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