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5.2. THE DEFINITE INTEGRAL 197

Theorem 5.2.4 (Order Property) If f and g are continuous on [a, b] and
f (x) ¤ g(x) for all x in [a, b], then
b b
f (x)dx ¤ g(x)dx.
a a

Proof. Suppose that f and g are continuous on [a, b] and f (x) ¤ g(x) for
all x in [a, b]. Let P = {a = x0 < x1 < x2 < · · · < xn = b} be a partition of
[a, b]. For each i there exists numbers ci , c— , di and d— such that
i
i
f (ci ) = absolute minimum of f on [xi’1 , xi ],
f (di ) = absolute maximum of f on [xi’1 , xi ],
g(c— ) = absolute minimum of g on [xi’1 , xi ],
i
g(d— ) = absolute maximum of g on [xi’1 , xi ].
i
By the assumption that f (x) ¤ g(x) on [a, b], we get

f (ci ) ¤ g(c— ) and f (di ) ¤ g(d— ).
i
i


Hence
Lf ¤ Lg and Uf ¤ Ug .
It follows that
b b
f (x)dx ¤ g(x)dx.
a a

This completes the proof of this theorem.

Theorem 5.2.5 (Mean Value Theorem for Integrals) If f is continuous
on [a, b], then there exists some point c in [a, b] such that
b
f (x)dx = f (c)(b ’ a).
a

Proof. Suppose that f is continuous on [a, b], and a < b. Let
m = absolute minimum of f on [a, b], and
M = absolute maximum of f on [a, b].
Then, by Theorem 5.2.4,
b b b
m(b ’ a) ¤ m dx ¤ f (x)dx ¤ M dx = M (b ’ a)
a a a
198 CHAPTER 5. THE DEFINITE INTEGRAL

and

b
1
m¤ f (x)dx ¤ M.
b’a a


By the intermediate value theorem for continuous functions, there exists some
c such that
b
1
f (c) = f (x)dx
b’a a


and

b
f (x)dx = f (c)(b ’ a).
a


For a = b, take c = a. This completes the proof of this theorem.

De¬nition 5.2.2 The number f (c) given in Theorem 5.2.6 is called the av-
erage value of f on [a, b], denoted fav [a, b]. That is

b
1
f (x)dx.
fav [a, b] =
b’a a




Theorem 5.2.6 (Fundamental Theorem of Calculus, First Form) Suppose
that f is continuous on some closed and bounded interval [a, b] and
x
g(x) = f (t)dt
a


for each x in [a, b]. Then g(x) is continuous on [a, b], di¬erentiable on (a, b)
and for all x in (a, b), g (x) = f (x). That is

x
d
f (t)dt = f (x).
dx a
5.2. THE DEFINITE INTEGRAL 199

Proof. Suppose that f is continuous on [a, b] and a < x < b. Then

1
[g(x + h) ’ g(x)]
g (x) = lim
h
h’0
x+h x
1
f (t)dt ’ f (t)dt
= lim
ha
h’0 a
x x+h x
1
f (t)dt ’
= lim f (t)dt + f (t)dt (Why?)
ha
h’0 x a
x+h
1
= lim f (t)dt
hx
h’0
1
[f (c)(x + h ’ x)]
= lim by Theorem 5.2.5)
h
h’0
= lim f (c)
h’0

for some c between x and x + h.
Since f is continuous on [a, b] and c is between x and x + h, it follows
that
g (x) = lim f (c) = f (x)
h’0

for all x such that a < x < b.
At the end points a and b, a similar argument can be used for one sided
derivatives, namely,

g(x + h) ’ g(x)
g (a+ ) = lim
h
+
h’0
g(x + h) ’ g(x)
g (b’ ) = lim .
h
h’0’

We leave the end points as an exercise. This completes the proof of this
theorem.

Theorem 5.2.7 (Fundamental Theorem of Calculus, Second Form) If f
and g are continuous on a closed and bounded interval [a, b] and g (x) = f (x)
on [a, b], then
b
f (x)dx = g(b) ’ g(a).
a

We use the notation: [g(x)]b = g(b) ’ g(a).
a
200 CHAPTER 5. THE DEFINITE INTEGRAL

Proof. Let f and g be continuous on the closed and bounded interval [a, b]
and for each x in [a, b], let
x
G(x) = f (t)dt.
a

Then, by Theorem 5.2.6, G (x) = f (x) on [a, b]. Since G (x) = g(x) for all x
on [a, b], there exists some constant C such that
G(x) = g(x) + C
for all x on [a, b]. Since G(a) = 0, we get C = ’g(a). Then
b
f (x)dx = G(b)
a
= g(b) + C
= g(b) ’ g(a).
This completes the proof of Theorem 5.2.7.

Theorem 5.2.8 (Leibniz Rule) If ±(x) and β(x) are di¬erentiable for all
x and f is continuous for all x, then
β(x)
d
f (t)dt = f (β(x)) · β (x) ’ f (±(x)) · ± (x).
dx ±(x)

Proof. Suppose that f is continuous for all x and ±(x) and β(x) are di¬er-
entiable for all x. Then
β(x) 0 β(x)
d d
f (t)dt = f (t)dt + f (t)dt
dx dx
±(x) ±(x) 0

β(x) ±(x)
d
f (t)dt ’
= f (t)dt
dx 0 0

β(x) ±(x)
d d(β(x)) d d(±(x))
· ’
= f (t)dt f (x)dt
d(β(x)) dx d(±(x)) dx
0 0

= f (β(x)) β (x) ’ f (±(x))± (x) (by Theorem 5.2.6)
This completes the proof of Theorem 5.2.8.
5.2. THE DEFINITE INTEGRAL 201

Example 5.2.1 Compute each of the following de¬nite integrals and sketch
the area represented by each integral:

4 π
2
(i) x dx (ii) sin x dx
0 0

π/2 10
ex dx
(iii) cos x dx (iv)
’π/2 0

π/3 π/2
(v) tan x dx (vi) cot x dx
0 π/6

π/4 3π/4
(vii) sec x dx (viii) csc x dx
’π/4 π/4

1 1
(xi) sinh x dx (x) cosh x dx
0 0



We note that each of the functions in the integrand is positive on the re-
spective interval of integration, and hence, represents an area. In order to
compute these de¬nite integrals, we use the Fundamental Theorem of Cal-
culus, Theorem 5.2.2. As in Chapter 4, we ¬rst determine an anti-derivative
g(x) of the integrand f (x) and then use
b
f (x)dx = g(b) ’ g(a) = [g(x)]b .
a
a




graph




4
4
x3 64
2
=
(i) x dx =
3 3
0 0
202 CHAPTER 5. THE DEFINITE INTEGRAL

graph



π
sin x dx = [’ cos x]π = 1 ’ (’1) = 2
(ii) 0
0




graph



π/2
π/2
cos x dx = [sin x]’π/2 = 1 ’ (’1) = 2
(iii)
’π/2




graph



10
ex dx = [ex ]10 = e10 ’ e0 = e10 ’ 1
(iv) 0
0




graph



π/3
π
π/3
tan x dx = [ln | sec x|]0 = ln sec
(v) = ln 2
3
0




graph
5.2. THE DEFINITE INTEGRAL 203

π/2
1
π/2
cot x dx = [ln | sin x|]π/6 = ln(1) ’ ln
(vi) = ln 2
2
π/6




graph




√ √
π/4
π/4
sec x dx = [ln | sec x + = ln | 2 + 1| ’ ln | 2 ’ 1|
(vii) tan x|]’π/4
’π/4




graph



3π/4
3π/4
csc x dx = [’ ln | csc x + cot x|]π/4
(viii)
π/4
√ √
= ’ ln | 2 ’ 1| + ln | 2 + 1|



graph



1
sinh x dx = [cosh x]1 = cosh 1 ’ cosh 0 = cosh 1 ’ 1
(ix) 0
0




graph
204 CHAPTER 5. THE DEFINITE INTEGRAL

1
cosh x dx = [sinh x]1 = sinh 1
(x) 0
0




graph




Example 5.2.2 Evaluate each of the following integrals:

10 π/2
1
(i) dx (ii) sin(2x)dx
x
1 0

π/6 2
(x4 ’ 3x2 + 2x ’ 1)dx
(iii) cos(3x)dx (iv)
0 0

3 4
(v) sinh(4x)dx (vi) cosh(2x)dx
0 0




d 1
(ln |x|) = ,
(i) Since
dx x
10
1
dx = [ln |x|]10 = ln(10)
1
x
1



’1
d
(ii) Since cos(2x) = sin(2x),
dx 2
π/2
π/2
’1 11
sin 2x dx = cos(2x) = + = 1.
2 22
0 0



π/6
π/6
1 1 π 1
(iii) cos(3x) = sin(3x) = sin =.
3 3 2 3
0 0
5.2. THE DEFINITE INTEGRAL 205

2
2
1
(x ’ 3x + 2x ’ 1)dx = x5 ’ x3 + x2 ’ x
4 2
(iv)
5
0 0
32
’8+4’2 ’0
=
5
2
=.
5
3
3
1 1 1
cosh(4x) = cosh(12) ’ cosh(0)
(v) sinh(4x)dx =
4 4 4
0 0
1
= (cosh(12) ’ 1)
4
4
4
1 1
(vi) cosh(2x)dx = sinh(2x) = cosh(8)
2 2
0 0



Example 5.2.3 Verify each of the following:
4 3 4
2 2
x2 dx
(i) x dx = x dx +
0 0 3

4 4
2
x3 dx
(ii) x dx <
1 1
x
d
(t2 + 3t + 1)dt = x2 + 3x + 1
(iii)
dx 0

x3
d
cos(t)dt = 3x2 cos(x3 ) ’ 2x cos(x2 ).
(iv)
dx x2

2
(v) If f (x) = sin x, then fav [0, π] = .
π

4
4
x3 64
2
=
(i) x dx =
3 3
0 0
3 4
3 4
x3 x3
2 2
x dx + x dx = +
3 3
0 3 0 3
206 CHAPTER 5. THE DEFINITE INTEGRAL

27 64 27 64
’0 + ’
= = .
3 3 3 3
Therefore,
4 3 4
2 2
x2 dx.
x dx = x dx +
1 0 3

4
4
x3 64 1
2
’ = 21
=
(ii) x dx =
3 3 3
1 1
4
4
x4 1
3
64 ’
=
x dx =
4 4
1 1
4 4
2
x3 dx. We observe that x2 < x3 on (1, 4].
Therefore, x dx <
1 1
x
x
t3 t2
2
(iii) (t + 3t + 1)dt = +3 +t
3 2
0 0


x3 3 2
= + x +x
3 2
x3 3 2
d
= x2 + 3x + 1.
+ x +x
dx 3 2

x3
d d 3
[sin t]x2
(iv) cos tdt = x
dx dx
x2


d
[sin(x3 ) ’ sin(x2 )]
=
dx

= cos(x3 ) · 3x2 ’ cos(x2 ) · 2x

= 3x2 cos(x3 ) ’ 2x cos(x2 ).
Using the Leibniz Rule, we get
x3
d
= cos(x3 ) · 3x2 ’ cos(x2 ) · 2x
cos tdt
dx x2

= 3x2 cos x3 ’ 2x cos x2 .
5.2. THE DEFINITE INTEGRAL 207

(v) The average value of sin x on [0, π] is given by
π
1
1
[’ cos x]π
sin x dx = 0
π’0 π
0
1
= [’(’1) + 1]
π
2
=.
π

Basic List of Inde¬nite Integrals:

xn+1
1
x dx = x4 + c
3 n
1. 2. x dx = + c, n = 1
4 n+1

1
dx = ln |x| + c sin x dx = ’ cos x + c
3. 4.
x

’1
5. sin(ax)dx = cos(ax) + c 6. cos x dx = sin x + c
a

1
tan xdx = ln | sec x| + c
7. cos(ax) dx = sin(ax) + c 8.
a

1
ln | sec(ax)| + c cot x dx = ln | sin x| + c
9. tan(ax) dx = 10.
a

1
ex dx = ex + c
ln | sin(ax)| + c
11. cot(ax) dx = 12.
a

1 ax
e’x dx = ’e’x + c eax dx =
13. 14. e +c
a

15. sinh x dx = cosh x + c 16. cosh x dx = sinh x + c


tanh x dx = ln | cosh x| + c coth x dx = ln | sinh x| + c
17. 18.

1 1
19. sinh(ax) dx = cosh(ax) + c 20. cosh(ax) dx = sinh(ax) + c
a a
208 CHAPTER 5. THE DEFINITE INTEGRAL


1 1
ln | cosh ax| + c ln | sinh(ax)| + c
21. tanh(ax) dx = 22. coth(ax) dx =
a a

sec x dx = ln | sec x + tan x| + c csc x dx = ’ ln | csc x + cot x| + c
23. 24.

1
ln | sec(ax) + tan(ax)| + c
25. sec(ax) dx =
a

’1
ln | csc(ax) + cot(ax)| + c
26. csc(ax) dx =
a

1
sec2 x dx = tan x + c sec2 (ax) dx =
27. 28. tan(ax) + c
a

’1
csc2 x dx = ’ cot x + c csc2 (ax) dx =
29. 30. cot(ax) + c
a

tan2 x dx = tan x ’ x + c cot2 x dx = ’ cot x ’ x + c
31. 32.

1 1
sin2 x dx = cos2 x dx =
(x ’ sin x cos x) + c
33. 34. (x + sin x cos x) + c
2 2

csc x dx = ’ csc x + c
35. sec x tan x dx = sec x + c 36.




Exercises 5.2 Using the preceding list of inde¬nite integrals, evaluate the
following:

5 3π/2 3π/2
1
1. dt 2. sin x dx 3. cos x dx
t
1 0 0

10 π/10 π/6
x
4. e dx 5. sin(5x) dx 6. cos(5x) dx
0 0 0
5.2. THE DEFINITE INTEGRAL 209

π/6 1 2
’x
e3x dx
7. cot(3x) dx 8. e dx 9.
’1
π/12 0

2 4 1
10. sinh(2x) dx 11. cosh(3x) dx 12. tanh(2x) dx
0 0 0

2 π/6 π/6
13. coth(3x) dx 14. sec(2x)dx 15. csc(2x) dx
1 π/12 π/12

π/8 π/6 π/4
2 2
tan2 x dx
16. sec (2x) dx 17. csc (2x) 18.
0 π/12 0

π/4 π π/2
2 2
cos2 x dx
19. cot x dx 20. sin xdx 21.
’π/2
π/6 0

π/4 π/4 2
e’3x dx
22. sec x tan x dx 23. csc x cot x dx 24.
π/6 π/6 0



Compute the average value of each given f on the given interval.

’π
26. f (x) = x1/3 , [0, 8]
25. f (x) = sin x, ,π
2

’π π
28. f (x) = sin2 x, [0, π]
27. f (x) = cos x, ,
22

30. f (x) = e’x , [’2, 2]
29. f (x) = cos2 x, [0, π]

Compute g (x) without computing the integrals explicitly.

4x3
x
(1 + t2 )2/3 dt
31. g(x) = 32. g(x) = arctan(x) dx
x2
0

x2 arcsinh x
3 1/3
(1 + t2 )3/2 dt
33. g(x) = (1 + t ) dt 34. g(x) =
x3 arcsin x
210 CHAPTER 5. THE DEFINITE INTEGRAL

x sin 3x
1
(1 + t2 )1/2 dt
35. g(x) = dt 36. g(x) =
t
1 sin 2x

sin(x3 ) 4x
1
3 1/3
37. g(x) = (1 + t ) dt 38. dt
1 + t2
sin(x2 ) x

x3 ex
2t dt
39. arcsin(x) dx 40.
x2 ln x




5.3 Integration by Substitution
Many functions are formed by using compositions. In dealing with a com-
posite function it is useful to change variables of integration. It is convenient
to use the following di¬erential notation:
If u = g(x), then du = g (x) dx.
The symbol “du” represents the “di¬erential of u,” namely, g (x)dx.

Theorem 5.3.1 (Change of Variable) If f, g and g are continuous on an
open interval containing [a, b], then
b g(b)
f (g(x)) · g (x) dx =
(i) f (u)du
a g(a)



(ii) f (g(x))g (x) dx = f (u)du,

where u = g(x) and du = g (x) dx.
Proof. Let f, g, and g be continuous on an open interval containing [a, b].
For each x in [a, b], let
x
F (x) = f (g(x))g (x)dx
a

and
g(x)
G(x) = f (u)du.
g(a)
5.3. INTEGRATION BY SUBSTITUTION 211

Then, by Leibniz Rule, we have

F (x) = f (g(x))g (x),

and
G (x) = f (g(x))g (x)
for all x on [a, b].
It follows that there exists some constant C such that

F (x) = G(x) + C

for all x on [a, b]. For x = a we get

0 = F (a) = G(a) + C = 0 + C

and, hence,
C = 0.
Therefore, F (x) = G(x) for all x on [a, b], and hence
b
f (g(x))g (x)dx = F (b)
a
= G(b)
g(b)
= f (u)du.
g(a)


This completes the proof of this theorem.

Remark 18 We say that we have changed the variable from x to u through
the substitution u = g(x).


Example 5.3.1

2 6
1 1 1
sin udu = [’ cos u]6 = (1 ’ cos 6),
(i) sin(3x) dx = 0
03 3 3
0
1
where u = 3x, du = 3 dx, dx = du.
3
212 CHAPTER 5. THE DEFINITE INTEGRAL

2 4
3
2
(ii) 3x cos(x ) dx = cos u du
2
0 0
3
[sin u]4
= 0
2
3
= sin 4,
2
3
where u = x2 , du = 2x dx, 3x dx = du.
2

3 9
1 1 1
x2
du = [eu ]9 = (e9 ’ 1),
eu
(iii) e x dx = 0
2 2 2
0 0
1
where u = x2 , du = 2x dx, x dx = dx.
2


De¬nition 5.3.1 Suppose that f and g are continuous on [a, b]. Then the
area bounded by the curves y = f (x), y = g(x), y = a and x = b is de¬ned
to be A, where
b
|f (x) ’ g(x)| dx.
A=
a

If f (x) ≥ g(x) for all x in [a, b], then
b
(f (x) ’ g(x)) dx.
A=
a

If g(x) ≥ f (x) for all x in [a, b], then
b
(g(x) ’ f (x)) dx.
A=
a



Example 5.3.2 Find the area, A, bounded by the curves y = sin x, y =
cos x, x = 0 and x = π.



graph
5.3. INTEGRATION BY SUBSTITUTION 213

π π
We observe that cos x ≥ sin x on 0, and sin x ≥ cos x on , π . There-
4 4
fore, the area is given by
π
| sin x ’ cos x| dx
A=
0
π/4 π
(cos x ’ sin x) dx + (sin x ’ cos x)dx
=
0 π/4
π/4
= [sin x + cos x]0 + [’ cos x ’ sin x]π
π/4
√ √ √ √
2 2 2 2
’1 + 1+
= + +
2 2 2 2

= 2 2.

Example 5.3.3 Find the area, A, bounded by y = x2 , y = x3 , x = 0 and
x = 2.



graph



We note that x3 ¤ x2 on [0, 1] and x3 ≥ x2 on [1, 2]. Therefore, by de¬nition,
1 2
2 3
(x3 ’ x2 ) dx
(x ’ x ) dx +
A=
0 1
1 2
13 14 14 13
x’ x x’
= + x
3 4 4 3
0 1
11 8 11
’ + 4’ ’ ’
=
34 3 43
1 4 1
= ++
12 3 12
3
= .
2

Example 5.3.4 Find the area bounded by y = x3 and y = x. To ¬nd the
interval over which the area is bounded by these curves, we ¬nd the points
of intersection.
214 CHAPTER 5. THE DEFINITE INTEGRAL

graph




x3 = x ” x3 ’ x = 0 ” x(x2 ’ 1) = 0
” x = 0, x = 1, x = ’1.

The curve y = x is below y = x3 on [’1, 0] and the curve y = x3 is below
the curve y = x on [0, 1]. The required area is A, where
0 1
3
(x ’ x3 ) dx
(x ’ x) dx +
A=
’1 0
1
0
1 2 x4
14 12
x’ x x’
= +
4 2 2 4
’1 0
11 11
’ ’
= +
24 24
1
=
2

Exercises 5.3 Find the area bounded by the given curves.

1. y = x2 , y = x3 2. y = x4 , y = x3

3. y = x2 , y = 4. y = 8 ’ x2 , y = x2
x

’π π
5. y = 3 ’ x2 , y = 2x 6. y = sin x, y = cos x, x = ,x =
2 2
π
7. y = x2 + 4x, y = x 8. y = sin 2x, y = x, x =
2
π
9. y 2 = 4x, x ’ y = 0 10. y = x + 3, y = cos x, x = 0, x =
2


Evaluate each of the following integrals:
5.3. INTEGRATION BY SUBSTITUTION 215



11. sin 3x dx 12. cos 5x dx

2
ex x dx x sin(x2 ) dx
13. 14.


x2 tan(x3 + 1) dx sec2 (3x + 1) dx
15. 16.


csc2 (2x ’ 1) dx x sinh(x2 ) dx
17. 18.


x2 cosh(x3 + 1) dx
19. 20. sec(3x + 5) dx


x tanh(x2 + 1) dx
csc(5x ’ 7) dx
21. 22.


x2 coth(x3 ) dx sin3 x cos x dx
23. 24.


tan5 x sec2 x dx cot3 x csc2 x dx
25. 26.


sec3 x tan x dx csc3 x cot x dx
27. 28.

(arcsin x)4 (arctan x)3

29. dx 30. dx
1 + x2
2
1’x
1 π/6
x2
31. xe dx 32. sin(3x)dx
0 0

π/4 3
1
33. cos(4x) dx 34. dx
(3x + 1)
0 0

π/2 π/6
3
cos3 (3x) sin 3x dx
35. sin x cos x dx 36.
0 0
216 CHAPTER 5. THE DEFINITE INTEGRAL

5.4 Integration by Parts
The product rule of di¬erentiation yields an integration technique known as
integration by parts. Let us begin with the product rule:

du(x) dv(x)
d
(u(x)v(x)) = v(x) + u(x) .
dx dx dx
On integrating each term with respect to x from x = a to x = b, we get
b b b
d du(x) dv(x)
(u(x)v(x)) dx = v(x) dx + u(x) dx.
dx dx dx
a a a

By using the di¬erential notation and the fundamental theorem of calculus,
we get
b b
[u(x)v(x)]b = v(x)u (x) dx + u(x)v (x) dx.
a
a a

The standard form of this integration by parts formula is written as
b
b
[u(x)v(x)]b ’ v(x)u (x) dx
(i) u(x)v (x) dx = a
a
a
and

udv = uv ’
(ii) vdu

We state this result as the following theorem:

Theorem 5.4.1 (Integration by Parts) If u(x) and v(x) are two functions
that are di¬erentiable on some open interval containing [a, b], then
b
b
[u(x)v(x)]b ’ v(x)u (x) dx
(i) u(x)v (x) dx = a
a
a
for de¬nite integrals and


udv = uv ’
(ii) vdu

for inde¬nite integrals.
5.4. INTEGRATION BY PARTS 217

Proof. Suppose that u and v are di¬erentiable on some open interval con-
taining [a, b]. For each x on [a, b], let
x x
F (x) = u(x)v (x)dx + v(x)u (x)dx.
a a

Then, for each x on [a, b],

F (x) = u(x)v (x) + v(x)u (x)
d
= (u(x)v(x)).
dx
Hence, there exists some constant C such that for each x on [a, b],

F (x) = u(x)v(x) + C.

For x = a, we get
F (a) = 0 = u(a)v(a) + C
and, hence,
C = ’u(a)v(a).
Then,
b b
u(x)v (x)dx + v(x)u (x)dx = F (b)
a a
= u(b)v(b) + C
= u(b)v(b) ’ u(a)v(a).

Consequently,
b b
u(x)v (x)dx = [u(b)v(b) ’ u(a)v(a)] ’ v(x)u (x)dx.
a a

This completes the proof of Theorem 5.4.1.

Remark 19 The “two parts” of the integrand are “u(x)” and “v (x)dx” or
“u” and “dv”. It becomes necessary to compute u (x) and v(x) to make the
integration by parts step.

Example 5.4.1 Evaluate the following integrals:
218 CHAPTER 5. THE DEFINITE INTEGRAL


xe’x dx
(i) x sin x dx (ii) (iii) (ln x) dx


x2 ex dx
(iv) arcsin x dx (v) arccos x dx (vi)



(i) We let u = x and dv = sin x dx. Then du = dx and

v(x) = sin x dx

= ’ cos x + c.

We drop the constant c, since we just need one v(x). Then, by the
integration by parts theorem, we get

x sin x dx = udv

= uv ’ vdu

= x(’ cos x) ’ (’ cos x) dx

= ’x cos x + sin x + c.



(ii) We let u = x, du = dx, dv = e’x dx, v = e’x dx = ’e’x . Then,


xe’x dx = x(’e’x ) ’ (’e’x ) dx

= ’xe’x ’ e’x + c.

1
(iii) We let u = (ln x), du = dx, dv = dx, v = x. Then,
x
1
ln x dx = x ln x ’ x· dx
x
= x ln x ’ x + c.
5.4. INTEGRATION BY PARTS 219

1
(iv) We let u = arcsin x, du = √ dx, dv = dx, v = x. Then,
1 ’ x2
x
arcsin x dx = x arcsin x ’ √ dx.
1 ’ x2
To evaluate the last integral, we make the substitution y = 1 ’ x2 . Then,
dy = ’2xdx and x dx = (’1/2)du and hence
(’1/2)du
x
√ dx =
u1/2
1 ’ x2
1
u’1/2 du
=’
2
= ’u1/2 + c

= ’ 1 ’ x2 + c.
Therefore,

arcsin x dx = x arcsin x ’ 1 ’ x2 + c.


(v) Part (v) is similar to part (iv) and is left as an exercise.

(vi) First we let u = x2 , du = 2x dx, dv = ex dx, v = ex dx = ex . Then,

x2 ex dx = x2 ex ’ 2xex dx

= x 2 ex ’ 2 xex dx.

To evaluate the last integral, we let u = x, du = dx, dv = ex dx, v = ex .
Then
xex dx = xex ’ ex dx

= xex ’ ex + c.
Therefore,

x2 ex dx = x2 ex ’ 2(xex ’ ex + c)

= x2 ex ’ 2xex + 2ex ’ 2c
= ex (x2 ’ 2x + 2) + D.
220 CHAPTER 5. THE DEFINITE INTEGRAL

Example 5.4.2 Evaluate the given integrals in terms of integrals of the
same kind but with a lower power of the integrand. Such formulas are called
the reduction formulas. Apply the reduction formulas for n = 3 and n = 4.

sinn x dx cscm+2 x dx cosn x dx
(i) (ii) (iii) (iv)

secm+2 x dx

(i) We let
u = (sin x)n’1 , du = (n ’ 1)(sin x)n’2 cos x dx

sin x dx = ’ cos x.
dv = sin x dx, v =

Then
sinn x dx = (sin x)n’1 (sin x dx)

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

= ’(sin x)n’1 cos x + (n ’ 1) (sin x)n’2 (1 ’ sin2 x) dx

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

sinn x dx.
’ (n ’ 1)

We now use algebra to solve the integral as follows:

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

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

’1 n’1
sinn x dx = (sin x)n’1 cos x + sinn’2 x dx . (1)
n n
We have reduced the exponent of the integrand by 2. For n = 3, we get
’1 2
sin3 x dx = (sin x)2 cos x + sin x dx
3 3
’1 ’2
(sin x)2 cos x
= cos x + c.
3 3
5.4. INTEGRATION BY PARTS 221

For n = 2, we get

’1 1
sin2 x dx = (sin x) cos x + 1 dx
2 2
’1 x
= sin x cos x + + c
2 2
1
= (x ’ sin x cos x) + c.
2
For n = 4, we get

’1 3
sin4 x dx = (sin x)3 cos x + sin2 x dx
4 4
’1 31
(sin x)3 cos x + · (x ’ sin x cos x) + c.
=
4 42
In this way, we have a reduction formula by which we can compute the
integral of any positive integral power of sin x. If n is a negative integer,
then it is useful to go in the direction as follows:
Suppose n = ’m, where m is a positive integer. Then, from equation
(1) we get

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

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