3 2 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 25 / 37
  • 77. About those limits Since √ √ 2 e 2(ln 3) = e ln 3 = e ln 3 = 3 we have √ ln 8 8√ 1 √ e 2x e 2x + 1 dx = u + 1 du ln 3 2 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 26 / 37
  • 78. An exponential example Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution Let u = e 2x , so du = 2e 2x dx. We have √ ln 8 8√ 1 √ e 2x e 2x + 1 dx = u + 1 du ln 3 2 3 Now let y = u + 1, dy = du. So 8√ 9 9 1 1 √ 1 u + 1 du = y dy = y 1/2 dy 2 3 2 4 2 4 9 1 2 1 19 = · y 3/2 = (27 − 8) = 2 3 4 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 27 / 37
  • 79. About those fractional powers We have 93/2 = (91/2 )3 = 33 = 27 43/2 = (41/2 )3 = 23 = 8 so 9 9 1 1 2 3/2 1 19 y 1/2 dy = · y = (27 − 8) = 2 4 2 3 4 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 28 / 37
  • 80. An exponential example Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution Let u = e 2x , so du = 2e 2x dx. We have √ ln 8 8√ 1 √ e 2x e 2x + 1 dx = u + 1 du ln 3 2 3 Now let y = u + 1, dy = du. So 8√ 9 9 1 1 √ 1 u + 1 du = y dy = y 1/2 dy 2 3 2 4 2 4 9 1 2 1 19 = · y 3/2 = (27 − 8) = 2 3 4 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 29 / 37
  • 81. Another way to skin that cat Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution Let u = e 2x + 1, V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
  • 82. Another way to skin that cat Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution Let u = e 2x + 1,so that du = 2e 2x dx. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
  • 83. Another way to skin that cat Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution Let u = e 2x + 1,so that du = 2e 2x dx. Then √ ln 8 9√ 1 √ e 2x e 2x + 1 dx = u du ln 3 2 4 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
  • 84. Another way to skin that cat Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution Let u = e 2x + 1,so that du = 2e 2x dx. Then √ ln 8 9√ 1 √ e 2x e 2x + 1 dx = u du ln 3 2 4 9 1 = u 3/2 3 4 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
  • 85. Another way to skin that cat Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution Let u = e 2x + 1,so that du = 2e 2x dx. Then √ ln 8 9√ 1 √ e 2x e 2x + 1 dx = u du ln 3 2 4 1 3/2 9 = u 3 4 1 19 = (27 − 8) = 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
  • 86. A third skinned cat Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution Let u = e 2x + 1, so that u 2 = e 2x + 1 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
  • 87. A third skinned cat Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution Let u = e 2x + 1, so that u 2 = e 2x + 1 =⇒ 2u du = 2e 2x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
  • 88. A third skinned cat Example √ ln 8 Find √ e 2x e 2x + 1 dx ln 3 Solution Let u = e 2x + 1, so that u 2 = e 2x + 1 =⇒ 2u du = 2e 2x dx Thus √ ln 8 3 3 1 3 19 √ = u · u du = u = ln 3 2 3 2 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
  • 89. A Trigonometric Example Example Find 3π/2 θ θ cot5 sec2 dθ. π 6 6 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 32 / 37
  • 90. A Trigonometric Example Example Find 3π/2 θ θ cot5 sec2 dθ. π 6 6 Before we dive in, think about: What “easy” substitutions might help? Which of the trig functions suggests a substitution? V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 32 / 37
  • 91. Solution θ 1 Let ϕ = . Then dϕ = dθ. 6 6 3π/2 π/4 θ θ cot5 sec2 dθ = 6 cot5 ϕ sec2 ϕ dϕ π 6 6 π/6 π/4 sec2 ϕ dϕ =6 π/6 tan5 ϕ V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 33 / 37
  • 92. Solution θ 1 Let ϕ = . Then dϕ = dθ. 6 6 3π/2 π/4 θ θ cot5 sec2 dθ = 6 cot5 ϕ sec2 ϕ dϕ π 6 6 π/6 π/4 sec2 ϕ dϕ =6 π/6 tan5 ϕ Now let u = tan ϕ. So du = sec2 ϕ dϕ, and π/4 1 sec2 ϕ dϕ −5 6 =6 √ u du π/6 tan5 ϕ 1/ 3 1 1 3 =6 − u −4 √ = [9 − 1] = 12. 4 1/ 3 2 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 33 / 37
  • 93. Graphs 3π/2 π/4 θ θ cot 5 sec 2 dθ 6 cot5 ϕ sec2 ϕ dϕ π 6 6 π/6 y y θ ϕ 3π π ππ 2 64 The areas of these two regions are the same. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 35 / 37
  • 94. Graphs π/4 1 −5 6 cot5 ϕ sec2 ϕ dϕ √ 6u du π/6 1/ 3 y y ϕ u ππ 1 1 64 √ 3 The areas of these two regions are the same. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 36 / 37
  • 95. Summary If F is an antiderivative for f , then: f (g (x))g (x) dx = F (g (x)) If F is an antiderivative for f , which is continuous on the range of g , then: b g (b) f (g (x))g (x) dx = f (u) du = F (g (b)) − F (g (a)) a g (a) Antidifferentiation in general and substitution in particular is a “nonlinear” problem that needs practice, intuition, and perserverance The whole antidifferentiation story is in Chapter 6 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 37 / 37