This document covers the evaluation of improper integrals and trigonometric substitutions. It describes techniques such as inverse substitutions, the p-test for convergence, and the comparison test, providing examples for clarity. It also includes a summary of methods for determining whether certain integrals converge or diverge.

1. Section 6.3: Inverse Substitutions
Section 6.5: Improper Integrals
1

2. Week 5 Learning Objectives
— Evaluate integrals involving square roots of quadratic
expressions using:
— Appropriate trigonometric substitutions
— Completion of the square
— Decide if a given definite integral is improper.
— Evaluating improper integrals by finding the suitable
limit(s).
— Use p-test to decide whether an improper integral is
convergent or divergent
— Use the comparison test to decide whether an improper
integral is convergent or divergent.
2

3. Example: Evaluate
ò -
2
/
1
3
/
1
2
2
1
1
dx
x
x
Can’t be solved using
the techniques seen earlier
Goal: Get rid of the radical
3
Recall: 1 − sin& ' = cos& '

4. Example Continued
4
Trick:
Let ! = sin &, for −
,
-
≤ & ≤
,
-
Thus:
This is called the inverse
substitution which can be
made provided &ℎ0& sin & is
a one-to-one function.
Therefore, we have to restrict t
to the interval −
,
-
,
,
-
t
t
t
t
x cos
cos
cos
sin
1
1 2
2
2
=
=
=
-
=
-
Since cos & is positive
on −
,
-
,
,
-

5. Example Continued
— Thus the integral reduces to
!
"
#
"
$
1
&$ 1 − &$
(& =
Since & = sin - ,
(& = cos - (-
= !
123456
"
#
7
8
csc$ & (& = 9
− cot &
123456
"
#
7
8
= − cot
;
6
+ cot(arcsin 1/3)
5
When & = 1/3, then 1/3 = sin -,
That is for what angle t, sine is 1/3?
Take arcsine of both sides to get:
arcsin 1/3 = arcsin(sin -) which yields:
When & =
"
$
then
7
8
= sin -,
arcsin 1/2 = arcsin(sin -) which
yields:
!
123456
"
#
7
8
1
sin$ - cos -
cos - (-

6. 6
Example

7. 7
Example

8. Example: Evaluate
Trick: Let ! = 2tan ', )*+ − /2 < ' < p/2.
Thus:
dx
x
ò + 2
4
1
4 + !2 = 4 + 4 tan2 ' = 2| sec '| = 2 sec '
8
1 + tan2 8 = sec2 8

9. The reference Triangle
This substitution can be realized as follows:
tan $ =
&
2
And
sec $ =
4 + &-
2
So, ∫
/
0123
4& = ∫
- 5673 8
- 567 8
4$ = ln | sec $ + tan $| + ;
Since x = 2 tan $
4& = 2 sec- $4$
Note: The reference triangle becomes very useful in back-substituting
Back-substituting: ∫
/
0123
4& = ln
0123
-
+
2
-
+ ;.
9

10. 10
Examples

11. 11

12. Example: Evaluate
Trick: Let 2" = sec ', )*+ 0 < ' < p/2.
Thus:
ò -
2
1
2
1
4
1
dx
x
t
t
t
t
x tan
tan
tan
1
sec
1
4 2
2
2
=
=
=
-
=
-
12
sec/ ' − 1 = tan/ '

13. Example Continued
When % = 1, then 2 1 = sec -, so - =
/
3
When % = 2, then 2 2 = sec -, so - = arcsec 4 13
-
2x
1
4%4 − 1
6
7
4
1
4%4 − 1
8% =
1
2
6
9
:
;<=>?= @
1
tan -
sec - tan - 8-
= A
1
2
ln | sec - + tan - |
9
:
;<=>?= @
=
1
2
ln 4 + 15 − ln 2 + 3

14. Summary of Trig Substitutions:
14

15. Your Turn: Evaluate
Trick: First complete the square, and use x - 2 = tan t.
ò +
-
3
1
2
5
4
1
dx
x
x
1
)
2
(
1
4
4
5
4 2
2
2
+
-
=
+
+
-
=
+
- x
x
x
x
x
15
Solution:
!
"
#
1
%& − 4% + 5
+% = ln 2 + 1 − ln 2 − 1

16. 16
Example

17. 17

18. Improper Integrals
Example 1:
Is the area bounded above by the curve: 2
1
x
y =
below by the x-axis, and to the left by ! = 1, finite?
Let A(t) denote the area of the region bounded
by % = 1/!' below the x-axis and between x =1 and x=t
Then, ( ) =
∫
+
, +
-. /! = 0
−
+
- +
,
= 1 −
+
,
18

19. dx
x
dx
x
b
b ò
ò ¥
®
¥
=
1
2
1
2
1
lim
1
If we consider different values of t, we observe that as t gets larger
larger, the area gets closer and closer to 1.
Therefore, the area of the shaded region as t tends to
infinity is given by
19

20. Type I Improper Integral
20

21. Example 2:
21
!
"
#
1
%
&% = lim
+→#
!
"
+
1
%
&% = lim
+→#
-
ln %
"
+
= lim
+→#
(ln 0 − ln 1) = ∞
Determine whether following improper integral
is divergent or convergent:
Therefore, The improper integral
is divergent.

22. Example 3:
Evaluate:
dx
x
ò
¥
¥
-
+1
1
2
What does this integral represent?
22

23. The area of the infinite region
under the curve of
y=
"
#$%"
and above the / − 1/23.
23

24. How to compute this area?
∫
"#
# $
%&'$
() = ∫
"#
* $
%&'$
() + ∫
*
# $
%&'$
()
∫
"#
* $
%&'$
= lim
0→"#
∫
0
* $
%&'$
()
lim
0→"#
2
arctan )
0
*
=
lim
0→"#
(arctan 0 − arctan ;) =
=
>
.
lim
0→#
@
*
0
1
)> + 1
()
= lim
0→#
2
arctan )
*
0
= lim
0→#
(arctan ; − BCD;BE0) =
F
2
Therefore, ∫
"#
# $
%&'$
() =
=
>
+
=
>
= F
24

25. Your Turn!
— Determine whether following improper integral is divergent or
convergent:
!
"#
$
%&'(%
25
Hint: use integration by parts to evaluate the integral
Solution: ∫
"#
$
%&'(% = −1. The improper integral is convergent.
!
"#
$
%&'(% = lim
>→"#
!
>
$
%&' (%
@ = % (A = &'(%
(@ = (%. A = &'
! %&'(% = %&' − ∫ &'(%
= lim
>→"#
B
(%&' − &')
>
$
= lim
>→"#
(−1 − E&> + &>) Note: lim
>→"#
E&> = lim
>→"#
E/&">= lim
>→"#
1/e"H=0

26. Example 4
— For what values of p is the the integral ∫
"
# "
$% &'
convergent?
26
We have seen already for the case where p=1, the improper integral is divergent.
So, let us assume that ( ≠ 1. Then,
,
"
#
1
'- &' = lim
2→#
,
"
2
'4-&' = lim
2→#
5
'4-6"
−( + 1
"
2
= lim
2→#
1
1 − (
1
9-4" − 1
Power Rule
for integration By the FTC (II)

27. — If ! > #, then $ − 1 > 0, so as ( → ∞, (+,-
→ ∞ and
-
./01 → 0. Therefore,
3
-
4
1
5+
65 = lim
.→4
1
1 − $
1
(+,-
− 1 =
1
1 − $
−1 =
1
$ − 1
.
So the improper integral is convergent for ! > #.
— If ! < #, then $ − 1 < 0, so as ( → ∞,
-
./01 = (-,+
→ ∞,
so the the improper integral is divergent.
27

28. To Summarize
— The improper integral p-test:
For ! > 0,
28
∫
%
& '
() *+ is convergent if , > 1 and divergent if , ≤ 1.
Note: You can use the p-test for integrals of the above form.
However, you must specify the value of p.

29. Example 5:
— Evaluate the following integral:
!
"
#
$ + 1 "
$'
($
Clearly, we can write
)*+ ,
)-
=
),*")*+
)-
= $
/
, + 2$1
/
, + $1
-
,.
Thus, ∫
"
# )*+ ,
)-
($ = ∫
"
# +
) 4
5/
,
+
"
)//, +
+
)-/, ($
= !
"
#
1
$ 4
1+
"
($ + !
"
#
2
$+/"
($ + !
"
#
1
$'/"
($
29
P=-1/2<1,
the integral is divergent P=1/2<1,
The integral is divergent
P=3/2>1,
The integral is
convergent
Conclusion: The integral is divergent.
Note: if one of the improper integrals diverges, then the whole integral is divergent.

30. Example 6:
— Find the area in the 4th quadrant bounded between
the x-axis, the y-axis, and the curve y = ln x.
That is, we need to evaluate the following integral:
!
"
#
| ln '| ('
PROBLEM! At ' = 0, , = ln ' has a vertical asymptote!
This takes us to Type II Improper integrals.
30

31. Type II Improper Integrals
31

32. Back to Example 5:
32
!
"
#
|ln '| (' = lim
,→".
− !
,
#
ln ' ('
Integrating by parts, we let
0 = ln ' (1 = ('
(0 =
#
2
(' 1 = '
Thus,
lim
,→".
− ∫
,
#
ln ' (' = lim
,→".
|
−['56 ' ,
# − ∫
,
#
'
#
2
(']
lim
,→".
∫
,
#
ln ' (' = lim
,→".
|
−['56 ' − '] ,
#
= lim
,→".
−(1 ln 1 − 1 − : ln : + :) = lim
,→".
− (: − : ln : − 1)
Recall: lim
,→".
: ln : = “0 ? ∞” (An indeterminate form). To evaluate this limit, rewrite
lim
,→".
: ln : = lim
,→".
BC ,
#/,
= lim
,→".
#/,
E#/,F = lim
,→".
−: = 0.
H
Since ln ' < 0 on the interval (0,1]

33. — Thus,
!
"
#
− ln ' (' = lim
,→".
−(0 − 0 ln 0 − 1) = 0 − 0 − 1 = 1.
Therefore, the area in the 4th quadrant bounded between
the x-axis, the y-axis, and the curve y = ln x is 1.
33
Note: We can also say that the improper integral ∫
"
#
| ln '| (' is convergent.

34. Example 6
— Determine whether ∫
"
#
$ sec ( )( converges or diverges.
— We know that sec x is improper at x =
*
+
, thus,
,
"
*
+
sec ( )( = lim
1→
*
+
3 ,
"
1
sec ( )(
= lim
1→
#
$
3 ln 5
sec ( + tan ( "
9
= lim
1→
#
$
3 ln sec t + tan t − ln sec 0 + tan 0 = ∞
34
Recall: lim
1→
#
$
3 sec = = ∞
and
lim
1→
*
+
3tan t = ∞

35. Is the integral: dx
x
ò -
3
0
3
/
2
)
1
(
1
convergent?
If yes, find its value.
Your Turn:
Hint: The function has a vertical asymptote at x = 1. So, write
∫
"
# $
%&$ '/)dx= ∫
"
$ $
%&$ '/)dx+ ∫
$
# $
%&$ '/)dx
35

36. Partial Solution:
36

37. A Comparison Test for Improper
Integral
— When we are presented with an improper integral, one of
the main questions we need to answer is whether the
integral is convergent or divergent. Consequently, we have a
test to determine the convergence or divergence of an
improper integral.
37
Note: The comparison test is very useful when the improper integral is
impossible to evaluate.
A similar theorem applies for
Type II improper integrals. (See example 9)

38. Example 7
— Show that ∫
"
#
$%&'
() is convergent.
— Clearly, we cannot evaluate this integral directly.
Therefore, trying to compare it to an integral that is
easily evaluated is the best approach to answering this
question.
— ∫
"
#
$%&'
()= ∫
"
*
$%&'
()+ ∫
*
#
$%&'
()
— ∫
"
*
$%&'
() = Number (it is a definite integral)
— To evaluate ∫
*
#
$%&'
(), we observe that )2
≥ ) for all
) ≥ 1. Therefore, −)2 ≤ ) and thus $%&'
≤ $%&.
38

39. Example Continued
— The integral of !"#
is easy to evaluate:
— ∫
%
&
!"#
'( = lim
-→&
∫
%
-
!"#
'( = lim
-→&
|
−!"#
%
-
— = lim
-→&
|
−!"#
%
-
= lim
-→&
−!"-
+ !"%
= !"%
.
— Since !"#3
≤ !"#
then ∫
%
&
!"#3
'( is convergent, thus
— ∫
5
&
!"#3
'( is also convergent by the comparison test.
39
Convergent

40. Example 8
— Determine whether the following integral is divergent or
convergent:
!
"
#
1 + &'(
)
*)
— We observe that 1 + &'(
> 1, that is
",-./
(
>
"
(
when ) ≥ 1.
— In addition, we know that ∫
"
# "
(
*) is divergent by the p-test,
as 7 = 1.
— Thus, by the comparison test ∫
"
# ",-./
(
*) is divergent.
40

41. Example 9
— Determine whether ∫
"
# $
%&%'
() is convergent or
divergent.
— Notice that this is an improper integral of type I
(upper limit is infinity) and type II (
$
%&%'
has a
vertical asymptote at ) = 0).
— Thus, we need to split up the integral:
∫
"
$ $
%&%'
() + ∫
$
# $
%&%'
() = -$ + -.
— Since ) + )/
> ), on 0,1 , then ) + )/ > ), thus,
1
) + )/
<
1
) 41

42. — Therefore, !" < ∫
%
" "
&
'( = 2
— On 1, ∞ , we observe that ( + (9 > (9, so
( + (9 > (9, that is
1
( + (9
<
1
(9
— Therefore, !< < ∫
"
= "
&
>
?
'(
— Since !" and !< both converge,
Then, ∫
%
= "
>
'( is convergent. 42
Since
"
&@&>
< 1/ (, then by the
Comparison Test, !" is convergent
as ∫
%
"
1/ ( '( is convergent.
Since
"
&@&>
< 1/ (9, then by the
Comparison Test, !< is convergent
as ∫
"
= "
&>/? '( is convergent, F =
9
<
> 1.