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# Differential and Integral Calculus (Part - 4) CA Foundation Notes | EduRev

## CA Foundation : Differential and Integral Calculus (Part - 4) CA Foundation Notes | EduRev

``` 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 dicult 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
1u
2
. 1=
p
1 u
2
is still a composite function that is dicult 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 dicult 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
1u
2
. 1=
p
1 u
2
is still a composite function that is dicult 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 (difult 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 dicult 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
1u
2
. 1=
p
1 u
2
is still a composite function that is dicult 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 (difult 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 (difult 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 dicult 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
1u
2
. 1=
p
1 u
2
is still a composite function that is dicult 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 (difult 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 (difult 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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