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8. Prove the Mean Value Theorem, Theorem 5.2.6.
(Hint: Let
m = absolute minimum of f on [a, b];
M = absolute minimum of f on [a, b];
b
1
f (x) dx;
fav [a, b] =
b’a a
b
m(b ’ a) ¤ f (x)dx ¤ M (b ’ a).
a

Then m ¤ fav [a, b] ¤ M . By the intermediate value theorem for contin-
uous functions, there exists some c on [a, b] such that f (c) = fav [a, b].)
9. Prove the Fundamental Theorem of Calculus, First Form, Theorem 5.2.6.
(Hint:
g(x + h) ’ g(x)
g (x) = lim
h
h’0
x+h x
1
f (t)dt ’ f (t)dt
= lim
h
h’0 a a
x x+h x
1
f (t)dt ’
= lim f (t) dx + f (t)dt
h
h’0 a x a
x+h
1
= lim f (t)dt
h
h’0 x
f (c), (for some c, x ¤ c ¤ x + h; )
= lim
h’0
= f (x)
where x ¤ c ¤ x + h, by Theorem 5.2.6.)
10. Prove the Leibniz Rule, Theorem 5.2.8.
(Hint:
β(x) β(x) ±(x)
f (t)dt ’
f (t)dt = f (t)dt
±(x) a a

for some a. Now use the chain rule of di¬erentiation.)
250 CHAPTER 5. THE DEFINITE INTEGRAL

11. Prove that if f and g are continuous on [a, b] and g is nonnegative, then
there is a number c in (a, b) for which
b b
f (x)g(x) dx = f (c) g(x) dx.
a a

(Hint: If m and M are the absolute minimum and absolute maximum of
f on [a, b], then mg(x) ¤ f (x)g(x) ¤ M g(x). By the Order Property,
b b b
g(x) dx ¤ f (x)g(x) ¤ M
m g(x) dx
a a a
b b
f (x)g(x) dx
a
m¤ ¤M if g(x) dx = 0 .
b
g(x) dx 0
a

By the Intermediate Value Theorem, there is some c such that
b
f (x)g(x) dx
a
f (x) = or
b
g(x) dx
a
b b
f (x)g(x) dx = f (c) g(x) dx.
a a
b
g(x) dx = 0, then g(x) ≡ 0 on [a, b] and all integrals are zero.)
If
a


Remark 20 The number f (c) is called the weighted average of f on [a, b]
with respect to the weight function g.



5.7 Volumes of Revolution
One simple application of the Riemann integral is to de¬ne the volume of a
solid.

Theorem 5.7.1 Suppose that a solid is bounded by the planes with equations
x = a and x = b. Let the cross-sectional area perpendicular to the x-axis at
x be given by a continuous function A(x). Then the volume V of the solid is
given by
b
V= A(x) dx.
a
5.7. VOLUMES OF REVOLUTION 251

Proof. Let P = {a = x0 < x1 < x2 < · · · < xn = b} be a partition of [a, b].
For each i = 1, 2, 3, · · · , n, let
Vi = volume of the solid between the planes with equations x = xi’1 and
x = xi ,
mi = absolute minimum of A(x) on [xi’1 , xi ],
Mi = absolute maximum of A(x) on [xi’1 , xi ],
∆xi = xi ’ xi’1 .
Then
Vi
¤ Mi .
mi ∆xi ¤ Vi ¤ Mi ∆xi , mi ¤
∆xi
Since A(x) is continuous, there exists some ci such that xi’1 ¤ ci ¤ Mi and
Vi
mi ¤ A(ci ) = ¤ Mi
∆xi
Vi = A(ci )∆xi
n
V= A(ci )∆xi .
i=1

It follows that for each partition P of [a, b] there exists a Riemann sum that
equals the volume. Hence, by de¬nition,
b
V= A(x) dx.
a


Theorem 5.7.2 Let f be a function that is continuous on [a, b]. Let R
denote the region bounded by the curves x = a, x = b, y = 0 and y = f (x).
Then the volume V obtained by rotating R about the x-axis is given by
b
π(f (x))2 dx.
V=
a

Proof. Clearly, the volume of the rotated solid is between the planes with
equations x = a and x = b. The cross-sectional area at x is the circle
generated by the line segment joining (x, 0) and (x, f (x)) and has area A(x) =
π(f (x))2 . Since f is continuous, A(x) is a continuous function of x. Then by
Theorem 5.7.1, the volume V is given by
b
π(f (x))2 dx.
V=
a
252 CHAPTER 5. THE DEFINITE INTEGRAL

Theorem 5.7.3 Let f and R be de¬ned as in Theorem 5.7.2. Assume that
f (x) > 0 for all x ∈ [a, b], either a ≥ 0 or b ¤ 0, so that [a, b] does not
contain 0. Then the volume V generated by rotating the region R about the
y-axis is given by
b
V= (2πxf (x)) dx.
a

Proof. The line segment joining (x, 0) and (x, f (x)) generates a cylinder
whose area is A(x) = 2πxf (x). We can see this if we cut the cylinder
vertically at (’x, 0) and ¬‚attening it out. By Theorem 5.7.1, we get
b
V= 2πxf (x) dx.
a



Theorem 5.7.4 Let f and g be continuous on [a, b] and suppose that f (x) >
g(x) > 0 for all x on [a, b]. Let R be the region bounded by the curves
x = a, x = b, y = f (x) and y = g(x).

(i) The volume generated by rotating R about the x-axis is given by
b
π[(f (x))2 ’ (g(x))2 ] dx.
a



(ii) If we assume R does not cross the y-axis, then the volume generated by
rotating R about the y-axis is given by
b
2πx[f (x) ’ g(x)]dx.
V=
a



(iii) If, in part (ii), R does not cross the line x = c, then the volume generated
by rotating R about the line x = c is given by
b
2π|c ’ x|[f (x) ’ g(x)]dx.
V=
a


Proof. We leave the proof as an exercise.
5.7. VOLUMES OF REVOLUTION 253

Remark 21 There are other various horizontal or vertical axes of rotation
that can be considered. The basic principles given in these theorems can be
used. Rotations about oblique lines will be considered later.

Example 5.7.1 Suppose that a pyramid is 16 units tall and has a square
base with edge length of 5 units. Find the volume of V of the pyramid.



graph



We let the y-axis go through the center of the pyramid and perpendicular
to the base. At height y, let the cross-sectional area perpendicular to the
y-axis be A(y). If s(y) is the side of the square A(y), then using similar
triangles, we get
16 ’ y 5
s(y)
(16 ’ y)
= , s(y) =
5 16 16
25
(16 ’ y)2 .
A(y) =
256
Then the volume of the pyramid is given by
16 16
25
(16 ’ y)2 dy
A(y)dy =
256
0 0
16
25 (16 ’ y)3
=
’3
256 0
3
25 (16) (25)(16)
= =
256 3 3
400
= cubic units.
3
1
(base side)2 · height
Check : V =
3
1
(25) · 16
=
3
400
= .
3
254 CHAPTER 5. THE DEFINITE INTEGRAL

Example 5.7.2 Consider the region R bounded by y = sin x, y = 0, x = 0
and x = π. Find the volume generated when R rotated about

(iii) y = ’2
(i) x-axis (ii) y-axis (iv) y = 1
(v) x = π (vi) x = 2π.


(i) By Theorem 5.7.2, the volume V is given by
π
π sin2 x dx
V=
0
π
1
(x ’ sin x cos x)
=π·
2 0
π2
=.
2




graph




(ii) By Theorem 5.7.3, the volume V is given by (integrating by parts)
π
V= 2πx sin x dx ; (u = x, dv = sin x dx)
0
= 2π[’x cos x + sin x]π
0
= 2π[π]
= 2π 2 .




graph
5.7. VOLUMES OF REVOLUTION 255

(iii) In this case, the volume V is given by


π
π(sin x + 2)2 dx
V=
0
π
π[sin2 x + 4 sin x + 4] dx
=
0
π
1
(x ’ sin x cos x) ’ 4 cos x + 4x

2 0
1
=π π + 8 + 4π
2
9
= π 2 + 8π.
2




graph




(iv) In this case,

π
π[12 ’ (1 ’ sin x)2 ] dx.
V=
0




graph
256 CHAPTER 5. THE DEFINITE INTEGRAL



π
π[1 ’ 1 + 2 sin x ’ sin2 x]dx
V=
0
π
1
= π ’2 cos x ’ (x ’ sin x cos x)
2 0
1
= π 4 ’ (π)
2
π(8 ’ π)
= .
2


(v)
π
(2π(π ’ x) sin x] dx
V=
0
π
[π sin x ’ x sin x] dx
= 2π
0
= 2π[’π cos x + x cos x ’ sin x]π
0
= 2π[2π ’ π]
= 2π 2 .




graph




(vi)
π
2π(2π ’ x) sin x dx
V=
0
= 2π[’2π cos x + x cos x ’ sin x]π
0
= 2π[4π ’ π]
= 6π 2 .
5.7. VOLUMES OF REVOLUTION 257

graph




Example 5.7.3 Consider the region R bounded by the circle (x ’ 4)2 + y 2 =
4. Compute the volume V generated when R is rotated around
(i) y = 0 (ii) x = 0 (iii) x = 2



graph



(i) Since the area crosses the x-axis, it is su¬cient to rotate the top half to
get the required solid.
6 6
2
[4 ’ (x ’ 4)2 ] dx
V= πy dx = π
2 2
6
1 88 32
= π 4x ’ (x ’ 4)3 = π 16 ’ ’ = π.
3 33 3
2
This is the volume of a sphere of radius 2.
(ii) In this case,
6 6
x[ 4 ’ (x ’ 4)2 ]dx ; x ’ 4 = 2 sin t
V= 2πx(2y) dx = 4π
2 2
dx = 2 cos tdt
π/2
= 4π (4 + 2 sin t)(2 cos t)(2 cos t)dt
’π/2
π/2
(16 cos2 t + 8 cos2 t sin t) dx
= 4π
’π/2
π/2
1 8
= 4π 16 · (t + sin t cos t) ’ cos3 t
2 3 ’π/2

= 4π[8(π)]
= 32π 2
258 CHAPTER 5. THE DEFINITE INTEGRAL

(iii) In this case,
6
2π(x ’ 2)2y dx
V=
2
6
(x ’ 2) 4 ’ (x ’ 4)2 dx ; x ’ 4 = 2 sin t
= 4π
2
dx = 2 cos tdt
π/2
= 4π (2 + 2 sin t)(2 cos t)(2 cos t)dt
’π/2
π/2
(8 cos2 t + 8 cos2 t sin t)dt
= 4π
’π/2
π/2
8
= 4π 4(t + sin t cos t) ’ cos3 t
3 ’π/2

= 4π[4π]
= 16π 2


Exercises 5.7

1. Consider the region R bounded by y = x and y = x2 . Find the volume
generated when R is rotated around the line with equation

(i) x = 0 (ii) y = 0 (iii) y = 1 (iv) x = 1
(vi) x = ’1 (vii) y = ’1
(v) x = 4 (viii) y = 2

2. Consider the region R bounded by y = sin x, y = cos x, x = 0, x =
π
. Find the volume generated when R is rotated about the line with
2
equation
π
(i) x = 0 (ii) y = 0 (iii) y = 1 (iv) x =
2
3. Consider the region R bounded by y = ex , x = 0, x = ln 2, y = 0. Find
the volume generated when R is rotated about the line with equation

(iv) y = ’2
(i) y = 0 (ii) x = 0 (iii) x = ln 2
(v) y = 2 (iv) x = 2
5.7. VOLUMES OF REVOLUTION 259

4. Consider the region R bounded by y = ln x, y = 0, x = 1, x = e. Find
the volume generated when R is rotated about the line with equation

(i) y = 0 (ii) x = 0 (iii) x = 1 (v) x = e
(vi) y = ’1
(v) y = 1

5. Consider the region R bounded by y = cosh x, y = 0, x = ’1, x =
1. Find the volume generated when R is rotated about the line with
equation

(iv) y = ’1
(i) y = 0 (ii) x = 2 (iii) x = 1
(v) y = 6 (vi) x = 0

6. Consider the region R bounded by y = x, y = x3 . Find the volume
generated when R is rotated about the line with equation

(iii) x = ’1
(i) y = 0 (ii) x = 0 (iv) x = 1
(vi) y = ’1
(v) y = 1

7. Consider the region R bounded by y = x2 , y = 8 ’ x2 . Find the volume
generated when R is rotated about the line with equation

(iii) y = ’4
(i) y = 0 (ii) x = 0 (iv) y = 8
(v) x = ’2 (vi) x = 2

8. Consider the region R bounded by y = sinh x, y = 0, x = 0, x = 2. Find
the volume generated when R is rotated about the line with equation

(iv) x = ’2
(i) y = 0 (ii) x = 0 (iii) x = 2
(v) y = ’1 (vi) y = 10

9. Consider the region R bounded by y = x, y = 4, x = 0. Find the
volume generated when R is rotated about the line with equation

(i) y = 0 (ii) x = 0 (iii) x = 16 (iv) y = 4

10. Compute the volume of a cone with height h and radius r.
260 CHAPTER 5. THE DEFINITE INTEGRAL

5.8 Arc Length and Surface Area
The Riemann integral is useful in computing the length of arcs. Let f and
f be continuous on [a, b]. Let C denote the arc
C = {(x, f (x)) : a ¤ x ¤ b}.
Let P = {a = x0 < x1 < x2 < . . . < xn = b} be a partition of [a, b]. For each
i = 1, 2, . . . , n, let



graph




∆xi = xi ’ xi’1
∆yi = f (xi ) ’ f (xi’1 )
∆si = (f (xi ) ’ f (xi’1 ))2 + (xi ’ xi )2
||∆|| = max {∆xn }.
1¤i¤n

Then ∆si is the length of the line segment joining the two points (xi’1 , f (xi’1 ))
and (xi , f (xi )). Let
n
A(P ) = ∆si .
i=1
Then A(P ) is called the polygonal approximation of C with respect to
the portion P .

De¬nition 5.8.1 Let C = {(x, f (x)) : x ∈ [a, b]} where f and f are con-
tinuous on [a, b]. Then the arc length L of the arc C is de¬ned by
n
(f (xi ) ’ f (xi’1 ))2 + (xi ’ xi’1 )2 .
L = lim Ap = lim
||∆||’0 ||∆||’0
i=1


Theorem 5.8.1 The arc length L de¬ned in De¬nition 5.8.1 is given by
b
(f (x))2 + 1 dx.
L=
a
5.8. ARC LENGTH AND SURFACE AREA 261

Proof. By the Mean Value Theorem, for each i = 1, 2, . . . , n,
f (xi ) ’ f (xi’1 ) = f (ci )(xi ’ xi’1 )
for some ci such that xi’1 < ci < xi . Therefore, each polynomial approxima-
tion Ap is a Riemann Sum of the continuous function
(f (x))2 + 1
n
(f (ci ))2 + 1 ∆xi ,
A(P ) =
i=1

for some ci such that xi’1 < ci < xi .
By the de¬nition of the Riemann integral, we get
b
(f (x))2 + 1 dx.
L=
a



Example 5.8.1 Let C = {(x, cosh x) : 0 ¤ x ¤ 2}. Then the arc length L
of C is given by
2
1 + sinh2 x dx
L=
0
2
= cosh x dx
0
= [sinh x]2
0
= sinh 2.

2 3/2
: 0 ¤ x ¤ 4 . Then the arc length
Example 5.8.2 Let C = x, x
3
L of the curve C is given by
2
4
2 3 1/2
·x
L= 1+ dx
32
0
4
(1 + x)1/2 dx
=
0
4
2
(1 + x)3/2
=
3 0
2√
= [5 5 ’ 1].
3
262 CHAPTER 5. THE DEFINITE INTEGRAL

De¬nition 5.8.2 Let C be de¬ned as in De¬nition 5.8.1.

(i) The surface area Sx generated by rotating C about the x-axis is given by
b
2π|f (x)| (f (x))2 + 1 dx.
Sx =
a



(ii) The surface area Sy generated by rotating C about the y-axis
b
2π|x| (f (x))2 + 1 dx.
Sy =
a



Example 5.8.3 Let C = {(x, cosh x) : 0 ¤ x ¤ 4}.

(i) Then the surface area Sx generated by rotating C around the x-axis is
given by
4
2π cosh x 1 + sinh2 x dx
Sx =
0
4
cosh2 x dx
= 2π
0
4
1
= 2π (x + sinh x cosh x)
2 0
= π[4 + sinh 4 cosh 4].


(ii) The surface area Sy generated by rotating the curve C about the y-axis
is given by
4
2πx 1 + sinh2 x dx
Sy =
0
4
= 2π x cosh x dx ; (u = x, dv = cosh x dx)
0
= 2π[x sinh x ’ cosh x]4
0
= 2π[4 sinh 4 ’ cosh 4 + 1]
5.8. ARC LENGTH AND SURFACE AREA 263

Theorem 5.8.2 Let C = {(x(t), y(t)) : a ¤ t ¤ b}. Suppose that x (t) and
y (t) are continuous on [a, b].

(i) The arc length L of C is given by
b
(x (t))2 + (y (t))2 dt.
L=
a


(ii) The surface area Sx generated by rotating C about the x-axis is given by
b
2π|y(t)| (x (t))2 + (y (t))2 dt.
Sx =
a


(iii) The surface area Sy generated by rotating C about the y-axis is given by
b
2π|x(t)| (x (t))2 + (y (t))2 dt.
Sy =
a


Proof. The proof of this theorem is left as an exercise.

π
Example 5.8.4 Let C = (et sin t, et cos t) : 0 ¤ t ¤ . Then
2

(x (t))2 + (y (t))2 dt
ds =
(et (sin t + cos t))2 + (et (cos t ’ sin t))2 dt
=
= {e2t (sin2 t + cos2 t + 2 sin t cos t + cos2 t + sin2 t ’ 2 cos t sin t)}1/2 dt

= et 2 dt.


(i) The arc length L of C is given by

√t
π/2
L= 2e dt
0
√ π/2
= 2 et 0

= 2 eπ/2 ’ 1 .
264 CHAPTER 5. THE DEFINITE INTEGRAL

(ii) The surface area Sx obtained by rotating C about the x-axis is given by
√t
π/2
t
Sx = 2π(e cos t)( 2e dt)
0
√ π/2
e2t cos tdt
= 2 2π
0
π/2
√ e2t
= 2 2π (2 cos t + sin t)
5 0
√ eπ 2
(1) ’
= 2 2π
5 5

2 2π π
(e ’ 2).
=
5

(iii) The surface area Sy obtained by rotating C about the y-axis is given by
√t
π/2
t
Sy = 2π(e sin t)( 2e dt)
0
√ π/2
e2t sin tdt
= 2 2π
0
π/2
√ e2t
[2 sin t ’ cos t]
= 2 2π
5
√0
√ 2eπ 1 2 2π
(2eπ + 1).
= 2 2π + =
5 5 5


Exercises 5.8 Find the arc lengths of the following curves:
1. y = x3/2 , 0 ¤ x ¤ 4
12
(x + 2)3/2 , 0 ¤ x ¤ 1
2. y =
3
π
3. C = (4(cos t + t sin t), 4(sin t ’ t cos t)) : 0 ¤ t ¤
2
π
4. x(t) = a(cos t + t sin t), y(t) = a(sin t ’ t cos t), 0 ¤ t ¤
2
5. x(t) = cos3 t, y(t) = sin3 t, 0 ¤ t ¤ π/2
5.8. ARC LENGTH AND SURFACE AREA 265

12
x , 0¤t¤1
6. y =
2
7. x(t) = t3 , y(t) = t2 , 0 ¤ t ¤ 1

8. x(t) = 1 ’ cos t, y(t) = t ’ sin t, 0 ¤ t ¤ 2π

9. In each of the curves in exercises 1-8, set up the integral that represents
the surface area generated when the given curve is rotated about

(a) the x-axis
(b) the y-axis

10. Let C = {(x, cosh x) : ’1 ¤ x ¤ 1}

(a) Find the length of C.
(b) Find the surface area when C is rotated around the x-axis.
(c) Find the surface area when C is rotated around the y-axis.

In exercises 11“20, consider the given curve C and the numbers a and b.
Determine the integral that represents:
(a) Arc length of C
(b) Surface area when C is rotated around the x-axis.
(c) Surface area when C is rotated around the y-axis.
(d) Surface area when C is rotated around the line x = a.
(e) Surface area when C is rotated around the line y = b.

11. C = {(x, sin x) : 0 ¤ x ¤ π}; a = π, b = 1
π
12. C = (x, cos x) : 0 ¤ x ¤ ; a = π, b = 2
3
13. C = {(t, ln t) : 1 ¤ t ¤ e}; a = 4, b = 3

14. C = {(2 + cos t, sin t) : 0 ¤ t ¤ π}; a = 4, b = ’2
π
; a = π, b = ’3
15. C = (t, ln sec t) : 0 ¤ t ¤
3
1
16. C = {(2x, cosh 2x) : 0 ¤ x ¤ 1}; a = ’2, b =
2
266 CHAPTER 5. THE DEFINITE INTEGRAL

π π
17. C = (cos t, 3 + sin t) : ’ ¤t¤ ; a = 2, b = 5
2 2
π
18. C = (et sin 2t, et cos 2t) : 0 ¤ t ¤ ; a = ’1, b = 3
4
19. C = {(e’t , et ) : 0 ¤ t ¤ ln 2}; a = ’1, b = ’4

20. C = {(4’t , 4t ) : 0 ¤ t ¤ 1}; a = ’2, b = ’3
Chapter 6

Techniques of Integration

6.1 Integration by formulae
There exist many books that contain extensive lists of integration, di¬eren-
tiation and other mathematical formulae. For our purpose we will use the
list given below.

1. af (u)du = a f (u)du

n n
2. ai fi (u) du = ai fi (u)du
i=1 i=1

un+1
n
+ C, n = ’1
3. u du =
n+1

u’1 du = ln |u| + C
4.

e6au
eau du =
5. +C
a

abu
bu
6. a du = + C, a > 0, a = 1
b ln a

ln |u|du = u ln |u| ’ u + C
7.

267
268 CHAPTER 6. TECHNIQUES OF INTEGRATION

’ cos(au)
8. sin(au)du = +C
a
sin(au)
9. cos(au)du = +C
a
ln | sec(au)|
10. tan(au)du = +C
a
ln | sin(au)|
11. cot(au)du = +C
a
ln | sec(au) + tan(au)|
12. sec(au)du = +C
a
ln | csc(au) ’ cot(au)|
13. csc(au)du = +C
a
cosh(au)
14. sinh(au)du = +C
a
sinh(au)
15. cosh(au)du = +C
a
ln | cosh(au)|
16. tanh(au)du = +C
a
ln | sinh(au)|
17. coth(au)du = +C
a
2
arctan(eau ) + C
18. sech (au)du =
a
2
arctanh (eau ) + C
19. csch (au) du =
a
u sin(au) cos(au)
sin2 (au)du = ’
20. +C
2 2a
u sin(au) cos(au)
cos2 (au)du =
21. + +C
2 2a
tan(au)
tan2 (au)du = ’u+C
22.
a
6.1. INTEGRATION BY FORMULAE 269

cot(au)
cot2 (au)du = ’ ’u+C
23.
a
tan(au)
sec2 (au)du =
24. +C
a
cot(au)
csc2 (au)du = ’
25. +C
a
u sinh(2au)
sinh2 (au)du = ’ +
26. +C
2 4a
u sinh(2au)
cosh2 (au)du =
27. + +C
2 4a
tanh(au)
tanh2 (au)du = u ’
28. +C
a
coth(au)
coth2 (au)du = u ’
29. +C
a
tanh(au)
sech 2 (au)du =
30. +C
a
’ coth(au)
csch 2 (au)du =
31. +C
a
sec(au)
32. sec(au) tan(au)du = +C
a
csc(au)
csc(au) cot(au)du = ’
33. +C
a
sech (au)
sech (au) tanh(au)du = ’
34. +C
a
csch (au)
csch (au) coth(au)du = ’
35. +C
a
du 1 u
36. = arctan +C
a2 + u2 a a
a+u
du 1 u 1
+C
37. = arctanh +C = ln
a2 ’ u2 a’u
a a 2a
270 CHAPTER 6. TECHNIQUES OF INTEGRATION

du u

38. +C
= arcsinh
a
a2 + u2
du u
√ + C, |a| > |u|
39. = arcsin
a
2 ’ u2
a
du u
√ + C, |u| > |a|
40. = arccosh
a
2 ’ a2
u
du 1 u
√ + C, |u| > |a|
41. = arcsec
a a
u u2 ’ a2
du 1 u
√ = ’ arcsech + C, |a| > |u|
42.
a a
u a2 ’ u2
du u
1
√ = ’ arccsch
43. +C
a a
u a2 + u2

u du
√ = a2 + u2 + C
44.
2 + u2
a

u du
= ’ ln a2 ’ u2 + C, |a| > |u|
45. 2 ’ u2
a

u du
√ = a2 + u2 + C
46.
2 + u2
a

u du
√ = ’ a2 ’ u2 + C, |a| > |u|
47.
2 ’ u2
a

u du
√ = u2 ’ a2 + C, |u| > |a|
48.
u2 ’ a2
1√
1 ’ a2 u2 + C, |a||u| < 1
49. arcsin(au)du = u arcsin(au) +
a
1√
1 ’ a2 u2 + C, |a||u| < 1
arccos(au)du = u arccos(au) ’
50.
a
1
ln(1 + a2 u2 ) + C
arctan(au)du = u arctan(au) ’
51.
2a
1
ln(1 + a2 u2 ) + C
52. arccot (au)du = uarccot (au) +
2a
6.1. INTEGRATION BY FORMULAE 271


1
arcsec (au)du = uarcsec (au) ’ ln au + a2 u2 ’ 1 + C, au > 1
53.
a

1
arccsc (au)du = uarccsc (au) + ln au + a2 u2 ’ 1 + C, au > 1
54.
a
1√
arcsinh (au)du = uarcsinh (au) ’ 1 + a2 u2 + C
55.
a
1√
’1 + a2 u2 + C, |a||u| > 1
arccosh (au)du = uarccosh (au) ’
56.
a
1
ln(’1 + a2 u2 ) + C, |a||u| = 1
57. arctanh (au)du = uarctanh (au) +
2a
1
ln(’1 + a2 u2 ) + C, |a||u| = 1
58. arccoth (au)du = uarccoth (au) +
2a
1
arcsin(au) + C, |a||u| < 1
59. arcsech (au)du = uarcsech (au) +
a

1
arccsch (au)du = uarccsch (au) + ln au + a2 u2 + 1 + C
60.
a
eau [a sin(bu) ’ b cos(bu)]
au
61. e sin(bu)du = +C
a2 + b2
eau [a cos(bu) + b sin(bu)]
au
62. e cos(bu)du = +C
a2 + b2
’1 n’1
sinn (u)du = sinn’1 (u) cos(u) + sinn’2 (u)du
63.
n n
n’1
1
cosn (u)du = cosn’1 (u) sin(u) + cosn’2 (u)du
64.
n n
tann’1 (u)
n
tann’2 (u)du

65. tan (u)du =
n’1
cotn’1 (u)
n
cotn’2 (u)du
cot (u)du = ’ ’
66.
n’1
n’2
1
secn (u)du = secn’2 (u) tan(u) + secn’2 (u)du
67.
n’1 n’1
272 CHAPTER 6. TECHNIQUES OF INTEGRATION

’1 n’2
cscn (u)du = cscn’2 (u) cot(u) + cscn’2 (u)du
68.
n’1 n’1
sin[(m ’ n)u] sin[(m + n)u]
+ C, m2 = n2

69. sin(mu)sin(nu)du =
2(m ’ n) 2(m + n)
sin[(m ’ n)u] sin[(m + n)u]
+ C, m2 = n2
70. cos(mu) cos(nu)du = +
2(m ’ n) 2(m + n)
cos[(m ’ n)u] cos[(m + n)u]
+ C, m2 = n2

71. sin(mu) cos(nu)du =
2(m ’ n) 2(m + n)


Exercises 6.1
1. De¬ne the statement that g(x) is an antiderivative of f (x) on the closed
interval [a, b]

2. Prove that if g(x) and h(x) are any two antiderivatives of f (x) on [a, b],
then there exists some constant C such that g(x) = ln(x) + C for all x
on [a, b].
In problems 3“30, evaluate each of the inde¬nite integrals.
4
x’3/5 dx
x5 dx
3. 4. dx 5.
x3

2
2/3 2
√ dx
6. 3x dx 7. 8. t t dt
x

t’1/2 + t3/2 dt (1 + x2 )2 dx t2 (1 + t)2 dt
10. 11.
9.

1
(1 + t2 )(1 ’ t2 )dt 13.
12. + sin t dt 14. (2 sin t + 3 cos t)dt
t1/2

3 sec2 t dt 2 csc2 x dx
15. 16. 17. 4 sec t tan t dt


18. 2 csc t cot t dt 19. sec t(sec t + tan t)dt
6.2. INTEGRATION BY SUBSTITUTION 273


sin x
csc t(csc t ’ cot t)dt
20. 21. dx
cos2 x

sin3 t ’ 3 cos3 t + 2
cos x
22. dx 23. dt 24. dt
sin2 x sin2 t cos2 t

tan2 t dt cot2 t dt (2 sec2 t + 1)dt
25. 26. 27.

2
28. dt 29. sinh t dt 30. cosh t dt
t


31. Determine f (x) if f (x) = cos x and f (0) = 2.

32. Determine f (x) if f (x) = sin x and f (0) = 1, f (0) = 2.

33. Determine f (x) if f (x) = sinh x and f (0) = 2, f (0) = ’3.

34. Prove each of the integration formulas 1“77.


6.2 Integration by Substitution
Theorem 6.2.1 Let f (x), g(x), f (g(x)) and g (x) be continuous on an in-
terval [a, b]. Suppose that F (u) = f (u) where u = g(x). Then

(i) f (g(x))g (x)dx = f (u)du = F (g(x)) + C

b u=g(b)
f (u)du = F (g(b)) ’ F (g(a)).
(ii) f (g(x))g (x)dx =
a u=g(u)


Proof. See the proof of Theorem 5.3.1.


Exercises 6.2 In problems 1“39, evaluate the integral by making the given
substitution.
274 CHAPTER 6. TECHNIQUES OF INTEGRATION


3x(x2 + 1)10 dx, u = x2 + 1 x sin(1 + x2 )dx, u = 1 + x2
1. 2.


√ 3x2
cos( t)
dx, u = 1 + x3
√ dt, x = t 4.
3.
(1 + x3 )3/2
t

2earcsin x 3earccos x
√ √
5. dx, u = arcsin x 6. dx
2 2
1’x 1’x

2 2
x 4x dx, u = 4x 10sin x cos x dx, u = sin x
7. 8.

(1 + ln x)10
4arctan x
dx, u = 4arctan x 10. dx, u = 1 + ln x
9. 2
1+x x

5arcsec x
(tan 2x)3 sec2 2x dx, u = tan 2x

11. dx, u = arcsec x 12.
x x2 ’ 1

(cot 3x)5 csc2 3x dx, u = cot 3x sin21 x cos x dx, u = sin x
13. 14.


cos5 x sin x dx, u = cos x (1 + sin x)10 cos x dx, u = 1 + sin x
15. 16.


sin3 x dx, u = cos x cos3 x dx, u = sin x
17. 18.


tan3 x dx, u = tan x cot3 x dx, u = cot x
19. 20.


sec4 x dx, u = tan x csc4 x dx, u = cot x
21. 22.


sin3 x cos3 x dx, u = sin x sin3 x cos3 x dx, u = cos x
23. 24.

sin(ln x)
tan4 x dx, u = tan x
25. 26. dx, u = ln x
x
6.2. INTEGRATION BY SUBSTITUTION 275

x cos(ln(1 + x2 ))
dx, u = ln(1 + x)2 28. tan3 x sec4 x dx, u = sec x
27. 2
1+x

dx
cot3 x csc4 x dx, u = csc x √
29. 30. , x = 2 sin t
4 ’ x2

dx dx
√ √
31. , x = 3 cos t 32. , x = 2 sinh t
2
9’x 2
4+x

dx
dx
√ , x = 3 cosh t 34.
33. , x = 2 tan t
4 + x2
x2 ’ 9

dx dx
√ , x = 2 sec t
35. , x = 2 tanh t 36.
4 ’ x2 x x2 ’ 4

2 +4) 2 +4
4esin(3x) cos(3x)dx, u = sin 3x x 3(x dx, u = 3x
37. 38.


3 etan 2x sec2 x dx, u = tan 2x
39. 40. x x + 2 dx, u = x + 2

Evaluate the following de¬nite integrals.
1 2
30
x(4 ’ x2 )1/2 dx
41. (x + 1) dx 42.
0 1

π/4 1
3 2
x3 (x2 + 1)3 dx
43. tan x sec x dx 44.
0 0

2 8
10
x2 (1 + x)1/2 dx
(x + 1)(x ’ 2) dx
45. 46.
0 0

π/6 π/4
47. sin(3x)dx 48. cos(2x)dx
0 0

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