Download Integration by inverse substitution

Roberto’s Notes on Integral Calculus
Chapter 2: Integration methods
Section 10
Integration by inverse substitution
by using the secant function
What you need to know already:
 How to use inverse trig substitution to
integrate a function involving a a2  b2 x2
form.
If we try the substitution bx  a sin  on the form b x  a we end up
with a negative under the root, which creates somewhat of a problem! We can avoid
the problem by using the same method, but a different trig identity.
2 2
2
Strategy for integrals that contain
b2 x2  a 2
For integrands involving a factor of this form:
1. Try the inverse substitution:
bx  a sec , bdx  a sec tan  d
2. Try to compute the new integral with suitable
What you can learn here:
 How to use inverse trig substitution to
integrate a function involving a b2 x2  a2
form.
3. If this works, change back to the original
variable, by using other trigonometric identities
and/or the triangle model, as necessary.
Same issues presented in the earlier note regarding domain and extended uses
apply here. All you need is practice, so here is an example.
Example:

x2
x2  9
We try the substitution:
x  3sec  , dx  3sec  tan  d ,
methods.
This time the relation is
Integral Calculus
Chapter 2: Integration methods
dx
3
 cos 
x
x
 sec  , so that x must represent the hypotenuse
3
Section 10: Integration by inverse substitution by using secant
Page 1
and 3 the adjacent side. This corresponds to the triangle representation
shown here.
x


2


x2  9


x 9
2


3  x  2   27
2
dx 
2
3


3 x  4 x  4  12  15

2
1
 x  2
2
9

9sec2 
9sec   9
2
3sec  tan  d

9sec2 
sec tan  d  9 sec3  d
tan 
dx
This changes our integral to:

2
3

3sec  tan 
9sec   9
2
d 
2
3

sec d 
2
ln sec  tan   c
3
Finally we switch back to x:

We saw earlier that the last integral above is given by:
1
  sec  tan   ln sec  tan    c
2
2
x2
ln

3
3
 x  2
3
2
9
c
By using this and the triangle representation of the substitution, remembering
that secant is the reciprocal of cosine, we conclude that:
9 x
 
2 3

Example:

x2  9
x
x2  9
 ln 
3
3
3
2
3 x 2  12 x  15

c


No other particular tricks to show you, so it is time for your integration
calisthenics!
dx
To find the general antiderivative of the given function, we first complete the
square:

Integral Calculus
2
3x  12 x  15
2
dx
x  2  3sec , dx  3sec tan  d
This leads to:
dx 

2

2
Now we make the inverse trig substitution:
3
x2

3 x  4 x  4  4  15
2
dx 
dx 

2


3 x  4 x  15
2
dx
Chapter 2: Integration methods
Section 10: Integration by inverse substitution by using secant
Page 2
Summary
 To deal with integrands that contain a square root of the form
b2 x 2  a 2 , we use the inverse substitution bx  a sec , bdx  a sec tan  d and hope that the
resulting integral can be computed.
 To return to the original variable, standard trigonometric identities and a suitable triangle model allow us to obtain an elegant formula for the integral.
Common errors to avoid
 Watch the algebra as you do the substitutions, both ways.
 Use the triangle model to return to the original variable, rather than constructing convoluted compositions of trig functions and their inverses.
Learning questions for Section I 2-10
Review questions:
1. Describe when and how to use the method of integration by inverse substitution by using the secant function.
Memory questions:
1. Which inverse substitution is used to integrate functions containing a factor of the form
Integral Calculus
Chapter 2: Integration methods
b2 x2  a2 ?
Section 10: Integration by inverse substitution by using secant
Page 3
Computation questions:
Evaluate the integrals proposed in questions 1-6 by using a suitable inverse substitution.
1.

2.

dx
x2  6 x
x3
x2  9
dx
3.

4.

x
x2  4x  1
x2
x2  a2

3/2
dx
dx
5.
  2x  4x 1
6.

dx
2
e x dx
3/2

2e2 x  4e x  1
3/2
What questions do you have for your instructor?
Integral Calculus
Chapter 2: Integration methods
Section 10: Integration by inverse substitution by using secant
Page 4