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FIRST YEAR CALCULUS
W W L CHEN
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W W L Chen, 1994, 2008.
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Chapter 10
TECHNIQUES OF INTEGRATION
10.1. Integration by Substitution
In this section, we discuss how we can use the Chain rule in differentiation to help solve problems in
integration. This technique is usually called integration by substitution. As we shall not prove any result
here, our discussion will be only heuristic.
We emphasize that the technique does not always work. First of all, we have little or no knowledge of
the antiderivatives of many functions. Secondly, there is no simple routine that we can describe to help
us find a suitable substitution even in the cases where the technique works. On the other hand, when
the technique does work, there may well be more than one suitable substitution!
Occasionally, the possibility of substitution may not be immediately obvious, and a certain amount of
trial and error does occur. The fact that one substitution does not appear to work does not mean that
the method fails. It may very well be the case that we have used a bad substitution.
INTEGRATION BY SUBSTITUTION – VERSION 1. If we make a substitution x = g(u), then
dx = g 0 (u) du, and
Z
Z
f (x) dx = f (g(u))g 0 (u) du.
Example 10.1.1. Consider the indefinite integral
Z
1
√
dx.
1 − x2
If we make a substitution x = sin u, then dx = cos u du, and
Z
Z
Z
1
cos u
√
p
du = du = u + C = sin−1 x + C.
dx =
1 − x2
1 − sin2 u
Chapter 10 : Techniques of Integration
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On the other hand, if we make a substitution x = cos v, then dx = − sin v dv, and
Z
Z
Z
1
sin v
√
dx = − √
dv = − dv = −v + C = − cos−1 x + C.
1 − x2
1 − cos2 v
Example 10.1.2. Consider the indefinite integral
Z
1
dx.
1 + x2
If we make a substitution x = tan u, then dx = sec2 u du, and
Z
1
dx =
1 + x2
Z
sec2 u
du =
1 + tan2 u
Z
du = u + C = tan−1 x + C.
On the other hand, if we make a substitution x = cot v, then dx = − csc2 v dv, and
Z
1
dx = −
1 + x2
Z
csc2 v
dv = −
1 + cot2 v
Z
dv = −v + C = − cot−1 x + C.
Example 10.1.3. Consider the indefinite integral
Z
√
x x + 1 dx.
If we make a substitution x = u2 − 1, then dx = 2u du, and
Z
Z
Z
Z
√
x x + 1 dx = 2(u2 − 1)u2 du = 2 u4 du − 2 u2 du
=
2 5 2 3
2
2
u − u + C = (x + 1)5/2 − (x + 1)3/2 + C.
5
3
5
3
On the other hand, if we make a substitution x = v − 1, then dx = dv, and
Z
Z
Z
Z
√
x x + 1 dx = (v − 1)v 1/2 dv = v 3/2 dv − v 1/2 dv
=
2 5/2 2 3/2
2
2
v
− v
+ C = (x + 1)5/2 − (x + 1)3/2 + C.
5
3
5
3
We can confirm that the indefinite integral is correct by checking that
√
d 2
2
(x + 1)5/2 − (x + 1)3/2 + C = x x + 1.
dx 5
3
INTEGRATION BY SUBSTITUTION – VERSION 2. Suppose that a function f (x) can be
written in the form f (x) = g(h(x))h0 (x). If we make a substitution u = h(x), then du = h0 (x) dx, and
Z
Z
Z
f (x) dx = g(h(x))h0 (x) dx = g(u) du.
Remark. Note that in Version 1, the variable x is initially written as a function of the new variable u,
whereas in Version 2, the new variable u is written as a function of x. The difference, however, is
minimal, as the substitution x = g(u) in Version 1 has to be invertible to enable us to return from the
new variable u to the original variable x at the end of the process.
Chapter 10 : Techniques of Integration
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Example 10.1.4. Consider the indefinite integral
Z
3
x2 ex dx.
Note first of all that the derivative of the function x3 is equal to 3x2 , so it is convenient to make the
substitution u = x3 . Then du = 3x2 dx, and
Z
3
x2 ex dx =
1
3
Z
3
3x2 ex dx =
1
3
Z
1 u
1 3
e + C = ex + C.
3
3
eu du =
3
A somewhat more complicated alternative is to note that the derivative of the function ex is equal to
3
3
3
3x2 ex , so it is convenient to make the substitution v = ex . Then dv = 3x2 ex dx, and
Z
3
x2 ex dx =
1
3
Z
3
3x2 ex dx =
1
3
Z
dv =
1
1 3
v + C = ex + C.
3
3
Example 10.1.5. Consider the indefinite integral
Z
x(x2 + 3)4 dx.
Note first of all that the derivative of the function x2 + 3 is equal to 2x, so it is convenient to make the
substitution u = x2 + 3. Then du = 2x dx, and
Z
x(x2 + 3)4 dx =
1
2
Z
2x(x2 + 3)4 dx =
1
2
Z
u4 du =
1 5
1 2
u +C =
(x + 3)5 + C.
10
10
Example 10.1.6. Consider the indefinite integral
Z
1
dx.
x log x
Note first of all that the derivative of the function log x is equal to 1/x, so it is convenient to make the
substitution u = log x. Then du = (1/x) dx, and
Z
1
dx =
x log x
Z
1
du = log |u| + C = log | log x| + C.
u
Example 10.1.7. Consider the indefinite integral
Z
tan3 x sec2 x dx.
Note first of all that the derivative of the function tan x is equal to sec2 x, so it is convenient to make
the substitution u = tan x. Then du = sec2 x dx, and
Z
3
2
tan x sec x dx =
Chapter 10 : Techniques of Integration
Z
u3 du =
1 4
1
u + C = tan4 x + C.
4
4
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Example 10.1.8. Consider the indefinite integral
Z
sin3 x cos3 x dx.
Note first of all that the derivative of the function sin x is equal to cos x, so it is perhaps convenient to
make the substitution u = sin x. Then du = cos x dx, and
Z
3
Z
3
3
Z
2
u (1 − u ) du =
sin x cos x dx =
(u3 − u5 ) du =
u4
u6
sin4 x sin6 x
−
+C =
−
+ C.
4
6
4
6
Alternatively, note that the derivative of the function cos x is equal to − sin x, so it is convenient to make
the substitution v = cos x. Then dv = − sin x dx, and
Z
Z
sin3 x cos3 x dx =
−(1 − v 2 )v 3 dv =
Z
(v 5 − v 3 ) dv =
v4
cos6 x cos4 x
v6
−
+ C0 =
−
+ C 0.
6
4
6
4
It can be checked that
cos6 x cos4 x
1
sin4 x sin6 x
−
=
−
+ .
4
6
6
4
12
Example 10.1.9. Recall Example 10.1.1. Since
Z
1
√
dx = sin−1 x + C,
1 − x2
we have
Z
0
1/2
√
1/2
π
1
1
= sin−1 − sin−1 0 = .
dx = sin−1 x
2
2
6
1−x
0
Note that we have in fact used the substitution x = sin u to show that
Z
Z
1
√
dx = du = u + C,
1 − x2
followed by an inverse substitution u = sin−1 x. Here, we need to make the extra step of substituting the
values x = 0 and x = 1/2 to the indefinite integral sin−1 x. Observe, however, that with the substitution
x = sin u, the variable x increases from 0 to 1/2 as the variable u increases from 0 to π/6. But then
Z
π/6
0
π/6
Z 1/2
π
1
√
du = u
= =
dx,
6
1 − x2
0
0
so it appears that we do not need the inverse substitution u = sin−1 x. Perhaps we can directly substitute
u = 0 and u = π/6 to the indefinite integral u.
DEFINITE INTEGRAL BY SUBSTITUTION – VERSION 1. Suppose that a substitution
x = g(u) satisfies the following conditions:
(a) There exist α, β ∈ R such that g(α) = A and g(β) = B.
(b) The derivative g 0 (u) > 0 for every u satisfying α < u < β.
Then dx = g 0 (u) du, and
Z
B
Z
β
f (x) dx =
A
Chapter 10 : Techniques of Integration
f (g(u))g 0 (u) du.
α
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Remark. If condition (b) above is replaced by the condition that the derivative g 0 (u) < 0 for every u
satisfying β < u < α, then the same conclusion holds if we adopt the convention that
Z
β
f (g(u))g 0 (u) du = −
α
Z
α
f (g(u))g 0 (u) du.
β
Example 10.1.10. To calculate the definite integral
1
Z
0
1
dx,
1 + x2
we can use the substitution x = tan u, so that dx = sec2 u du. Note that tan 0 = 0 and tan(π/4) = 1,
and that sec2 u > 0 whenever 0 < u < π/4. It follows that
Z
1
0
1
dx =
1 + x2
Z
π/4
0
sec2 u
du =
1 + tan2 u
π/4
Z
0
π/4
π
π
du = u
= −0= .
4
4
0
We can compare this to first observing Example 10.1.2, so that
1
Z
0
1
π
π
1
−1
dx = tan x = tan−1 1 − tan−1 0 = − 0 = .
2
1+x
4
4
0
Example 10.1.11. To calculate the definite integral
Z
3
√
x x + 1 dx,
0
we can use the substitution x = g(u) = u2 − 1, so that dx = 2u du. Note that g(1) = 0 and g(2) = 3,
and that g 0 (u) = 2u > 0 whenever 1 < u < 2. It follows that
Z
0
3
√
x x + 1 dx =
Z
2
2(u2 − 1)u2 du =
1
2 5 2 3
u − u
5
3
2
=
1
64 16
−
5
3
−
2 2
−
5 3
=
62 14
116
−
=
.
5
3
15
DEFINITE INTEGRAL BY SUBSTITUTION – VERSION 2. Suppose that a substitution
u = h(x) satisfies the following conditions:
(a) There exists a function g(u) such that f (x) = g(h(x))h0 (x) for every x ∈ [A, B].
(b) The derivative h0 (x) > 0 for every x satisfying A < x < B.
Then du = h0 (x) dx, and
Z
B
Z
B
f (x) dx =
A
g(h(x))h0 (x) dx =
A
Z
h(B)
g(u) du.
h(A)
Remark. If condition (b) above is replaced by the condition that the derivative h0 (x) < 0 for every x
satisfying A < x < B, then the same conclusion holds if we adopt the convention that
Z
h(B)
Z
h(A)
g(u) du = −
h(A)
Chapter 10 : Techniques of Integration
g(u) du.
h(B)
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Example 10.1.12. To calculate the definite integral
Z 1
x(x2 + 3)4 dx,
0
we can use the substitution u = h(x) = x2 + 3, so that du = 2x dx. Note that h(0) = 3 and h(1) = 4,
and that h0 (x) = 2x > 0 whenever 0 < x < 1. It follows that
Z
0
1
1
x(x + 3) dx =
2
2
4
Z
4
4
1 u5
1 1024 243
781
u dx =
=
−
=
.
2 5 3
2
5
5
10
4
3
We can compare this to first observing Example 10.1.4, so that
1
Z
0
1 2
x(x + 3) dx =
(x + 3)5
10
2
4
1
=
0
781
1024 243
−
=
.
10
10
10
Example 10.1.13. To calculate the definite integral
Z 4
1
dx,
x
log
x
2
we can use the substitution u = h(x) = log x, so that du = h0 (x) dx, where h0 (x) = 1/x > 0 whenever
2 < x < 4. Note also that h(2) = log 2 and h(4) = log 4. It follows that
Z
2
4
1
dx =
x log x
Z
log 4
log 2
log 4
log 4
1
du = log |u|
= log log 4 − log log 2 = log
= log 2.
u
log 2
log 2
10.2. Integration by Parts
Recall the Product rule for differentiation, that
(uv)0 = uv 0 + vu0 .
Integrating with respect to x, we obtain
Z
Z
Z
(uv)0 dx = uv 0 dx + vu0 dx.
Now the indefinite integral on the left hand side is of the form uv. Rewriting this equation, we have
Z
Z
0
uv dx = uv − vu0 dx.
(1)
Equation (1) is called
R the formula for integration by parts for indefinite integrals. RIt is very useful if the
indefinite integral vu0 dx is much easier to calculate than the indefinite integral uv 0 dx.
Example 10.2.1. Consider the indefinite integral
Z
xex dx.
Writing u = x and v 0 = ex , we have
Z
0
uv dx =
Chapter 10 : Techniques of Integration
Z
xex dx.
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Furthermore, v = ex and u0 = 1. It follows that
Z
Z
0
x
uv − vu dx = xe − ex dx = xex − ex + C.
Hence
Z
xex dx = xex − ex + C.
Example 10.2.2. Consider the indefinite integral
Z
log x dx.
Writing u = log x and v 0 = 1, we have
Z
uv 0 dx =
Z
log x dx.
Furthermore,
v=x
and
u0 =
1
.
x
It follows that
Z
uv −
Z
0
vu dx = x log x −
1
x dx = x log x − x + C.
x
Hence
Z
log x dx = x log x − x + C.
Example 10.2.3. Consider the indefinite integral
Z
ex sin x dx.
Writing u = ex and v 0 = sin x, we have
Z
uv 0 dx =
Z
ex sin x dx.
Furthermore, v = − cos x and u0 = ex . It follows that
Z
Z
uv − vu0 dx = −ex cos x + ex cos x dx.
Hence
Z
ex sin x dx = −ex cos x +
Z
ex cos x dx.
(2)
We now need to study the indefinite integral
Z
Chapter 10 : Techniques of Integration
ex cos x dx.
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Writing u = ex and v 0 = cos x, we have
Z
Z
0
uv dx =
W W L Chen, 1994, 2008
ex cos x dx.
Furthermore, v = sin x and u0 = ex . It follows that
Z
Z
0
x
uv − vu dx = e sin x − ex sin x dx.
Hence
Z
ex cos x dx = ex sin x −
Z
ex sin x dx.
(3)
It looks like we are back to the same old problem. However, if we combine (2) and (3), then we obtain
Z
Z
ex sin x dx = −ex cos x + ex sin x − ex sin x dx,
so that
Z
2
ex sin x dx = ex sin x − ex cos x = ex (sin x − cos x).
Adding an arbitrary constant, which we may in view of Proposition 9C, we have
Z
1
ex sin x dx = ex (sin x − cos x) + C.
2
Example 10.2.4. Consider the indefinite integral
Z
x3 cos x dx.
Writing u = x3 and v 0 = cos x, we have
Z
Z
0
uv dx =
x3 cos x dx.
Furthermore, v = sin x and u0 = 3x2 . It follows that
Z
Z
0
3
uv − vu dx = x sin x − 3 x2 sin x dx.
Hence
Z
x3 cos x dx = x3 sin x − 3
Z
x2 sin x dx.
(4)
We now need to study the indefinite integral
Z
x2 sin x dx.
Writing u = x2 and v 0 = sin x, we have
Z
0
uv dx =
Chapter 10 : Techniques of Integration
Z
x2 sin x dx.
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Furthermore, v = − cos x and u0 = 2x. It follows that
Z
Z
0
2
uv − vu dx = −x cos x + 2 x cos x dx.
Hence
Z
2
Z
2
x sin x dx = −x cos x + 2
x cos x dx.
Combining (4) and (5), we have
Z
Z
3
3
2
x cos x dx = x sin x + 3x cos x − 6 x cos x dx.
(5)
(6)
We now need to study the indefinite integral
Z
x cos x dx.
Writing u = x and v 0 = cos x, we have
Z
Z
0
uv dx =
x cos x dx.
Furthermore, v = sin x and u0 = 1. It follows that
Z
Z
uv − vu0 dx = x sin x − sin x dx.
Hence
Z
Z
x cos x dx = x sin x −
sin x dx.
(7)
Combining (6) and (7), we have
Z
Z
x3 cos x dx = x3 sin x + 3x2 cos x − 6x sin x + 6 sin x dx
= x3 sin x + 3x2 cos x − 6x sin x − 6 cos x + C.
The technique is also valid for definite integrals, in view of the first Fundamental theorem of integral
calculus. For definite integrals over the interval [A, B], we have
Z
B
x=B Z
uv dx = uv
−
B
0
A
x=A
vu0 dx.
(8)
A
Equation (8) is called the formula for integration by parts for definite integrals.
Example 10.2.5. Consider the definite integral
Z
π/2
x3 cos x dx.
0
Writing u = x3 and v 0 = cos x, we have
Z
π/2
0
Z
uv dx =
0
Chapter 10 : Techniques of Integration
π/2
x3 cos x dx.
0
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Furthermore, v = sin x and u0 = 3x2 . It follows that
x=π/2 Z
uv
−
π/2
π/2
Z
vu0 dx = x3 sin x
−3
0
x=0
π/2
x2 sin x dx.
0
0
Hence
Z
π/2
π/2
Z
x3 cos x dx = x3 sin x
−3
0
π/2
π3
−3
8
x2 sin x dx =
0
0
Z
π/2
x2 sin x dx.
(9)
0
We now need to study the definite integral
π/2
Z
x2 sin x dx.
0
Writing u = x2 and v 0 = sin x, we have
π/2
Z
π/2
Z
0
x2 sin x dx.
uv dx =
0
0
Furthermore, v = − cos x and u0 = 2x. It follows that
x=π/2
Z
π/2
−
uv
π/2
Z
vu0 dx = −x2 cos x
+2
0
x=0
π/2
x cos x dx.
0
0
Hence
Z
π/2
π/2
Z
x2 sin x dx = −x2 cos x
+2
0
π/2
Z
x cos x dx.
0
0
π/2
x cos x dx = 2
(10)
0
Combining (9) and (10), we have
Z
π/2
π3
x cos x dx =
−6
8
3
0
Z
π/2
x cos x dx.
(11)
0
We now need to study the definite integral
Z
π/2
x cos x dx.
0
Writing u = x and v 0 = cos x, we have
π/2
Z
uv 0 dx =
π/2
Z
x cos x dx.
0
0
Furthermore, v = sin x and u0 = 1. It follows that
x=π/2
Z
−
uv
x=0
π/2
π/2 Z
vu0 dx = x sin x
−
0
0
π/2
sin x dx.
0
Hence
Z
π/2
π/2 Z
x cos x dx = x sin x
−
0
Chapter 10 : Techniques of Integration
0
0
π/2
π
sin x dx = −
2
Z
π/2
sin x dx.
(12)
0
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Combining (11) and (12), we have
Z
π/2
x3 cos x dx =
0
π3
− 3π + 6
8
Z
π/2
sin x dx =
0
π/2
π3
π3
− 3π + 6 − cos x
=
− 3π + 6.
8
8
0
10.3. Trigonometric Integrals
In this section, we consider integrals involving the six trigonometric functions sin x, cos x, tan x, cot x,
sec x and csc x. If we consider differentiation formulas involving these functions, then we can divide
these into three groups: (a) sin x and cos x; (b) tan x and sec x; and (c) cot x and csc x. Note that the
derivative of any of these functions can be expressed in terms of the two functions in the group to which
it belongs. This division is also substantiated by integral formulas.
It follows that given any indefinite integral
Z
f (x) dx,
where the integrand f (x) involves trigonometric functions, it may be beneficial to try first to express
f (x) in terms of trigonometric functions from only one of these three groups.
Example 10.3.1. Consider the indefinite integral
Z
Z tan x
sec3 x cot x
tan x + sec3 x cot x
dx
=
+
dx
cos2 x
cos2 x
cos2 x
Z Z
Z
sin x
sec5 x
sin x
sec5 x
=
+
dx
=
dx
+
dx.
cos3 x
tan x
cos3 x
tan x
Note that we can also write
Z
sin x
dx =
cos3 x
Z
tan x sec2 x dx =
1
tan2 x + C.
2
However, the indefinite integral
Z
sec5 x
dx
tan x
does not appear to be so simple.
Let us consider first integrals involving sin x and cos x. Consider an integral of the form
Z
sinm x cosn x dx.
When m = 1, the integral is simple to evaluate. Clearly
Z
1
cosn+1 x + C
sin x cosn x dx = −
n+1
if n 6= −1,
and
Z
Chapter 10 : Techniques of Integration
sin x cos−1 x dx = − log | cos x| + C.
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When n = 1, the integral is also simple to evaluate. Clearly
Z
1
sinm x cos x dx =
sinm+1 x + C
m+1
W W L Chen, 1994, 2008
if m 6= −1,
and
Z
sin−1 x cos x dx = log | sin x| + C.
In the general case, we may use standard trigonometric formulas like
sin2 x + cos2 x = 1,
(13)
sin 2x = 2 sin x cos x,
2
(14)
2
cos 2x = cos x − sin x.
(15)
Note also that combining (13) and (15), we have
cos 2x = 2 cos2 x − 1 = 1 − 2 sin2 x.
(16)
Example 10.3.2. Consider the indefinite integral
Z
sin5 x dx.
Using (13), we can write
sin5 x = sin4 x sin x = (1 − cos2 x)2 sin x = (1 − 2 cos2 x + cos4 x) sin x,
so that
Z
sin5 x dx =
Z
(1 − 2 cos2 x + cos4 x) sin x dx
Z
Z
Z
2
= sin x dx − 2 cos x sin x dx + cos4 x sin x dx
= − cos x +
2
1
cos3 x − cos5 x + C.
3
5
Example 10.3.3. Consider the indefinite integral
Z
sin3 x cos3 x dx.
Using (13), we can write
sin3 x cos3 x = cos2 x sin3 x cos x = (1 − sin2 x) sin3 x cos x = sin3 x cos x − sin5 x cos x,
so that
Z
sin3 x cos3 x dx =
(sin3 x cos x − sin5 x cos x) dx
Z
Z
3
= sin x cos x dx − sin5 x cos x dx
=
Chapter 10 : Techniques of Integration
Z
1
1
sin4 x − sin6 x + C.
4
6
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Example 10.3.4. Consider the indefinite integral
Z
sin4 4x dx.
Using (16), we can write
1
1
(1 − cos 8x)2 = (1 − 2 cos 8x + cos2 8x)
4
4
1
1
3 1
1
=
1 − 2 cos 8x + (1 + cos 16x) = − cos 8x + cos 16x,
4
2
8 2
8
sin4 4x =
so that
Z
3 1
1
− cos 8x + cos 16x dx
8 2
8
Z
Z
Z
3
1
1
=
dx −
cos 8x dx +
cos 16x dx
8
2
8
1
1
3
sin 8x +
sin 16x + C.
= x−
8
16
128
sin4 4x dx =
Z Example 10.3.5. Consider the indefinite integral
Z
sin2 x cos4 x dx.
Using (14) and (16), we can write
1
1
1
(1 + cos 2x) sin2 2x = sin2 2x + cos 2x sin2 2x
8
8
8
1
1
1
1
1
2
=
(1 − cos 4x) + cos 2x sin 2x =
−
cos 4x + cos 2x sin2 2x,
16
8
16 16
8
sin2 x cos4 x = cos2 x(sin x cos x)2 =
so that
Z
1
1
1
2
sin x cos x dx =
−
cos 4x + cos 2x sin 2x dx
16 16
8
Z
Z
Z
1
1
1
=
dx −
cos 4x dx +
cos 2x sin2 2x dx
16
16
8
1
1
1
=
x−
sin 4x +
sin3 2x + C.
16
64
48
2
4
Z Let us consider next integrals involving tan x and sec x. Consider an integral of the form
Z
tanm x secn x dx.
When m = 1, the integral is simple to evaluate. Clearly
Z
1
tan x secn x dx = secn x + C
n
if n 6= 0,
and
Z
tan x dx = − log | cos x| + C.
Chapter 10 : Techniques of Integration
page 13 of 26
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When n = 2, the integral is also simple to evaluate. Clearly
Z
tanm x sec2 x dx =
1
tanm+1 x + C
m+1
if m 6= −1,
and
Z
tan−1 x sec2 x dx = log | tan x| + C.
In the general case, we may use standard trigonometric formulas like
1 + tan2 x = sec2 x.
(17)
Example 10.3.6. Consider the indefinite integral
Z
tan3 x dx.
Using (17), we can write
tan3 x = tan2 x tan x = (sec2 x − 1) tan x = sec2 x tan x − tan x,
so that
Z
tan3 x dx =
Z
(sec2 x tan x − tan x) dx
Z
Z
= sec2 x tan x dx − tan x dx
=
1
tan2 x + log | cos x| + C.
2
Example 10.3.7. Consider the indefinite integral
Z
tan4 x dx.
Using (17), we can write
tan4 x = tan2 x tan2 x = (sec2 x − 1) tan2 x = sec2 x tan2 x − tan2 x = sec2 x tan2 x − sec2 x + 1,
so that
Z
4
Z
(sec2 x tan2 x − sec2 x + 1) dx
Z
Z
Z
= sec2 x tan2 x dx − sec2 x dx + dx
tan x dx =
=
Chapter 10 : Techniques of Integration
1
tan3 x − tan x + x + C.
3
page 14 of 26
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W W L Chen, 1994, 2008
Example 10.3.8. Consider the indefinite integral
Z
sec3 x dx.
Writing u = sec x and v 0 = sec2 x, we have
Z
uv 0 dx =
Z
sec3 x dx.
Furthermore, v = tan x and u0 = tan x sec x. It follows that
Z
Z
uv − vu0 dx = sec x tan x − tan2 x sec x dx.
Hence
Z
Z
3
sec x dx = sec x tan x −
We now need to study the indefinite integral
Z
tan2 x sec x dx.
(18)
tan2 x sec x dx.
Using (17), we can write
tan2 x sec x = (sec2 x − 1) sec x = sec3 x − sec x,
so that
Z
2
tan x sec x dx =
Z
3
Z
(sec x − sec x) dx =
3
sec x dx −
Z
sec x dx.
(19)
Combining (18) and (19), we have
Z
Z
Z
sec3 x dx = sec x tan x − sec3 x dx + sec x dx,
so that
Z
1
1
sec x dx = sec x tan x +
2
2
3
Z
sec x dx =
1
1
sec x tan x + log | sec x + tan x| + C.
2
2
Example 10.3.9. Consider the indefinite integral
Z
tan2 x sec3 x dx.
Writing u = tan2 sec x and v 0 = sec2 x, we have
Z
Z
0
uv dx = tan2 x sec3 x dx.
Furthermore, v = tan x and u0 = 2 tan x sec3 x + tan3 x sec x. It follows that
Z
Z
0
3
uv − vu dx = tan x sec x − (2 tan2 x sec3 x + tan4 x sec x) dx.
Chapter 10 : Techniques of Integration
page 15 of 26
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W W L Chen, 1994, 2008
Hence
Z
2
3
Z
3
(2 tan2 x sec3 x + tan4 x sec x) dx
Z
Z
= tan3 x sec x − 2 tan2 x sec3 x dx − tan4 x sec x dx.
tan x sec x dx = tan x sec x −
(20)
We now need to study the indefinite integral
Z
tan4 x sec x dx.
Using (17), we can write (the reader should check this)
tan4 x sec x = tan2 x sec3 x − sec3 x + sec x,
so that
Z
tan4 x sec x dx =
Z
tan2 x sec3 x dx −
Z
sec3 x dx +
Z
sec x dx.
(21)
Combining (20) and (21), we have
Z
1
tan3 x sec x +
4
1
= tan3 x sec x +
4
tan2 x sec3 x dx =
Z
Z
1
1
sec3 x dx −
sec x dx
4
4
1
1
tan x sec x − log | tan x + sec x| + C.
8
8
Occasionally, it may be necessary to convert an expression involving tan x and sec x to one involving
sin x and cos x instead.
Example 10.3.10. Consider the indefinite integral
Z
tan4 7x
dx.
sec5 7x
Here the identity (17) does not help very much. However, we have
tan4 7x
= sin4 7x cos 7x,
sec5 7x
so that
Z
tan4 7x
dx =
sec5 7x
Z
sin4 7x cos 7x dx =
1
sin5 7x + C.
35
Let us consider finally integrals involving cot x and csc x. Consider an integral of the form
Z
cotm x cscn x dx.
When m = 1, the integral is simple to evaluate. Clearly
Z
cot x cscn x dx = −
Chapter 10 : Techniques of Integration
1
cscn x + C
n
if n 6= 0,
page 16 of 26
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W W L Chen, 1994, 2008
and
Z
cot x dx = log | sin x| + C.
When n = 2, the integral is also simple to evaluate. Clearly
Z
cotm x csc2 x dx = −
1
cotm+1 x + C
m+1
if m 6= −1,
and
Z
cot−1 x csc2 x dx = − log | cot x| + C.
The details are similar to the case of tan x and sec x.
10.4. Trigonometric Substitutions
In
techniques to handle integrals involving square roots of the form
√ this section,
√ we shall consider
√
a2 − b2 x2 , a2 + b2 x2 or b2 x2 − a2 . Without loss of generality, assume that a, b > 0.
Let us consider first the case
√
a2 − b2 x2 . If we use the substitution
a
sin θ,
b
x=
then
p
a2 − b2 x2 =
q
√
a2 (1 − sin2 θ) = a2 cos2 θ = a| cos θ|,
while
a
cos θ dθ.
b
dx =
Example 10.4.1. Consider the indefinite integral
Z
√
1
dx.
4 − x2
If we use the substitution x = 2 sin θ, then
p
4 − x2 = 2| cos θ|
and
dx = 2 cos θ dθ,
so that
Z
√
1
dx =
4 − x2
Z
1
2 cos θ dθ.
2| cos θ|
Suppose that cos θ > 0. Then
Z
1
√
dx =
4 − x2
Chapter 10 : Techniques of Integration
Z
dθ = θ + C = sin−1
x
2
+ C.
page 17 of 26
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W W L Chen, 1994, 2008
Example 10.4.2. Consider the indefinite integral
Z
p
x3 9 − 4x2 dx.
If we use the substitution x =
p
3
2
sin θ, then
9 − 4x2 = 3| cos θ|
and
dx =
3
cos θ dθ.
2
Suppose that cos θ > 0. Then
Z
Z
p
243
3
2
sin3 θ cos2 θ dθ
x 9 − 4x dx =
16
Z
243
=
(1 − cos2 θ) sin θ cos2 θ dθ
16
Z
Z
243
243
2
=
sin θ cos θ dθ −
sin θ cos4 θ dθ
16
16
243
81
cos5 θ + C.
= − cos3 θ +
16
80
Next, note that cos2 θ = 1 − sin2 θ = 1 − 94 x2 , so that
Z
x
3
p
9−
4x2
Let us consider next the case
81
dx = −
16
√
4
1 − x2
9
3/2
243
+
80
5/2
4 2
+ C.
1− x
9
a2 + b2 x2 . If we use the substitution
x=
a
tan θ,
b
then
q
p
√
a2 + b2 x2 = a2 (1 + tan2 θ) = a2 sec2 θ = a| sec θ|,
while
dx =
a
sec2 θ dθ.
b
Example 10.4.3. Consider the indefinite integral
Z
p
x2 1 + x2 dx.
If we use the substitution x = tan θ, then
p
1 + x2 = | sec θ|
and
dx = sec2 θ dθ.
Suppose that sec θ > 0. Then
Z
x2
Z
p
1 + x2 dx = tan2 θ sec3 θ dθ.
We have shown earlier that
Z
1
1
1
tan2 θ sec3 θ dθ = tan3 θ sec θ + tan θ sec θ − log | tan θ + sec θ| + C.
4
8
8
Chapter 10 : Techniques of Integration
page 18 of 26
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First Year Calculus
W W L Chen, 1994, 2008
Next, note that sec2 θ = 1 + tan2 θ = 1 + x2 , so that
Z
p
1
1
1
x2 1 + x2 dx = x3 (1 + x2 )1/2 + x(1 + x2 )1/2 − log |x + (1 + x2 )1/2 | + C.
4
8
8
Let us consider finally the case
√
b2 x2 − a2 . If we use the substitution
x=
a
sec θ,
b
then
p
p
p
b2 x2 − a2 = a2 (sec2 θ − 1) = a2 tan2 θ = a| tan θ|,
while
dx =
a
tan θ sec θ dθ.
b
Example 10.4.4. Consider the indefinite integral
Z √
x2 − 4
dx.
x
If we use the substitution x = 2 sec θ, then
p
x2 − 4 = 2| tan θ|
and
dx = 2 tan θ sec θ dθ.
Suppose that tan θ > 0. Then
Z √
x2 − 4
dx = 2
x
Z
Z
2
tan θ dθ = 2
Z
2
(sec θ − 1) dθ = 2
Z
2
sec θ dθ − 2
dθ = 2 tan θ − 2θ + C.
Next, note that
tan2 θ = sec2 θ − 1 =
1 2
x −1
4
θ = sec−1
and
x
2
,
so that
Z √
x2 − 4
dx = 2
x
1/2
x
x
p
1 2
+ C = x2 − 4 − 2 sec−1
+ C.
x −1
− 2 sec−1
4
2
2
10.5. Completing Squares
In
p this section, we shall consider techniques to handle integrals involving square roots of the form
αx2 + βx + γ, where α 6= 0. Our task is to show that such integrals can be reduced to integrals
discussed in the previous section.
Note that
β
γ
αx + βx + γ = α x + x +
α
α
2
2
Chapter 10 : Techniques of Integration
β
β2
=α x + x+ 2
α
4α
2
b2
+ γ−
4α
β
=α x+
2α
2
β2
+ γ−
4α
.
page 19 of 26
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First Year Calculus
W W L Chen, 1994, 2008
Suppose first of all that we use a substitution
y =x+
β
.
2α
Then dy = dx and
αx2 + βx + γ = αy 2 + δ,
where
δ=γ−
It now follows that
p
β2
.
4α
αx2 + βx + γ is of the form
p
 pa2 − b2 y 2
a2 + b2 y 2
p 2 2
b y − a2
if α < 0 and δ > 0,
if α > 0 and δ > 0,
if α > 0 and δ < 0.
Example 10.5.1. Consider the indefinite integral
Z
1
√
dx.
3 − 2x − x2
We have
3 − 2x − x2 = −(x2 + 2x − 3) = −(x2 + 2x + 1) + 4 = −(x + 1)2 + 4 = −y 2 + 4,
where we use the substitution y = x + 1. Note that α = −1 < 0 and δ = 4 > 0. Then
Z
Z
1
1
√
p
dx =
dx.
2
3 − 2x − x
4 − y2
We have shown earlier that
Z
1
p
4 − y2
dy = sin−1
y
2
+ C.
It follows that
Z
√
1
dx = sin−1
3 − 2x − x2
x+1
2
+ C.
Example 10.5.2. Consider the indefinite integral
Z √ 2
x − 4x
dx.
x−2
We have
x2 − 4x = (x2 − 4x + 4) − 4 = (x − 2)2 − 4 = y 2 − 4,
where we use the substitution y = x − 2. Note that α = 1 > 0 and δ = −4 < 0. Then
Z √ 2
Z p 2
x − 4x
y −4
dx =
dy.
x−2
y
Chapter 10 : Techniques of Integration
page 20 of 26
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First Year Calculus
W W L Chen, 1994, 2008
We have shown earlier that
Z p
y
p
y2 − 4
+ C.
dy = y 2 − 4 − 2 sec−1
y
2
It follows that
Z √ 2
p
p
x − 4x
x−2
x−2
dx = (x − 2)2 − 4 − 2 sec−1
+ C = x2 − 4x − 2 sec−1
+ C.
x−2
2
2
10.6. Partial Fractions
In this section, we shall consider techniques to handle integrals of the form
Z
p(x)
dx,
q(x)
where p(x) and q(x) are polynomials in x.
If the degree of p(x) is not smaller than the degree of q(x), then we can always find polynomials a(x)
and r(x) such that
p(x)
r(x)
= a(x) +
,
q(x)
q(x)
where r(x) = 0 or has degree smaller than the degree of q(x).
Example 10.6.1. Consider the indefinite integral
Z
x5 + 2x4 + 4x3 + x + 1
dx.
x2 + x + 1
Note that
x5 + 2x4 + 4x3 + x + 1
2x + 4
= (x3 + x2 + 2x − 3) + 2
,
2
x +x+1
x +x+1
so that
Z
x5 + 2x4 + 4x3 + x + 1
dx =
x2 + x + 1
Z
3
2
(x + x + 2x − 3) dx +
Z
2x + 4
dx.
+x+1
x2
It does not take a genius to work out the indefinite integral
Z
(x3 + x2 + 2x − 3) dx.
We can therefore restrict our attention to the case when the polynomial p(x) is of lower degree than
the polynomial q(x).
The first step is to factorize the polynomial q(x) into a product of irreducible factors. It is a fundamental result in algebra that a real polynomial q(x) can be factorized into a product of irreducible linear
factors and quadratic factors with real coefficients.
Chapter 10 : Techniques of Integration
page 21 of 26
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First Year Calculus
W W L Chen, 1994, 2008
Example 10.6.2. Suppose that q(x) = x4 − 4x3 + 5x2 − 4x + 4. Then q(x) can be factorized into a
product of irreducible linear factors in the form (x − 2)2 (x2 + 1).
Suppose that a linear factor (ax + b) occurs n times in the factorization of q(x). Then we write down
a decomposition
An
A2
A1
+ ... +
,
+
ax + b (ax + b)2
(ax + b)n
where the constants A1 , . . . , An will be determined later. Suppose that a quadratic factor (ax2 + bx + c)
occurs n times in the factorization of q(x). Then we write down a decomposition
A1 x + B1
An x + Bn
A2 x + B 2
+ ... +
,
+
ax2 + bx + c (ax2 + bx + c)2
(ax2 + bx + c)n
where the constants A1 , . . . , An and B1 , . . . , Bn will be determined later. We proceed to add all the
decompositions and equate their sum to
p(x)
,
q(x)
and then calculate all the constants by equating coefficients.
Example 10.6.3. Suppose that
p(x)
2x3 − 11x2 + 17x − 16
2x3 − 11x2 + 17x − 16
= 4
=
.
q(x)
x − 4x3 + 5x2 − 4x + 4
(x − 2)2 (x2 + 1)
We now write
c1
c2
c3 x + c4
2x3 − 11x2 + 17x − 16
=
+
+ 2
.
(x − 2)2 (x2 + 1)
x − 2 (x − 2)2
x +1
Now
c2
c3 x + c4
c1 (x − 2)(x2 + 1) + c2 (x2 + 1) + (c3 x + c4 )(x − 2)2
c1
+
+
=
,
x − 2 (x − 2)2
x2 + 1
(x − 2)2 (x2 + 1)
so that
c1 (x − 2)(x2 + 1) + c2 (x2 + 1) + (c3 x + c4 )(x − 2)2 = 2x3 − 11x2 + 17x − 16.
Note now that
c1 (x − 2)(x2 + 1) + c2 (x2 + 1) + (c3 x + c4 )(x − 2)2
= c1 (x3 − 2x2 + x − 2) + c2 (x2 + 1) + c3 (x3 − 4x2 + 4x) + c4 (x2 − 4x + 4)
= (c1 + c3 )x3 + (−2c1 + c2 − 4c3 + c4 )x2 + (c1 + 4c3 − 4c4 )x + (−2c1 + c2 + 4c4 ).
Equating coefficients, we have
c1
+ c3
=
2,
−2c1 + c2 − 4c3 + c4 = −11,
c1
−2c1 + c2
Chapter 10 : Techniques of Integration
+ 4c3 − 4c4 =
17,
+ 4c4 = −16.
page 22 of 26
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W W L Chen, 1994, 2008
This system has solution c1 = 1, c2 = −2, c3 = 1 and c4 = −3. Hence
2x3 − 11x2 + 17x − 16
x−3
1
2
+ 2
=
−
,
x4 − 4x3 + 5x2 − 4x + 4
x − 2 (x − 2)2
x +1
so that
Z
2x3 − 11x2 + 17x − 16
dx =
x4 − 4x3 + 5x2 − 4x + 4
Z
1
dx −
x−2
Z
2
dx +
(x − 2)2
Z
x−3
dx.
x2 + 1
We shall calculate the three indefinite integrals on the right hand side later.
To calculate the indefinite integrals that arise, note that these indefinite integrals are of the form
Z
A
dx,
(ax + b)k
(22)
or
Z
Ax + B
dx,
+ bx + c)k
(23)
(ax2
where A and B are constants and k is a positive integer. The integral (22) is simple. If k 6= 1, then we
have
Z
A
A
dx = −
+ C.
k
(ax + b)
(k − 1)a(ax + b)k−1
On the other hand, we have
Z
A
A
dx = log |ax + b| + C.
ax + b
a
The integral (23) is a bit more complicated. Note that
Z
Ax + B
A
dx =
(ax2 + bx + c)k
2a
Z
Z
Ab
2ax + b
1
dx
+
B
−
dx.
(ax2 + bx + c)k
2a
(ax2 + bx + c)k
The first integral on the right hand side is simple. If k 6= 1, then we have
Z
2ax + b
1
dx = −
+ C.
(ax2 + bx + c)k
(k − 1)(ax2 + bx + c)k−1
On the other hand, we have
Z
2ax + b
dx = log |ax2 + bx + c| + C.
ax2 + bx + c
It remains to study the integral
Z
1
dx.
(ax2 + bx + c)k
To do this, we may try the technique of completing squares as described in the previous section, and
then use a trigonometric substitution to find the integral.
Chapter 10 : Techniques of Integration
page 23 of 26
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First Year Calculus
W W L Chen, 1994, 2008
Example 10.6.4. Let us continue the discussion of Example 10.6.3. Note that
Z
Z
1
dx = log |x − 2| + C
x−2
and
1
1
dx = −
+ C.
2
(x − 2)
x−2
On the other hand, we have
Z
x−3
1
dx =
x2 + 1
2
Z
2x
dx − 3
x2 + 1
Z
1
dx.
x2 + 1
Clearly
Z
2x
dx = log |x2 + 1| + C.
+1
x2
Using the substitution x = tan θ, we have
Z
1
dx =
x2 + 1
Z
sec2 θ
dθ =
1 + tan2 θ
Z
dθ = θ + C = tan−1 x + C.
It follows that
Z
Z
Z
Z
1
2
x−3
2x3 − 11x2 + 17x − 16
dx =
dx −
dx
dx +
4
3
2
2
x − 4x + 5x − 4x + 4
x−2
(x − 2)
x2 + 1
2
1
= log |x − 2| +
+ log |x2 + 1| − 3 tan−1 x + C.
x−2 2
Example 10.6.5. Consider the indefinite integral
Z
x2 + x − 3
dx.
x3 − 2x2 − x + 2
Note first of all that
x3 − 2x2 − x + 2 = (x − 2)(x + 1)(x − 1),
so we consider partial fractions of the form
x2 + x − 3
c1
c2
c3
=
+
+
(x − 2)(x + 1)(x − 1)
x−2 x+1 x−1
c1 (x + 1)(x − 1) + c2 (x − 2)(x − 1) + c3 (x − 2)(x + 1)
=
.
(x − 2)(x + 1)(x − 1)
It follows that
c1 (x + 1)(x − 1) + c2 (x − 2)(x − 1) + c3 (x − 2)(x + 1) = x2 + x − 3.
(24)
We may equate coefficients and solve for c1 , c2 , c3 . Alternatively, substituting x = 2, −1, 1 into equation
(24), we get respectively 3c1 = 3, 6c2 = −3 and −2c3 = −1, so that c1 = 1, c2 = −1/2 and c3 = 1/2.
Hence
Z
Z
Z
Z
x2 + x − 3
1
1
1
1
1
dx =
dx −
dx +
dx
3
2
x − 2x − x + 2
x−2
2
x+1
2
x−1
1
1
= log |x − 2| − log |x + 1| + log |x − 1| + C.
2
2
Chapter 10 : Techniques of Integration
page 24 of 26
c
First Year Calculus
W W L Chen, 1994, 2008
Example 10.6.6. Consider the indefinite integral
Z
x6 − 2
dx.
x4 + x2
Note that
x2 − 2
x6 − 2
= x2 − 1 + 4
,
4
2
x +x
x + x2
so that
Z
x6 − 2
dx =
x4 + x2
Z
(x2 − 1) dx +
Z
x2 − 2
1
dx = x3 − x +
x4 + x2
3
Z
x2 − 2
dx.
x4 + x2
(25)
Next, we study the integral
Z
x2 − 2
dx.
x4 + x2
Note first of all that
x4 + x2 = x2 (x2 + 1),
so we consider partial fractions of the form
x2 − 2
c1
c2
c3 x + c4
c1 x(x2 + 1) + c2 (x2 + 1) + (c3 x + c4 )x2
=
+ 2+ 2
=
.
+ 1)
x
x
x +1
x2 (x2 + 1)
x2 (x2
It follows that
c1 x(x2 + 1) + c2 (x2 + 1) + (c3 x + c4 )x2 = x2 − 2.
Equating coefficients, we have
c1
+ c3
c2
=
0,
+ c4 =
1,
=
0,
c1
= −2.
c2
This system has solution c1 = 0, c2 = −2, c3 = 0 and c4 = 3. Hence
Z
x2 − 2
dx = −2
x4 + x2
Z
1
dx + 3
x2
Z
x2
1
2
dx = + 3 tan−1 x + C.
+1
x
(26)
Combining (25) and (26), we obtain
Z
x6 − 2
1
2
dx = x3 − x + + 3 tan−1 x + C.
4
2
x +x
3
x
Chapter 10 : Techniques of Integration
page 25 of 26
c
First Year Calculus
W W L Chen, 1994, 2008
Problems for Chapter 10
1. Evaluate each of the following indefinite integrals:
Z
Z
a)
sin x cos 7x dx
b)
e2x cos 3x dx
Z
Z
cos 2x
1
√
d)
dx
dx
e)
1 − sin 2x
16 − 3x + x2
Z
Z
1
x2
g)
dx
h)
dx
2
3
2
x + 4x − 4
x + 3x + 3x + 1
Z
Z
x2 + 3x − 1
j)
dx
k)
log(x6 ) dx
x4 + x3 + x2 + x
Z
Z
1
dx
n)
e2x cos x dx
m)
x2 − 5x + 4
√
Z 4
Z
√
x +x x+1
p)
dx
q)
x2 x − 1 dx
x
Z
Z
2
s)
e4x+2 dx
t)
xex dx
Z
Z
(log x)5
2x + 3
v)
dx
w)
dx
x
x2 + 3x − 4
Z
Z √
1
( x + 1)10
√
√
y)
dx
z)
dx
2
2
x
a −x
Z
Z
1
bb)
x5 ex dx
cc)
dx
2
x − 4x + 3
Z
x−4
dx
ee)
(x2 + 4)(x + 1)
2. Evaluate each of the following definite integrals:
Z 3
Z 4 √
x+1
√
a)
dx
x(1 + 2x2 )4 dx
b)
x
2
1
Z π/4
Z 1 p
cos x
d)
dx
e)
x x2 + 1 dx
(1 + sin x)2
0
0
Z π/4
Z π/2
√
g)
2 − 2 cos x dx
cos2 2x dx
h)
0
Z
j)
0
0
1/2
√
Z
c)
Z
f)
i)
x2 log x dx
x sec2 x dx
Z √
Z
l)
Z
o)
sin2 3x dx
(x3 +
Z
r)
cot x csc4 x dx
x
p
√
x) dx
x2 + 4 dx
Z
log x
dx
x
Z
sin−1 x
√
dx
x)
1 − x2
Z
1
aa)
dx
x2 + a2
Z
dd)
xex dx
u)
Z
2
x2 + 1
dx
(x + 1)4
4
e x
√ dx
x
c)
1
Z
f)
1
Z
i)
√
π/4
x cos 2x dx
0
x
dx
1 − x2
Chapter 10 : Techniques of Integration
page 26 of 26