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+
’m ’ 1
’1 m
cscm+2 x dx = (csc x)m cot x + (csc x)m dx . (2)
m+1 m+1

This gives us the reduction formula for part (iii). Also,

’1 n’2
cscn x dx = (csc xn’2 ) cot x + (csc xn’2 ) dx.
n’1 n’1
222 CHAPTER 5. THE DEFINITE INTEGRAL

(iii) We can derive a formula by a method similar to part (i). However, let
us make use of a trigonometric reduction formula to get it. Recall that
π π
’ x and cos ’ x = sin x. Then
cos x = sin
2 2
π π
cosn x dx = sinn ’x ’ x, du = ’dx
dx let u =
2 2
sinn (u)(’du)
=

sinn udu
=’
’1 n’1
(sin u)n’1 cos u + sinn’2 udu
=’ (by (1))
n n
1 π π
n’1
’x ’x
= sin cos
n 2 2
n’1 π π
n’2
’ ’x ’x
sin d
n 2 2
n’1
1
cosn x dx = (cos x)n’1 sin x + cosn’2 x dx . (3)
n n

To get part (iv) we replace n by ’m and get
’m ’ 1
1
cos’m x dx = (cos x)’m’1 sin x + cos’m’2 x dx
’m ’m
’1 m+1
secm x dx = (sec x)m tan x + secm+2 x dx.
m m
On solving for the last integral, we get

1 m
secm+2 x dx = (sec x)m tan x + secm x dx . (4)
m+1 m+1

n’2
1
secn x dx = secn’2 x tan x + secn’2 x dx.
Also,
n’1 n’1
In parts (ii), (iii) and (vi) we leave the cases for n = 3 and 4 as an exercise.
These are handled as in part (i).

Example 5.4.3 Develop the reduction formulas for the following integrals:
5.4. INTEGRATION BY PARTS 223


sinhn x dx coshn x dx
tann x dx cotn x dx
(i) (ii) (iii) (iv)




(i) First, we break tan2 x = sec2 x ’ 1 away from the integrand:


tann x dx = tann’2 x · tan2 x dx

tann’2 x(sec2 x ’ 1) dx
=

tann x dx = tann’2 x sec2 x dx ’ tann’2 x dx.


For the middle integral, we let u = tan x as a substitution.


tann x dx = un’2 du ’ tann’2 x dx
un’1
’ tann’2 x dx
=
n’1
(tan x)n’1
’ tann’2 x dx.
=
n’1


Therefore,


(tan x)n’1
n
tann’2 x dx n = 1 .

tan x dx = (5)
n’1

tan x dx = ln | sec x| + c for n = 1.
224 CHAPTER 5. THE DEFINITE INTEGRAL

π
’ x = cot x in (5).
(ii) We use the reduction formula tan
2
π π
cotn x dx = tann ’x let u = ’ x, du = ’dx
dx;
2 2
tann u(’du)
=’

tann u du
=’
tann’1 (u)
’ tann’2 u du , n = 1
=’
n’1
cotn’1 x
’ cotn’2 x(’dx), n = 1
=’
n’1
cotn’1 x
+ cotn’2 x dx, n = 1
=’
n’1
cot x dx = ln | sin x| + c, for n = 1.

Therefore,
cotn’1 x
n
cotn’2 x dx, n = 1
cot (x) dx = ’ + (6)
n’1

cot x dx = ln | sin x| + c.



sinhn x dx = (sinhn’1 x)(sinh x dx); u = sinhn’1 x, dv = sinh x dx
(iii)


= sinhn’1 x cosh x ’ cosh x · (n ’ 1) sinhn’2 x cosh xdx


= sinhn’1 x cosh x ’ (n ’ 1) sinhn’2 x(cosh2 x) dx


= sinhn’1 x cosh x ’ (n ’ 1) sinhn’2 x(1 + sinh2 x) dx


= sinhn’1 x cosh x ’ (n ’ 1) sinhn’2 x dx ’ (n ’ 1) sinhn x dx.
5.4. INTEGRATION BY PARTS 225

On bringing the last integral to the left, we get


sinhn x dx = sinhn’1 x cosh x ’ (n ’ 1) sinhn’2 x dx
n

n’1
1
sinhn x dx = sinhn’1 cosh x ’ sinhn’2 x dx . (7)
n n



coshn x dx = (coshn’1 x)(cosh x dx); u = coshn’1 x, dv = cosh x dx, v = sinh x
(iv)


= coshn’1 (x) sinh x ’ sinh x(n ’ 1) coshn’2 x sinh xdx


= coshn’1 x sinh x ’ (n ’ 1) coshn’2 x sinh2 x dx


= coshn’1 x sinh x ’ (n ’ 1) coshn’2 x(cosh2 x ’ 1) dx


= coshn’1 x sinh x ’ (n ’ 1) coshn x dx

+(n ’ 1) coshn’2 x dx

coshn x dx +(n ’ 1) coshn x dx = coshn’1 x sinh x

+(n ’ 1) coshn’2 x dx

coshn x dx = coshn’1 x sinh x + (n ’ 1) coshn’2 x dx
n


n’1
1
coshn x dx = coshn’1 x sinh x + coshn’2 x dx (8)
n n


Example 5.4.4 Develop reduction formulas for the following:
226 CHAPTER 5. THE DEFINITE INTEGRAL


xn ex dx xn ln x dx (ln x)n dx
(i) (ii) (iii)


xn sin x xn cos x dx eax sin(ln x) dx
(iv) (v) (vi)


eax cos(ln x) dx
(vii)

(i) We let u = xn , dv = ex dx, du = nxn’1 dx, v = ex . Then

xn ex dx = xn ex ’ ex (nxn’1 ) dx

= x n ex ’ n xn’1 ex dx.

Therefore,
xn ex dx = xn ex ’ n xn’1 ex dx . (9)


(ii) We let u = ln x, du = (1/x) dx, dv = xn dx, v = xn+1 /(n + 1). Then,
xn+1 xn+1 1
n
’ · dx
x ln x dx = (ln x)
n+1 n+1 x
xn+1 (ln x) 1
xn dx

=
n+1 n+1
xn+1 (ln x) xn+1

= + c.
(n + 1)2
n+1
Therefore,
xn+1
n
[(n + 1) ln(x) ’ 1] + c .
x ln x dx = (10)
(n + 1)2

1
(iii) We let u = (ln x)n , du = n(ln x)n’1 dx, dv = dx, v = x. Then,
x
1
(ln x)n dx = x(ln x)n ’ x · n(ln x)n’1 · dx
x
= x(ln x)n ’ n (ln x)n’1 dx
5.4. INTEGRATION BY PARTS 227

Therefore,

(ln x)n dx = x(ln x)n ’ n (ln x)n’1 dx . (11)


(iv) We let u = xn , du = nxn’1 dx, dv = sin x dx, v = ’ cos x. Then,



xn sin x dx = xn (’ cos x) ’ (’ cos x)nxn’1 dx
(—)
n n’1
= ’x cos x = n x cos x dx.


Again in the last integral we let u = xn’1 , du = (n ’ 1)xn’2 dx, dv =
cos x dx, v = sin x. Then


xn’1 cos x dx = xn’1 sin x ’ sin x(n ’ 1)xn’2 dx
(——)
= xn’1 sin x ’ (n ’ 1) xn’2 sin x dx.


By substitution, we get the reduction formula

xn sin x dx = ’xn cos x + n xn’1 sin x ’ (n ’ 1) xn’2 sin x dx

xn sin x dx = ’xn cos x + nxn’1 sin x ’ n(n ’ 1) xn’2 sin x dx
(12)


(v) We can use (——) and (—) in part (iv) to get the following:

xn’1 cos x dx = xn’1 sin x ’ (n ’ 1) xn’2 sin x dx by (——)

= xn’1 sin x ’ (n ’ 1) ’xn’2 cos x + (n ’ 2) xn’3 cos x dx by (—)
228 CHAPTER 5. THE DEFINITE INTEGRAL


xn’1 cos x dx = xn’1 x+(n’1)xn’2 cos x’(n’1)(n’2) xn’3 cos x dx.


If we replace n by n + 1 throughout the last equation, we get


xn cos x dx = xn sin x + nxn’1 cos x ’ n(n ’ 1) xn’2 cos x dx
(13)

1 ax
(vi) We let dv = eax dx, v = e , u = sin(bx), du = b cos(bx) dx. Then
a

1 ax b
eax sin(bx) dx = eax cos(bx) dx.
e sin(bx) ’ (— — —)
a a

1 ax
In the last integral, we let dv = eax dx, v = e , u = cos bx. Then
a

1 ax b
eax cos(bx) dx = eax sin bx dx (— — ——)
e cos bx +
a a

First we substitute (— — ——) into (— — —) and then solve for

eax sin bx dx.



1 ax b 1 ax b
eax sin bx dx = eax sin bx dx
e sin bx ’ e cos bx +
a aa a
eax b2
= 2 (a sin bx ’ b cos bx) ’ 2 ax sin bx dx
a a
eax
b2 ax
e sin bx dx = 2 (a sin bx ’ b cos bx dx)
1+ 2
a a

eax
ax
(a sin bx ’ b cos bx) + c .
e sin bx dx = 2 (14)
a + b2
5.4. INTEGRATION BY PARTS 229

(vii) We start with (— — ——) and substitute in (14) without the constant c and
get
1 ax b
eax cos bx dx = eax sin bx dx
e cos bx +
a a
b eax
1 ax
(a sin bx ’ b cos bx) + c
= e cos bx +
a a2 b2
a
b2
ax 1 1
b sin bx ’
=e cos bx + 2 cos bx + c
a + b2
a a
eax
=2 [b sin b + a cos bx] + c.
a + b2
Therefore,

eax
ax
e cos bx dx = 2 [b sin bx + a cos bx] + c . (15)
a + b2


Exercises 5.4 Evaluate the following integrals and check your answers by
di¬erentiation. You may use the reduction formulas given in the examples.

dx
xe’2x dx x3 ln x
1. 2. 3.
x(ln x)4

(ln x)3 dx e2x sin 3x dx e3x cos 2x dx
4. 5. 6.


x2 sin 2x dx x2 cos 3x dx
7. 8. 9. x ln(x + 1) dx


10. arcsin(2x) dx 11. arccos(2x) dx 12. arctan(2x) dx


sec3 x dx sec5 x dx tan5 x dx
13. 14. 15.


x2 ln x dx x3 sin x dx x3 cos x dx
16. 17. 18.
230 CHAPTER 5. THE DEFINITE INTEGRAL


x(ln x)3 dx
19. x sinh x dx 20. x cosh x dx 21.


sin3 x dx
22. x arctan x dx 23. xarccot x dx 24.


cos3 x dx sin4 x dx cos4 x dx
25. 26. 27.


sinh2 x dx cosh2 x dx sinh3 x dx
28. 29. 30.


x2 sinh x dx x2 cosh x dx x3 sinh x dx
31. 32. 33.


x3 e’x dx
x3 cosh x dx x2 e2x dx
34. 35. 36.


37. x sin(3x) dx 38. x cos(x + 1)dx 39. x ln(x + 1)dx


x 2x dx x 102x dx x2 103x dx
40. 41. 42.


x2 (ln x)3 dx
43. 44. arcsinh (3x)dx 45. arccosh (2x)dx


46. arctanh (2x)dx 47. arccoth (3x)dx 48. xarcsec x dx


50. xarccsc x dx



5.5 Logarithmic, Exponential and Hyperbolic
Functions
With the Fundamental Theorems of Calculus it is possible to rigorously de-
velop the logarithmic, exponential and hyperbolic functions.
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS231

De¬nition 5.5.1 For each x > 0 we de¬ne the natural logarithm of x, de-
noted ln x, by the equation
x
1
ln(x) = dt , x > 0.
t
1



Theorem 5.5.1 (Natural Logarithm) The natural logarithm, ln x, has the
following properties:
1
d
(ln x) = > 0 for all x > 0.
(i)
dx x
The natural logarithm is an increasing, continuous and di¬erentiable
function on (0, ∞).

(ii) If a > 0 and b > 0, then ln(ab) = ln(a) + ln(b).

(iii) If a > 0 and b > 0, then ln(a/b) = ln(a) + ln(b).

(iv) If a > 0 and n is a natural number, then ln(an ) = n ln a.

(v) The range of ln x is (’∞, ∞).

(vi) ln x is one-to-one and has a unique inverse, denoted ex .

Proof.

(i) Since 1/t is continuous on (0, ∞), (i) follows from the Fundamental The-
orem of Calculus, Second Form.

(ii) Suppose that a > 0 and b > 0. Then
ab
1
ln(ab) = dt
t
1
a ab
1 1
= dt + dt
t t
1 a
b
1 1 1
= ln a + adu ; u= t, du = dt
1 au a a
= ln a + ln b.
232 CHAPTER 5. THE DEFINITE INTEGRAL

(iii) If a > 0 and b > 0, then

(a) 1
a b
ln = dt
b t
1
a
a
1 b b
1
b
= dt + dt; u = t, du = dt
1t t a a
a
a 1
1 1 a
= dt + dt
au
t b
1 b b
a b
1 1
dt ’
= du
t u
1 1
= ln a ’ ln b.


(iv) If a > 0 and n is a natural number, then
an
1
n
dt ; t = un , dt = nun’1 du
ln(a ) =
t
1
a
1
· nun’1 du
= n
1u
a
1
=n du
u
1
= n ln a

as required.

(v) From the partition {1, 2, 3, 4, · · · }, we get the following inequality using
upper and lower sum approximations:



graph




13 111 11
= + + < ln 4 < 1 + + .
12 234 23
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS233

Hence, ln 4 > 1. ln(4n ) = n ln 4 > n and ln 4’n = ’n ln 4 < ’n. By
the intermediate value theorem, every interval (’n, n) is contained in
the range of ln x. Therefore, the range of ln x is (’∞, ∞), since the
derivative of ln x is always positive, ln x is increasing and hence one-to-
one. The inverse of ln x exists.

(vi) Let e denote the number such that ln(e) = 1. Then we de¬ne y = ex if
and only if x = ln(y) for x ∈ (’∞, ∞), y > 0.

This completes the proof.

De¬nition 5.5.2 If x is any real number, we de¬ne y = ex if and only if
x = ln y.


Theorem 5.5.2 (Exponential Function) The function y = ex has the fol-
lowing properties:

d
(ex ) = ex .
(i) e0 = 1, ln(ex ) = x for every real x and
dx
(ii) ea · eb = ea+b for all real numbers a and b.

ea
(iii) b = ea’b for all real numbers a and b.
e
(iv) (ea )n = ena for all real numbers a and natural numbers n.

Proof.

(i) Since ln(1) = 0, e0 = 1. By de¬nition y = ex if and only if x = ln(y) =
ln(ex ). Suppose y = ex . Then x = ln y. By implicit di¬erentiation, we
get
1 dy dy
= y = ex .
1= ,
y dx dx
Therefore,
d
(ex ) = ex .
dx
234 CHAPTER 5. THE DEFINITE INTEGRAL

(ii) Since ln x is increasing and, hence, one-to-one,

ea · eb = ea+b ”
ln(ea · eb ) = ln(ea+b ) ”
ln(ea ) + ln(eb ) = a + b ”
a + b = a + b.

It follows that for all real numbers a and b,

ea · eb = ea+b .
ea
= ea’b ”
(iii)
eb
ea
= ln(ea’b ) ”
ln
eb

ln(ea ) ’ ln(eb ) = a ’ b ”

a ’ b = a ’ b.


It follows that for all real numbers a and b,

ea
= ea’b .
b
e


(ea )n = ena ”
(iv)

ln((ea )n ) = ln(ena ) ”

n ln(ea ) = na ”

na = na.

Therefore, for all real numbers a and natural numbers n, we have

(ea )n = ena .
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS235

De¬nition 5.5.3 Suppose b > 0 and b = 1. Then we de¬ne the following:
(i) For each real number x, bx = ex ln b .
ln x
(ii) y = logb x = .
ln b

Theorem 5.5.3 (General Exponential Function) Suppose b > 0 and b = 1.
Then
(i) ln(bx ) = x ln b, for all real numbers x.
d
(bx ) = bx ln b, for all real numbers x.
(ii)
dx
(iii) bx1 · bx2 = bx1 +x2 , for all real numbers x1 and x2 .
bx1
(iv) x2 = bx1 ’x2 , for all real numbers x1 and x2 .
b
(v) (bx1 )x2 = bx1 x2 , for all real numbers x1 and x2 .
bx
x
+ c.
(vi) b dx =
ln b

Proof.

(i) ln(bx ) = ln(ex ln b ) = x ln b

dx d
(ex ln b ) = ex ln b · (ln b)
(ii) (b ) = (by the chain rule)
dx dx

= bx ln b.

(iii) bx1 · bx2 = ex1 ln b · ex2 ln b

= e(x1 ln b+x2 ln b)

= e(x1 +x2 ) ln b

= b(x1 +x2 )
236 CHAPTER 5. THE DEFINITE INTEGRAL

bx1 ex1 ln b
(iv) = x2 ln b
bx2 e

= ex1 ln b’x2 ln b

= e(x1 ’x2 ) ln b

= b(x1 ’x2 ) .

(v) By De¬nition 5.5.3 (i), we get
x1 )
(bx1 )x2 = ex2 ln(b
x1 ln b )
= ex2 ln(e

= ex2 ·x1 ln b

= e(x1 x2 ) ln b

= bx1 x2 .


(vi) Since
dx
(b ) = bx ln b,
dx
we get

bx (ln b) dx = bx + c,

bx dx = bx + c,
ln b
bx
x
e dx = + D,
ln b
where D is some constant. This completes the proof.

Theorem 5.5.4 If u(x) > 0 for all x, and u(x) and v(x) are di¬erentiable
functions, then we de¬ne
y = (u(x))v(x) = ev(x) ln(u(x)) .
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS237

Then y is a di¬erentiable function of x and

dy d u (x)
(u(x))v(x) = (u(x))v(x) v (x) ln(u(x)) + v(x)
= .
dx dx u(x)

Proof. This theorem follows by the chain rule and the product rule as follows

d d v ln u u u1
[uv ] = ] = ev ln u v ln u + v = uv v ln u + v
[e .
dx dx u u



Theorem 5.5.5 The following di¬erentiation formulas for the hyperbolic
functions are valid.

d d
(i) (sinh x) = cosh x (ii) (cosh x) = sinh x
dx dx
d d
(tanh x) = sech2 x (coth x) = ’csch2 x
(iii) (iv)
dx dx
d d
(sech x) = ’sech x tanh x (csch x) = ’csch x coth x
(v) (vi)
dx dx

Proof. We use the de¬nitions and properties of hyperbolic functions given
in Chapter 1 and the di¬erentiation formulas of this chapter.

ex ’ e’x ex + e’x
d d
(i) (sinh x) = = = cosh x.
dx dx 2 2

ex + e’x ex ’ e’x
d d
(ii) (cosh x) = = = sinh x.
dx dx 2 2

(cosh x)(cosh x) ’ sinh(sinh x)
d d sinh x
(iii) (tanh x) = =
(cosh x)2
dx dx cosh x

cosh2 x ’ sinh2 x 1
= sech2 x
=
= 2 2
(cosh x) cosh x)
238 CHAPTER 5. THE DEFINITE INTEGRAL

d d
(tanh x)’1 = ’1(tanh x)’2 · sech2 x
(iv) (coth x) =
dx dx

cosh2 x 1 1
=’ · =’
sinh2 x cosh2 x sinh2 x

= ’csch2 x.

d d
(cosh x)’1 = ’1(cosh x)’2 · sinh x
(v) (sech x) =
dx dx

= ’ sech x tanh x.

d d
(sinh x)’1 = ’1(sinh x)’2 · cosh x
(vi) (csch x) =
dx dx

= ’ coth x csch x.


This completes the proof.


Theorem 5.5.6 The following integration formulas are valid:

(i) sinh x dx = cosh x + c (ii) cosh x dx = sinh x + c


coth xdx = ln | sinh x| + c
(iii) tanh x dx = ln(cosh x) + c (iv)

x
sech x dx = 2 arctan(ex ) + c csch x dx = ln tanh
(v) (vi) +c
2

Proof. Each formula can be easily veri¬ed by di¬erentiating the right-hand
side to get the integrands on the left-hand side. This proof is left as an
exercise.


Theorem 5.5.7 The following di¬erentiation and integration formulas are
valid:
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS239

d 1 dx
(arcsinh x) = √ √
(i) (ii) = arcsinh x + c
dx 1 + x2 1 + x2

dx
d 1
(arccosh x) = √ √
(iv) = arccosh x + c
(iii)
dx x2 ’ 1 x2 ’ 1

d 1 1
, |x| < 1
(v) (arctanh x) = (vi) dx = arctanh x + c
1 ’ x2 1 ’ x2
dx

Proof. This theorem follows directly from the following de¬nitions:

√ √
(1) arcsinh x = ln(x + 1 + x2 ) x2 ’ 1)
(2) arccosh x = ln(x +

1 1+x
, |x| < 1.
(3) arctanh x = ln
1’x
2

The proof is left as an exercise.


Exercises 5.5

1. Prove Theorem 5.5.6.

2. Prove Theorem 5.5.7.

3. Show that sinh mx and cosh mx are linearly independent if m = 0. (Hint:
Show that the Wronskian W (sinh mx, cosh mx) is not zero if m = 0.)

4. Show that emx and e’mx are linearly independent if m = 0.

5. Show that solution of the equation y ’ m2 y = 0 can be expressed as
y = c1 emx + c2 e’mx .

6. Show that every solution of y ’ m2 y = 0 can be written as y =
A sinh mx + B cosh mx.

7. Determine the relation between c1 and c2 in problem 5 with A and B in
problem 6.

8. Prove the basic identities for hyperbolic functions:
240 CHAPTER 5. THE DEFINITE INTEGRAL

(i) sinh(x + y) = sinh x cosh y + cosh x sinh y.
(ii) sinh(x ’ y) = sinh x cosh y ’ cosh x sinh y.
(iii) cosh(x + y) = cosh x cosh y + sinh x sinh y.
(iv) cosh(x ’ y) = cosh x cosh y ’ sinh x sinh y.
(v) sinh 2x = 2 sinh x cosh x.
(vi) cosh2 x + sinh2 x = 2 cosh2 x ’ 1 = 1 + 2 sinh2 x = cosh 2x.
(vii) cosh2 x ’ sinh2 x = 1, 1 ’ tanh2 x = sech2 x, coth2 x ’ 1 = csch2 x.

9. Eliminate the radical sign using the given substitution:
√ √
a2 + x2 , x = a sinh t a2 ’ x2 , x = tanh t
(i) (ii)

x2 ’ a2 , x = a cosh t.
(iii)

10. Compute y in each of the following:

(ii) y = 4 tanh(5x) ’ 6 coth(3x)
(i) y = 2 sinh(3x) + 4 cosh(2x)

(iv) y = 3 sinh2 (4x + 1)
(iii) y = x sech (2x) + x2 csch (5x)

(v) y = 4 cosh2 (2x ’ 1) (vi) y = sinh(2x) cosh(3x)

11. Compute y in each of the following:

3 2
(i) y = x2 e’x (ii) y = 2x (iii) y = (x2 + 1)sin(2x)
3 +1)
(iv) y = log10 (x2 + 1) (v) y = log2 (sec x + tan x)(vi) y = 10(x

12. Compute y in each of the following:


(i) y = x ln x ’ x (ii) y = ln(x + x2 ’ 4) (iii) y = ln(x + 4 + x2 )

1 1+x
(iv) y = ln (v) y = arcsinh (3x) (vi) y = arccosh (3x)
1’x
2
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS241

13. Evaluate each of the following integrals:

2
x3 ex dx x2 ln(x + 1) dx
(i) sinh(3x) dx (ii) (iii)

2
x4x dx
(iv) x sinh 2x dx (v) x cosh 3x dx (vi)


14. Evaluate each of the following integrals:


(i) arcsinh x dx (ii) arccosh x dx (iii) arctanh x dx

dx dx dx
√ √ √
(iv) (v) (vi)
4 ’ x2 x2 ’ 4
4 + x2

15. Logarithmic Di¬erentiation is a process of computing derivatives by ¬rst
taking logarithms and then using implicit di¬erentiation. Find y in each
of the following, using logarithmic di¬erentiation.

(x2 + 1)3 (x2 + 4)10 3 +1)
(ii) y = (x2 + 4)(x
(i) y = 2
(x + 2)5 (x2 + 3)4
3 +1)
(iii) y = (sin x + 3)(4 cos x+7) (iv) y = (3 sinh x + cos x + 5)(x
2 3 +1)
(v) y = (ex + 1)(2x+1) (vi) y = x2 (x2 + 1)(x

In problems 16“30, compute f (x) each f (x).
x2
x
sinh3 (t)dt cosh5 (t)dt
16. f (x) = 17. f (x) =
1 x

cosh x sech x
2 3/2
(1 + t3 )1/2 dt
18. f (x) = (1 + t ) dt 19. f (x) =
sinh x tanh x

2
(ln x)2 ex
(1 + 4t2 )π dt
(4 + t2 )5/2 dt
20. f (x) = 21. f (x) =
2
ex
ln x
242 CHAPTER 5. THE DEFINITE INTEGRAL


ecos x 3x
1 1
22. f (x) = dt 23. f (x) = dt
(1 + t2 )3/2 (4 + t2 )5/2
esin x 2x

53x log3 x
2 3/2
(1 + 5t3 )1/2 dt
24. f (x) = (1 + 2t ) dt 25. f (x) =
42x log2 x

3
4x
arccosh x
1 2
et dt
26. f (x) = dt 27. f (x) =
(1 + t2 )3/2 2
2x
arcsinh x

5cos x cosh(x3 )
’t2 3
e’t dt
28. f (x) = e dt 29. f (x) =
4sin x sinh(x2 )

arccoth x
sin(t2 )dt
30. f (x) =
arctanh x

In problems 31“40, evaluate the given integrals.

earctan x earcsin x
esin 2x cos 2x dx

31. dx 32. dx 33.
1 + x2 1 ’ x2

e2x
2 x3
ex cos(1 + 2ex )dx
dx 36.
34. x e dx 35.
1 + e2x

4arcsec x
3x 2 3x cos x

37. e sec (2 + e )dx 38. 10 sin x dx 39. dx
x x2 ’ 1

2 +3
x 10x
40. dx




5.6 The Riemann Integral
In de¬ning the de¬nite integral, we restricted the de¬nition to continuous
functions. However, the de¬nite integral as de¬ned for continuous functions
is a special case of the general Riemann Integral de¬ned for bounded functions
that are not necessarily continuous.
5.6. THE RIEMANN INTEGRAL 243

De¬nition 5.6.1 Let f be a function that is de¬ned and bounded on a
closed and bounded interval [a, b]. Let P = {a = x0 < x1 < x2 < · · · <
xn = b} be a partition of [a, b]. Let C = {ci : xi’1 ¤ ci ¤ xi , i = 1, 2, · · · , n}
be any arbitrary selection of points of [a, b]. Then the Riemann Sum that is
associated with P and C is denoted R(P ) and is de¬ned by
R(P ) = f (c1 )(x1 ’ x0 ) + f (c2 )(x2 ’ x1 ) + · · · + f (cn )(xn + xn’1 )
n
f (ci )(xi ’ xi’1 ).
=
i=1

Let ∆xi = xi ’ xi’1 , i = 1, 2, · · · , n. Let ||∆|| = max {∆xi }. We write
1¤i¤n

n
R(P ) = f (ci )∆xi .
i=1

We say that
n
lim f (ci )∆xi = I
||∆||’0
i=1
if and only if for each > 0 there exists some δ > 0 such that
n
f (ci )∆xi ’ I <
i=1

whenever ||∆|| < δ for all partitions P and all selections C that de¬ne the
Riemann Sum.
If the limit I exists as a ¬nite number, we say that f is (Riemann) inte-
grable and write
b
I= f (x) dx.
a
Next we will show that if f is continuous, the Riemann integral of f is
the de¬nite integral de¬ned by lower and upper sums and it exists. We ¬rst
prove two results that are important.

De¬nition 5.6.2 A function f is said to be uniformly continuous on its
domain D if for each > 0 there exists δ > 0 such that if |x1 ’ x2 | < δ, for
any x1 and x2 in D, then
|f (x1 ) ’ f (x2 )| < .
244 CHAPTER 5. THE DEFINITE INTEGRAL

De¬nition 5.6.3 A collection C = {U± : U± is an open interval} is said to
cover a set D if each element of D belongs to some element of C.

Theorem 5.6.1 If C = {U± : U± is an open interval} covers a closed and
bounded interval [a, b], then there exists a ¬nite subcollection B = {U±1 , U±2 , · · · , U±n }
of C that covers [a, b].

Proof. We de¬ne a set A as follows:

A = {x : x ∈ [a, b] and [a, x] can be covered by a ¬nite subcollection of C}.

Since a ∈ A, A is not empty. A is bounded from above by b. Then A has a
least upper bound, say lub(A) = p. Clearly, p ¤ b. If p < b, then some U±
in C contains p. If U± = (a± , b± ), then a± < p < b± . Since p = ub(A), there
exists some point a— of A between a± and p. There exists a subcollection
B = {U±1 , · · · , U±n } that covers [a, a— ]. Then the collection
B1 = {U±1 , · · · , U±n , U± } covers [a, b± ). By the de¬nition of A, A must
contain all points of [a, b] between p and b± . This contradicts the assump-
tion that p = ub(A). So, p = b and b ∈ A. It follows that some ¬nite
subcollection of C covers [a, b] as required.

Theorem 5.6.2 If f is continuous on a closed and bounded interval [a, b],
then f is uniformly continuous on [a, b].
Proof. Let > 0 be given. If p ∈ [a, b], then there exists δp > 0 such
that |f (x) ’ f (p)| < /3, whenever p ’ δp < x < p + δp . Let Up =
1
1
p ’ δp , p + δp . Then C = {Up : p ∈ [a, b]} covers [a, b]. By The-
3 3
orem 5.6.1, some ¬nite subcollection B = {Up1 , Up2 , . . . , Upn } of C covers
1
[a, b]. Let δ = min{δpi : i = 1, 2, · · · , n}. Suppose that |x1 ’ x2 | < δ for
3
any two points x1 and x2 of [a, b]. Then x1 ∈ Upi and x2 ∈ Upj for some pi
and pj . We note that

|pi ’ pj | = |(pi ’ x1 ) + (x1 ’ x2 ) + (x2 ’ pj )|
¤ |pi ’ xi | + |x1 ’ x2 | + |x2 ’ pj |
1 1
< δpi + δ + δpj
3 3
¤ max{δpi , δpj }.
5.6. THE RIEMANN INTEGRAL 245

It follows that both pi and pj are either in Upi or Upj . Suppose that pi and
pj are both in Upi . Then

|x2 ’ pi | = |(x2 ’ x1 )| + (x1 ’ pi )|
¤ |x2 ’ x1 | + |x1 ’ pi |
1
< δ + δpi
3
< δpi .

So, x1 , x2 , pi and pj are all in Upi . Then

|f (x1 ) ’ f (x2 )| = |(f (x1 ) ’ f (pi )) + (f (pi ) ’ f (x2 ))|
¤ |f (x1 ) ’ f (pi )| + |f (pi ) ’ f (x2 )|
< +
3 3
<.

By De¬nition 5.6.2, f is uniformly continuous on [a, b].

Theorem 5.6.3 If f is continuous on [a, b], then f is (Riemann) integrable
and the de¬nite integral and the Riemann integral have the same value.
Proof. Let P = {a = x0 < x1 < x2 < . . . < xn = b} be a partition of [a, b]
and C = {ci : xi’1 ¤ ci ¤ xi , i = 1, 2, . . . , n} be an arbitrary selection. For
each i = 1, 2, . . . , n let
mi = absolute minimum of f on [xi’1 , xi ] obtained at c— , f (c— ) = mi ;
i
i
——
Mi = absolute maximum of f on [xi’1 , xi ] obtained at ci ) = Mi ;
m = absolute minimum of f on [a, b];
M = absolute maximum of f on [a, b];
n
R(P ) = f (ci )∆xi ,
i=1
Then for each i = 1, 2, . . . , n, we have
n
n
f (c— )(xi ’ xi’1 ) ¤ f (ci )(xi ’ xi’1 )
m(b ’ a) ¤ i
i=1
i=1
n
f (c—— )(xi ’ xi’1 ) ¤ M (b ’ a).
¤ i
i=1
246 CHAPTER 5. THE DEFINITE INTEGRAL

We recall that
n n
n
f (c—— )∆xi .
f (c— )∆xi , R(P ) = f (ci )∆xi , U (P ) =
L(P ) = i
i
i=1 i=1
i=1

We note that L(P ) and U (P ) are also Riemann sums and for every partition
P , we have
L(P ) ¤ R(P ) ¤ U (P ).
To prove the theorem, it is su¬cient to show that

lub{L(P )} = glb{U (P )}.

Since f is uniformly continuous, by Theorem 5.6.2, for each > 0 there is
some δ > 0 such that |f (x)’f (y)| < whenever |x’y| < δ for x and y in
b’a
[a, b]. Consider all partitions P , selections C = {ci }, C — = {c— }, C —— = {c—— }
i
i
such that
δ
||∆|| = max (xi ’ xi’1 ) < .
3
1¤i¤n

Then, for each i = 1, 2, . . . , n

|f (c—— ) ’ f (c— )| <
i
i
b’a
|f (c— ) ’ f (ci )| <
i
b’a
|f (c—— ) ’ f (ci )| <
i
b’a
n
(f (c—— ) ’ f (c— ))∆xi
|U (P ) ’ L(P )| = i
i
i=1
m
|f (c—— ) ’ f (c— )|∆xi
¤ i
i
i=1
n
< ∆xi
b’a i=1
=.

It follows that

lub{L(P )} = lim R(P )p = glb{U (P )} = I.
||∆||’0
5.6. THE RIEMANN INTEGRAL 247

By de¬nition of the de¬nite integral, I equals the de¬nite integral of f (x)
from x = a to x = b, which is also the Riemann integral of f on [a, b]. We
write
b
I= f (x) dx.
a
This proves Theorem 5.6.2 as well as Theorem 5.2.1.

Exercises 5.6
1. Prove Theorem 5.2.3. (Hint: For each partition P = {a = x0 < x1 <
. . . < xn = b] of [a, b],
g(b) ’ g(a) = [g(xn ) ’ g(xn’1 )] + [g(xn’1 ) ’ g(xn’2 )] + . . . + [g(x1 ) ’ g(x0 )]
n
[g(xi ) ’ g(xi’1 )]
=
i=1
n
g (ci )(xi ’ xi’1 )
= (by Mean Value Theorem)
i=1
n
f (ci )(xi ’ xi’1 )
=
i=1
= R(P )
for some selection C = {ci : xi’1 < ci < xi , i = 1, 2, · · · , n}.)
2. Prove Theorem 5.2.3 on the linearity property of the de¬nite integral.
(Hint:
n
b
[Af (ci ) + Bg(ci )] · [xi ’ xi’1 )
[Af (x) + bg(x)] dx = lim
||∆||’0
a i=1
n n
= lim A f (ci )∆xi + B g(ci )∆xi
||∆||’0
i=1 i=1
n
n
+ B lim g(ci )∆xi
=A lim f (ci )∆xi
||∆||’0
||∆||’0
i=1
i=1
b b
=A f (x)dx + B g(x)dx.)
a a
248 CHAPTER 5. THE DEFINITE INTEGRAL

3. Prove Theorem 5.2.4.
(Hint: [a, b] = [a, c] ∪ [c, b]. If P = {a = x0 < x1 < . . . < xn = b} is a
partition of [a, b], then for some i, P1 = {a = x0 < . . . < xi’1 < c < xi <
. . . < xn = b} yields a partition of [a, b]; {a < x0 < · · · < xi’1 < c} is a
partition of [a, c] and {c < xi < · · · < xn = b} is a partition of [c, b]. The
addition of c to the partition does not increase ||∆||.)
4. Prove Theorem 5.2.5.
(Hint: For each partition P and selection C we have
n n
f (ci )(xi ’ xi’1 ) ¤ g(ci )(xi ’ xi’1 ).)
i=1 i=1


5. Prove that if f is continuous on [a, b] and f (x) > 0 for each x ∈ [a, b],
then
b
f (x) dx > 0.
a
(Hint: There is some c in [a, b] such that f (c) is the absolute minimum
of f on [a, b] and f (c) > 0. Then argue that

0 < f (c)(b ’ a) ¤ L(P ) ¤ U (P )

for each partition P .)
6. Prove that if f and g are continuous on [a, b], f (x) > g(x) for all x in
[a, b], then
b b
f (x) dx > g(x) dx.
a a
(Hint: By problem 5,
b
(f (x) ’ g(x)) dx > 0.
a

Use the linearity property to prove the statement.)
7. Prove that if f is continuous on [a, b], then
b
b
f (x) dx ¤ |f (x)|dx.
a
a
5.6. THE RIEMANN INTEGRAL 249

(Hint: Recall that ’|f (x)| ¤ f (x) ¤ |f (x)| for all x ∈ [a, b]. Use problem
5 to conclude the result.)

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