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Page 1
13 Integration by Multiple Techniques
Whenever we encounter a complex integral, often the rst thing to do is to use substitution to
eliminate the composite function(s).
Example: I =
R
x
3
e
1=x
dx:
Solution: Note that e
1=x
is a composite fucntion. The substitution to change it into a simple,
elementary function is u = 1=x, du = u
0
dx = x
2
dx or x
2
dx = du = d( 1=x). Thus,
I =
Z
x
3
e
1=x
dx =
Z
(1=x)e
1=x
d( 1=x) =
Z
ue
u
du = [ue
u
e
u
] + C = e
1=x
[1 1=x] + C:
Page 2
13 Integration by Multiple Techniques
Whenever we encounter a complex integral, often the rst thing to do is to use substitution to
eliminate the composite function(s).
Example: I =
R
x
3
e
1=x
dx:
Solution: Note that e
1=x
is a composite fucntion. The substitution to change it into a simple,
elementary function is u = 1=x, du = u
0
dx = x
2
dx or x
2
dx = du = d( 1=x). Thus,
I =
Z
x
3
e
1=x
dx =
Z
(1=x)e
1=x
d( 1=x) =
Z
ue
u
du = [ue
u
e
u
] + C = e
1=x
[1 1=x] + C:
Example: I =
R
cos(
p
x)dx:
Solution: Note that cos(
p
x) is a composite fucntion. The substitution to change it into a
simple, elementary function is u =
p
x, du = u
0
dx =
1
2
p
x
dx or dx = 2
p
xdu = 2udu. However,
note that x = (
p
x)
2
,
I =
Z
cos(
p
x)dx =
Z
cos(
p
x)d(
p
x)
2
=
Z
cos(u)du
2
= 2
Z
ucos(u)du
= 2
Z
udsin(u) = 2[usin(u) + cos(u)] + C = 2[
p
xsin(
p
x) + cos(
p
x)] + C:
Example: I =
R
sec(x)dx =
R
(1=cos(x))dx:
Solution: This is a dicult integral. Note that 1=cos(x) is a composite fucntion. One substitu-
tion to change it into a simple, elementary function is u = cos(x). However, x = cos
1
(u) and
dx = x
0
du =
du
p
1 u
2
. 1=
p
1 u
2
is still a composite function that is dicult to deal with. Here
is how we can deal with it. It involves trigonometric substitution followed by partial fractions.
I =
Z
dx
cos(x)
=
Z
cos(x)dx
cos
2
(x)
=
Z
dsin(x)
1 sin
2
(x)
=
Z
du
1 u
2
=
Z
[
1
u 1
1
u + 1
]
du
2
= ln
r
j
u + 1
u 1
j + C = lnj
u + 1
p
1 u
2
j + C = lnj
sin(x) + 1
cos(x)
j + C = lnjtan(x) + sec(x)j + C;
where u = sin(x) and
p
1 u
2
= cos(x) were used!
More Exercises on Integration by Multiple techniques:
1. Substitution followed by Integration by Parts
Example:
(i)
Z
x
3
sin(x
2
)dx =
1
2
Z
x
2
sin(x
2
)dx
2
=
1
2
Z
u sin udu =
1
2
Z
ud cosu
=
1
2
[sin u u cosu] + C =
1
2
[sin(x
2
) x
2
cos(x
2
)] + C:
Page 3
13 Integration by Multiple Techniques
Whenever we encounter a complex integral, often the rst thing to do is to use substitution to
eliminate the composite function(s).
Example: I =
R
x
3
e
1=x
dx:
Solution: Note that e
1=x
is a composite fucntion. The substitution to change it into a simple,
elementary function is u = 1=x, du = u
0
dx = x
2
dx or x
2
dx = du = d( 1=x). Thus,
I =
Z
x
3
e
1=x
dx =
Z
(1=x)e
1=x
d( 1=x) =
Z
ue
u
du = [ue
u
e
u
] + C = e
1=x
[1 1=x] + C:
Example: I =
R
cos(
p
x)dx:
Solution: Note that cos(
p
x) is a composite fucntion. The substitution to change it into a
simple, elementary function is u =
p
x, du = u
0
dx =
1
2
p
x
dx or dx = 2
p
xdu = 2udu. However,
note that x = (
p
x)
2
,
I =
Z
cos(
p
x)dx =
Z
cos(
p
x)d(
p
x)
2
=
Z
cos(u)du
2
= 2
Z
ucos(u)du
= 2
Z
udsin(u) = 2[usin(u) + cos(u)] + C = 2[
p
xsin(
p
x) + cos(
p
x)] + C:
Example: I =
R
sec(x)dx =
R
(1=cos(x))dx:
Solution: This is a dicult integral. Note that 1=cos(x) is a composite fucntion. One substitu-
tion to change it into a simple, elementary function is u = cos(x). However, x = cos
1
(u) and
dx = x
0
du =
du
p
1 u
2
. 1=
p
1 u
2
is still a composite function that is dicult to deal with. Here
is how we can deal with it. It involves trigonometric substitution followed by partial fractions.
I =
Z
dx
cos(x)
=
Z
cos(x)dx
cos
2
(x)
=
Z
dsin(x)
1 sin
2
(x)
=
Z
du
1 u
2
=
Z
[
1
u 1
1
u + 1
]
du
2
= ln
r
j
u + 1
u 1
j + C = lnj
u + 1
p
1 u
2
j + C = lnj
sin(x) + 1
cos(x)
j + C = lnjtan(x) + sec(x)j + C;
where u = sin(x) and
p
1 u
2
= cos(x) were used!
More Exercises on Integration by Multiple techniques:
1. Substitution followed by Integration by Parts
Example:
(i)
Z
x
3
sin(x
2
)dx =
1
2
Z
x
2
sin(x
2
)dx
2
=
1
2
Z
u sin udu =
1
2
Z
ud cosu
=
1
2
[sin u u cosu] + C =
1
2
[sin(x
2
) x
2
cos(x
2
)] + C:
(ii)
Z
xtan
1
p
xdx =
Z
(
p
x)
2
tan
1
p
xd(
p
x)
2
=
Z
u
2
tan
1
udu
2
=
1
2
Z
tan
1
udu
4
=
1
2
[u
4
tan
1
u
Z
u
4
dtan
1
u] =
1
2
[u
4
tan
1
u
Z
u
4
1 + 1
1 + u
2
du] =
1
2
[u
4
tan
1
u
Z
(u
2
1+
1
1 + u
2
)du]
=
1
2
[u
4
tan
1
u
u
3
3
+ u tan
1
u] + C =
1
2
[(x
2
1)tan
1
p
x
x
3=2
3
+
p
x] + C:
Edwards/Penney 5
th
ed 9.3 Problems.
(15)
Z
x
p
x + 3dx; u = x + 3 (17)
Z
x
5
p
x
3
+ 1dx; u = x
3
+ 1
(23)
Z
sec
1
p
xdx; u =
p
x (29)
Z
x
3
cos(x
2
)dx; u = x
2
(25)
Z
tan
1
p
xdx; u =
p
x
(31)
Z
ln x
x
p
x
dx; u =
p
x (37)
Z
e
p
x
dx; u =
p
x
2. Substitution followed by Integration by Partial Fractions
Example:
(i)
Z
cos xdx
sin
2
x sin x 6
=
Z
d sin x
sin
2
x sin x 6
=
Z
du
(u 3)(u + 2)
=
1
5
lnj
u 3
u + 2
j+C =
1
5
lnj
sin x 3
sin x + 2
j+C
Edwards/Penney 5
th
ed 9.5 Problems (difult ones!)
(40)
Z
sec
2
t
tan
3
t + tan
2
t
dt (37)
Z
e
4t
(e
2t
1)
3
dt (39)
Z
1 + ln t
t(3 + 2 ln t)
2
dt
2. Integration by Parts followed by Partial Fractions
Page 4
13 Integration by Multiple Techniques
Whenever we encounter a complex integral, often the rst thing to do is to use substitution to
eliminate the composite function(s).
Example: I =
R
x
3
e
1=x
dx:
Solution: Note that e
1=x
is a composite fucntion. The substitution to change it into a simple,
elementary function is u = 1=x, du = u
0
dx = x
2
dx or x
2
dx = du = d( 1=x). Thus,
I =
Z
x
3
e
1=x
dx =
Z
(1=x)e
1=x
d( 1=x) =
Z
ue
u
du = [ue
u
e
u
] + C = e
1=x
[1 1=x] + C:
Example: I =
R
cos(
p
x)dx:
Solution: Note that cos(
p
x) is a composite fucntion. The substitution to change it into a
simple, elementary function is u =
p
x, du = u
0
dx =
1
2
p
x
dx or dx = 2
p
xdu = 2udu. However,
note that x = (
p
x)
2
,
I =
Z
cos(
p
x)dx =
Z
cos(
p
x)d(
p
x)
2
=
Z
cos(u)du
2
= 2
Z
ucos(u)du
= 2
Z
udsin(u) = 2[usin(u) + cos(u)] + C = 2[
p
xsin(
p
x) + cos(
p
x)] + C:
Example: I =
R
sec(x)dx =
R
(1=cos(x))dx:
Solution: This is a dicult integral. Note that 1=cos(x) is a composite fucntion. One substitu-
tion to change it into a simple, elementary function is u = cos(x). However, x = cos
1
(u) and
dx = x
0
du =
du
p
1 u
2
. 1=
p
1 u
2
is still a composite function that is dicult to deal with. Here
is how we can deal with it. It involves trigonometric substitution followed by partial fractions.
I =
Z
dx
cos(x)
=
Z
cos(x)dx
cos
2
(x)
=
Z
dsin(x)
1 sin
2
(x)
=
Z
du
1 u
2
=
Z
[
1
u 1
1
u + 1
]
du
2
= ln
r
j
u + 1
u 1
j + C = lnj
u + 1
p
1 u
2
j + C = lnj
sin(x) + 1
cos(x)
j + C = lnjtan(x) + sec(x)j + C;
where u = sin(x) and
p
1 u
2
= cos(x) were used!
More Exercises on Integration by Multiple techniques:
1. Substitution followed by Integration by Parts
Example:
(i)
Z
x
3
sin(x
2
)dx =
1
2
Z
x
2
sin(x
2
)dx
2
=
1
2
Z
u sin udu =
1
2
Z
ud cosu
=
1
2
[sin u u cosu] + C =
1
2
[sin(x
2
) x
2
cos(x
2
)] + C:
(ii)
Z
xtan
1
p
xdx =
Z
(
p
x)
2
tan
1
p
xd(
p
x)
2
=
Z
u
2
tan
1
udu
2
=
1
2
Z
tan
1
udu
4
=
1
2
[u
4
tan
1
u
Z
u
4
dtan
1
u] =
1
2
[u
4
tan
1
u
Z
u
4
1 + 1
1 + u
2
du] =
1
2
[u
4
tan
1
u
Z
(u
2
1+
1
1 + u
2
)du]
=
1
2
[u
4
tan
1
u
u
3
3
+ u tan
1
u] + C =
1
2
[(x
2
1)tan
1
p
x
x
3=2
3
+
p
x] + C:
Edwards/Penney 5
th
ed 9.3 Problems.
(15)
Z
x
p
x + 3dx; u = x + 3 (17)
Z
x
5
p
x
3
+ 1dx; u = x
3
+ 1
(23)
Z
sec
1
p
xdx; u =
p
x (29)
Z
x
3
cos(x
2
)dx; u = x
2
(25)
Z
tan
1
p
xdx; u =
p
x
(31)
Z
ln x
x
p
x
dx; u =
p
x (37)
Z
e
p
x
dx; u =
p
x
2. Substitution followed by Integration by Partial Fractions
Example:
(i)
Z
cos xdx
sin
2
x sin x 6
=
Z
d sin x
sin
2
x sin x 6
=
Z
du
(u 3)(u + 2)
=
1
5
lnj
u 3
u + 2
j+C =
1
5
lnj
sin x 3
sin x + 2
j+C
Edwards/Penney 5
th
ed 9.5 Problems (difult ones!)
(40)
Z
sec
2
t
tan
3
t + tan
2
t
dt (37)
Z
e
4t
(e
2t
1)
3
dt (39)
Z
1 + ln t
t(3 + 2 ln t)
2
dt
2. Integration by Parts followed by Partial Fractions
Example:
(i)
Z
(2x + 2) tan
1
xdx =
Z
tan
1
xd(x
2
+ 2x) = (x
2
+ 2x) tan
1
x
Z
(x
2
+ 2x)d tan
1
x
= (x
2
+ 2x) tan
1
x
Z
1 + x
2
+ 2x 1
1 + x
2
dx = (x
2
+ 2x) tan
1
x
Z
[dx +
d(1 + x
2
)
1 + x
2
dx
1 + x
2
]
= (x
2
+ 2x) tan
1
x x ln(1 + x
2
) + tan
1
x + C = (x + 1)
2
tan
1
x x ln(1 + x
2
) + C:
Li's Problems (difult ones!)
(Li)
Z
(2t + 1) lntdt
Page 5
13 Integration by Multiple Techniques
Whenever we encounter a complex integral, often the rst thing to do is to use substitution to
eliminate the composite function(s).
Example: I =
R
x
3
e
1=x
dx:
Solution: Note that e
1=x
is a composite fucntion. The substitution to change it into a simple,
elementary function is u = 1=x, du = u
0
dx = x
2
dx or x
2
dx = du = d( 1=x). Thus,
I =
Z
x
3
e
1=x
dx =
Z
(1=x)e
1=x
d( 1=x) =
Z
ue
u
du = [ue
u
e
u
] + C = e
1=x
[1 1=x] + C:
Example: I =
R
cos(
p
x)dx:
Solution: Note that cos(
p
x) is a composite fucntion. The substitution to change it into a
simple, elementary function is u =
p
x, du = u
0
dx =
1
2
p
x
dx or dx = 2
p
xdu = 2udu. However,
note that x = (
p
x)
2
,
I =
Z
cos(
p
x)dx =
Z
cos(
p
x)d(
p
x)
2
=
Z
cos(u)du
2
= 2
Z
ucos(u)du
= 2
Z
udsin(u) = 2[usin(u) + cos(u)] + C = 2[
p
xsin(
p
x) + cos(
p
x)] + C:
Example: I =
R
sec(x)dx =
R
(1=cos(x))dx:
Solution: This is a dicult integral. Note that 1=cos(x) is a composite fucntion. One substitu-
tion to change it into a simple, elementary function is u = cos(x). However, x = cos
1
(u) and
dx = x
0
du =
du
p
1 u
2
. 1=
p
1 u
2
is still a composite function that is dicult to deal with. Here
is how we can deal with it. It involves trigonometric substitution followed by partial fractions.
I =
Z
dx
cos(x)
=
Z
cos(x)dx
cos
2
(x)
=
Z
dsin(x)
1 sin
2
(x)
=
Z
du
1 u
2
=
Z
[
1
u 1
1
u + 1
]
du
2
= ln
r
j
u + 1
u 1
j + C = lnj
u + 1
p
1 u
2
j + C = lnj
sin(x) + 1
cos(x)
j + C = lnjtan(x) + sec(x)j + C;
where u = sin(x) and
p
1 u
2
= cos(x) were used!
More Exercises on Integration by Multiple techniques:
1. Substitution followed by Integration by Parts
Example:
(i)
Z
x
3
sin(x
2
)dx =
1
2
Z
x
2
sin(x
2
)dx
2
=
1
2
Z
u sin udu =
1
2
Z
ud cosu
=
1
2
[sin u u cosu] + C =
1
2
[sin(x
2
) x
2
cos(x
2
)] + C:
(ii)
Z
xtan
1
p
xdx =
Z
(
p
x)
2
tan
1
p
xd(
p
x)
2
=
Z
u
2
tan
1
udu
2
=
1
2
Z
tan
1
udu
4
=
1
2
[u
4
tan
1
u
Z
u
4
dtan
1
u] =
1
2
[u
4
tan
1
u
Z
u
4
1 + 1
1 + u
2
du] =
1
2
[u
4
tan
1
u
Z
(u
2
1+
1
1 + u
2
)du]
=
1
2
[u
4
tan
1
u
u
3
3
+ u tan
1
u] + C =
1
2
[(x
2
1)tan
1
p
x
x
3=2
3
+
p
x] + C:
Edwards/Penney 5
th
ed 9.3 Problems.
(15)
Z
x
p
x + 3dx; u = x + 3 (17)
Z
x
5
p
x
3
+ 1dx; u = x
3
+ 1
(23)
Z
sec
1
p
xdx; u =
p
x (29)
Z
x
3
cos(x
2
)dx; u = x
2
(25)
Z
tan
1
p
xdx; u =
p
x
(31)
Z
ln x
x
p
x
dx; u =
p
x (37)
Z
e
p
x
dx; u =
p
x
2. Substitution followed by Integration by Partial Fractions
Example:
(i)
Z
cos xdx
sin
2
x sin x 6
=
Z
d sin x
sin
2
x sin x 6
=
Z
du
(u 3)(u + 2)
=
1
5
lnj
u 3
u + 2
j+C =
1
5
lnj
sin x 3
sin x + 2
j+C
Edwards/Penney 5
th
ed 9.5 Problems (difult ones!)
(40)
Z
sec
2
t
tan
3
t + tan
2
t
dt (37)
Z
e
4t
(e
2t
1)
3
dt (39)
Z
1 + ln t
t(3 + 2 ln t)
2
dt
2. Integration by Parts followed by Partial Fractions
Example:
(i)
Z
(2x + 2) tan
1
xdx =
Z
tan
1
xd(x
2
+ 2x) = (x
2
+ 2x) tan
1
x
Z
(x
2
+ 2x)d tan
1
x
= (x
2
+ 2x) tan
1
x
Z
1 + x
2
+ 2x 1
1 + x
2
dx = (x
2
+ 2x) tan
1
x
Z
[dx +
d(1 + x
2
)
1 + x
2
dx
1 + x
2
]
= (x
2
+ 2x) tan
1
x x ln(1 + x
2
) + tan
1
x + C = (x + 1)
2
tan
1
x x ln(1 + x
2
) + C:
Li's Problems (difult ones!)
(Li)
Z
(2t + 1) lntdt
14 Applications of Integrals
The central idea in all sorts of application problems involving integrals is breaking the whole
into pieces and then adding the pieces up to obtain the whole.
14.1 Area under a curve.
We know the area of a rectangle is:
Area = height width;
(see Fig. 1(a)). This formula can not be applied to the area under the curve in Fig. 1(b)
because the top side of the area is NOT a straight, horizontal line. However, if we divide the
area into in nitely many rectangles with in nitely thin width dx, the in nitely small area of
the rectangle located at x is
dA = height infintely small width = f(x)dx:
Adding up the area of all the thin rectangles, we obtain the area under the curve
Area = A(b) A(a) =
A(b)
Z
0
dA =
b
Z
a
f(x)dx;
where A(x) =
R
x
a
f(s)ds is the area under the curve between a and x.
14.2 Volume of a solid.
We know the volume of a cylinder is
V olume = (cross section area) length;
(see Fig. 2(a)). This formula does not apply to the volume of the solid obtained by rotating
a curve around the x-axis (see Fig. 2(b)). This is because the cross-section area is NOT a
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FAQs on ICAI Notes: Differential and Integral Calculus- 4 - Mathematics for GRE Paper II
| 1. What are the applications of differential calculus? |  |
Ans. Differential calculus has various applications in fields like physics, engineering, economics, and computer science. It is used to determine rates of change, optimize functions, analyze motion, and solve optimization problems, among other things.
| 2. How is integral calculus used in real life? | |
Ans. Integral calculus is used in real-life applications such as calculating areas and volumes, determining population growth, analyzing fluid flow, and solving problems related to accumulation and averages.
| 3. What is the fundamental theorem of calculus? | |
Ans. The fundamental theorem of calculus states that the derivative and integral are inverse operations of each other. It establishes the connection between the concept of differentiation and integration and forms the basis of integral calculus.
| 4. What is the difference between differential and integral calculus? | |
Ans. Differential calculus deals with the study of rates of change and slope of curves, while integral calculus focuses on finding areas, volumes, and accumulation of quantities. Differential calculus involves differentiation, while integral calculus involves integration.
| 5. Can you give an example of how differential and integral calculus are used together? | |
Ans. One example of how differential and integral calculus are used together is in solving optimization problems. Differential calculus is used to find the critical points and determine whether they correspond to maximum or minimum values. Integral calculus is then used to calculate the optimal value by finding the area or volume under the curve.
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