1. Techniques of Integration Section 8-IT
1.1. Basic integration formulas. Integration is more difficult than derivation. The derivative of every rational function or trigonometric function is another function of the same
type. This is not true for integration. Just recall the integral
Z y
dx
.
x
1
The integral of this rational function f (x) = x−1 is not a rational function. In fact, we had
to give that function a name, natural logarithm, because it was a new type of function for
us. At least in this sense, it is more difficult to find integrals than derivatives. With time
people have developed several techniques that help us find integrals. Certain techniques
work only with certain types of integrals. The purpose of this section is to summarize few
of the most used integration techniques. Before we start, let us recall that integration is a
linear operation, meaning that for all constants a, b, and integrable functions f , g, holds
Z
Z
Z
a f (x) + b g(x) dx = a f (x) dx + b g(x) dx.
This property is used all the time to simplify the work needed to find integrals.
One could say that the two most used integration techniques are integration by substitution and integration by parts. The first technique has been studied in the textbook, and the
second one will be studied in the textbook. We can now say that integration by substitution is the integral version of the chain rule for derivation. Similarly, integration by parts
is the integral version of the product rule for derivation. Let us recall the integration by
substitution formula: Given appropriate functions f and g, holds,
Z
Z
0
f (g(x)) g (x) dx = f (u) du.
Having mentioned the two most useful integration techniques, we can now concentrate
on other integration techniques which apply to somehow more specific situations.
(a)
(b)
(c)
(d)
(e)
Completing the square.
Separating fractions.
Reducing improper fractions.
Multiplication by one.
Eliminating a square root.
Later on we give examples of each of the techniques mentioned in the list above. But
before we start, we review few basic integration formulas that we will need later on. We
summarize them in two sets, the first one is just the integral version of the derivative formulas
for simple functions, and the definition of the natural logarithm, namely,
Z
Z
dx
xn+1
n
+ c, n 6= −1,
= ln |x| + c,
x dx =
n+1
x
Z
Z
ekx
akx
ekx dx =
+ c,
akx dx =
+ c,
k
k ln(a)
Z
Z
sin(x) dx = − cos(x) + c,
cos(x) dx = sin(x) + c,
Z
Z
sinh(x) dx = cosh(x) + c,
cosh(x) dx = sinh(x) + c
Z
Z
x
x
dx
1
dx
√
= arcsin
+ c,
= arctan
+ c.
2
2
a
a +x
a
a
a2 − x2
The second set of integration formulas is obtained using integration by substitution and
the first two integrals in the first set. We assume that a, b, are real constants with b 6= 0,
1
2
and n > 1 is an integer.
Z
dx
1
= ln |a + bx| + c,
(a + bx)
b
Z
1
dx
=−
+ c,
(a + bx)n
b(n − 1) (a + bx)n−1
Z
1
x dx
= ln(1 + x2 ) + c,
1 + x2
2
Z
1
x dx
=−
+ c,
2
n
(1 + x )
2(n − 1) (1 + x2 )n−1
Substitute: u = a + bx.
Substitute: u = a + bx.
Substitute: u = 1 + x2 .
Substitute: u = 1 + x2 .

sin(x)
 Recall:
tan(x) =
;
cos(x)

Substitute:
u = cos(x).

cos(x)
 Recall:
cot(x) =
;
sin(x)

Substitute:
u = sin(x).
Z
tan(x) dx = − ln | cos(x)| + c,
Z
cot(x) dx = ln | sin(x)| + c,
We now start giving examples of the integration techniques listed in (a)-(e). The first
integration technique we listed is completing the square, and the idea is to rewrite a second
degree polynomial into a total square plus a constant term. As it is usually the case, the
idea is better illustrated in an example.
Z
2 dx
.
Example 1.1 (Completing the square): Evaluate
2
x − 6x + 10
Solution: The idea is to rewrite the second degree polynomial in the denominator as,
6
6 2 6 2
x+
−
+ 10.
x2 − 6x + 10 = x2 − 2
2
2
2
2
We just multiplied the linear term in x by 1 = , we reordered the factors in that term,
2
6 2
and then we added and subtracted the term
. The first three terms on the right hand
2
side form a complete square,
6 2 6 2
x2 − 6x + 10 = x −
−
+ 10
2
2
⇒
x2 − 6x + 10 = (x − 3)2 + 1.
Therefore, our original integral looks like,
Z
Z
dx
2 dx
=
2
.
2
x − 6x + 10
(x − 3)2 + 1
At this point it should not be hard to see that the substitution u = x − 3, which implies
du = dx transforms the integral into an integral we know how to evaluate,
Z
Z
du
2 dx
=2
= 2 arctan(u) + c.
x2 − 6x + 10
1 + u2
Substituting back the variable x we obtain,
Z
2 dx
= 2 arctan(x − 3) + c.
x2 − 6x + 10
C
3
The next integral is an example of the technique called separating fractions. We use the
linearity property of the integral to reduce an integral into two or more simpler integrals.
Z
3x + 2
√
dx.
Example 1.2 (Separating fractions): Evaluate
1 − x2
Solution: We use the linearity property of integration to split the integral,
Z
Z
Z
3x + 2
x dx
dx
√
dx = 3 √
+2 √
(1)
1 − x2
1 − x2
1 − x2
In the first term of the right-hand side we substitute u = 1 − x2 , then du = −2x dx, which
implies x dx = −du/2. Hence,
Z
Z
Z
p
1 du
3
3 u1/2
x dx
=3 − √ =−
u−1/2 du = −
3 √
+ c1 = −3 1 − x2 + c1 .
2 u
2
2 1/2
1 − x2
The second term in Eq. (1) can be computed using common integrals (see above),
Z
dx
2 √
= 2 arcsin(x) + c2 .
1 − x2
Denoting c = c1 + c2 , we conclude that
Z
p
3x + 2
√
dx = −3 1 − x2 + 2 arcsin(x) + c.
2
1−x
C
The next technique of integration is called reducing improper fractions, and the idea is
to use the long division of polynomials to compute integrals of certain rational functions.
Z
4x2 − 7
Example 1.3 (Reducing improper fractions): Evaluate
dx.
2x + 3
Solution: The degree of the polynomial in the numerator is greater or equal the degree of
the polynomial in the denominator. In this case it is convenient to do the long division:
2x − 3
2x + 3
4x2
−7
− 4x2 − 6x
− 6x − 7
6x + 9
⇒
4x2 − 7
2
= 2x − 3 +
.
2x + 3
2x + 3
2
Using the result of the long division given above, the integral is simple to compute,
Z
Z
Z
Z
4x2 − 7
2 dx
2 dx
2
dx = (2x − 3) dx +
= x − 3x +
.
2x + 3
2x + 3
2x + 3
The integral on the right hand side above can be computed with the substitution u = 2x+3,
which implies du = 2 dx, hence
Z
Z
2 dx
du
=
= ln |u| + c = ln |2x + 3| + c.
2x + 3
u
We then conclude that
Z
4x2 − 7
dx = x2 − 3x + ln |2x + 3| + c.
2x + 3
C
Another integration technique is called multiplication by one. The idea is to multiply
numerator and denominator of the integrad by the same factor, choosing the factor so that
the integral can be converted into an integral we know how to evaluate.
4
Z
Example 1.4 (Multiplication by one): Evaluate
π
p
1 − cos(x) dx.
0
p
Solution: For this integral we multiply by 1 = p
1 + cos(x)
, that is, we multiply numerator
1 + cos(x)
and denominator by the conjugate of the function we are integrating. Therefore, we get
p
Z πp
Z πp
1 − cos(x)
1 + cos(x)
dx
1 − cos(x) dx =
·p
1
1 + cos(x)
0
0
Z πp
1 − cos2 (x)
p
=
dx
1 + cos(x)
0
q
Z π
sin2 (x)
p
=
dx
1 + cos(x)
0
Z π
| sin(x)|
p
=
dx.
1 + cos(x)
0
√
In the last equation above we used again the fact that u2 = |u|. It is important to realize
that we are integrating in the interval x ∈ [0, π], and in that interval | sin(x)| = sin(x).
Therefore,
Z πp
Z π
sin(x)
p
1 − cos(x) dx =
dx.
1 + cos(x)
0
0
Now it should not be hard to realize that the substitution u = cos(x) will be useful to
evaluate the integral, since du = − sin(x) dx. Notice that we are computing a definite
integral, so when we perform the substitution the integral limits must be modified accordingly.
We get,
Z πp
Z −1
Z 1
√
−du
(1 + u)1/2 1
√
1 − cos(x) dx =
=
(1 + u)−1/2 du =
= 2( 2 − 0),
1/2
−1
1+u
0
1
−1
Z πp
√
1 − cos(x) dx = 2 2.
C
so we conclude that
0
There is a useful integral, that has already been discussed in earlier
Z sections, that can
be done using the multiplication by one method too. The integral
sec(x) dx is a really
non-trivial example of this technique, and we compute it again here emphasizing how the
multiplication by 1 was used.
Z
Example 1.5 (Multiplication by one): Evaluate
sec(x) dx.
sec(x) + tan(x)
, that is,
sec(x) + tan(x)
Z
Z
Z
sec(x) + tan(x)
sec2 (x) + sec(x) tan(x)
sec(x) dx = sec(x)
dx =
dx.
sec(x) + tan(x)
sec(x) + tan(x)
Solution: For this integral, we multiply by 1 =
This multiplication is useful since tan0 (x) = sec2 (x), and sec0 (x) = sec(x) tan(x). Therefore,
we substitute
u = sec(x) + tan(x) ⇒ du = sec(x) tan(x) + sec2 (x) dx,
which implies
Z
Z
sec(x) dx =
du
= ln |u| + c,
u
Z
so we conclude that
sec(x) dx = ln | sec(x) + tan(x)| + c.
C
5
The integral of the csc function can be done in a similar way, so we summarize this and
the previous result:
Z
Z
sec(x) dx = ln | sec(x) + tan(x)| + c,
csc(x) dx = − ln | csc(x) + cot(x)| + c.
The last technique of integration we introduce here is called eliminating a square root.
In the following example we use a trigonometric identity to eliminate a square root.
Z π/4 p
1 + cos(4x) dx.
Example 1.6 (Eliminating a square root): Evaluate
0
Solution: The function under the square root can be converted into a square using the
trigonometric identity
1
cos2 (θ) = 1 + cos(2θ) .
2
Therefore, using this identity for θ = 2x we get
1 + cos(4x) = 2 cos2 (2x).
Therefore, we get
Z π/4 p
Z
1 + cos(4x) dx =
0
0
π/4
p
2 cos2 (2x) dx
√
√ Z
= 2
π/4
| cos(2x)| dx.
0
In the last equation above we used the fact that u2 = |u|. It is important to realize that we
are integrating in the interval x ∈ [0, π/4], equivalently, 2x ∈ [0, π/2], and in that interval
| cos(2x)| = cos(2x). Therefore,
√
Z π/4 p
π/4
√ Z π/4
√ 1
2
1 + cos(4x) dx = 2
=
(1 − 0),
cos(2x) dx = 2 sin(2x)
2
2
0
0
0
√
Z π/4 p
2
so we conclude that
1 + cos(4x) dx =
.
C
2
0