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College of Engineering
and Architecture
Owned and Operated by the City Government of Urdaneta
THE INDEFINITE INTEGRAL
We shall begin our study of the inverse of differentiation. We have learned in differential
calculus that differentiation is the process of finding the derivative or differential of a given
function. This time, we reverse the process, that is, we find the function whose derivative or
differential is given. The inverse of differentiation is called antidifferentiation or integration.
OBJECTIVES. At the end, the student should be able to:
1.
Know the concept of the indefinite integral.
2.
Apply the basic integration formulas.
The symbol ∫ is used to denote the process of integration. The integral of 4𝑥 3 is 𝑥 4 + 𝐶 can
be expressed symbolically as
∫ 4𝑥 3 𝑑𝑥 = 𝑥 4 + 𝐶
The expression ∫ 4𝑥 3 𝑑𝑥 is called an indefinite integral. Generally, the indefinite integral of a
function f(x) is denoted by ∫ 𝑓(𝑥)𝑑𝑥 and defined as follows
∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶 if 𝑑𝐹(𝑥) = 𝑓(𝑥)𝑑𝑥 … 𝐸𝑞𝑛 (1.1)
We call the symbol ∫ the integral sign, f(x) the integrand, C the arbitrary constant of
integration, and F(x) + C the value of the indefinite integral ∫ 𝑓(𝑥)𝑑𝑥. The differential
dx indicates that x is the variable of integration.
1.1 The Indefinite Integral
We know that by differentiation, a derivative or differential is obtained from a given
function. Suppose the derivative (or differential) of a function is given, how can a function be
found? For example, what function has for its derivative 4𝑥 3 or for its differential 4𝑥 3 𝑑𝑥? We
𝑑
know that (𝑥 4 ) = 4𝑥 3 or 𝑑(𝑥 4 ) = 4𝑥 3 𝑑𝑥. Thus, the answer to the question above is 𝑥 4 .
𝑑𝑥
However, that is not the only possible answer. We may also give the following functions as
valid answers:
𝑥4 − 3
𝑥 4 + 15
𝑥4 + 8
Table below illustrate further the integral concept:
DIFFERENTIAL
INTEGRAL
𝑑(𝑥 6 ) = 6𝑥 5 𝑑𝑥
∫ 6𝑥 5 𝑑𝑥 = 𝑥 6 + 𝐶
𝑑(𝑥) = 𝑑𝑥
∫ 𝑑𝑥 = 𝑥 + 𝐶
𝑑(𝑒 𝑥 ) = 𝑒 𝑥 𝑑𝑥
∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝐶
𝑑(ln 𝑥) =
Notice that from the possible answers, the functions differ only by a constant. If we let C be
any constant, then we may write
𝑑𝑥
𝑥
𝑑(sin 𝑥) = cos 𝑥 𝑑𝑥
∫
𝑑𝑥
= ln 𝑥 + 𝐶
𝑥
∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶
4
𝑥 +𝐶
as the general answer.
The function 𝑥 4 + 𝐶 is called the antiderivative or the integral of4𝑥 3 . Antidifferentiation or
integration is the process of finding the function whose derivative or differential is given.
The symbol of differentiation 𝑑, and the symbol of integration ∫ are inverse to each
other. What happens if 𝑑 precedes ∫ or when it follows ∫ ? Let us suppose 𝑓(𝑥) and
𝐹(𝑥) are functions whose relationship is given by
𝑑𝐹(𝑥) = 𝑓(𝑥)𝑑𝑥 … (1)
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College of Engineering
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Integrating both sides of (1) gives
∫ 𝑑𝐹(𝑥) = ∫ 𝑓(𝑥)𝑑𝑥 … (2)
We have discussed in the preceding section that the process of finding the integral
of a function 𝑓(𝑥) requires our ability to guess another function 𝐹(𝑥) whose
derivative is 𝑓(𝑥) or whose differential is 𝑓(𝑥)𝑑𝑥. For example, we guess that
From 𝐸𝑞𝑛 (1.1) we see that ∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶 . Substituting this in (2), we
have
∫ 3𝑥 2 𝑑𝑥 = 𝑥 3 + 𝐶 because 𝑑(𝑥 3 + 𝐶) = 3𝑥 2 𝑑𝑥.
∫ 𝑑𝐹(𝑥) = 𝐹(𝑥) + 𝐶 … 𝐸𝑞𝑛 (1.2)

Integrating the differential of a function gives that function plus an
arbitrary constant.
To reduce the amount of guesswork, basic integration formulas, are set up to
facilitate integration. The first six formulas are given below. Note that the letters 𝑎,
𝑛, and 𝐶 are constants and 𝑓(𝑥) and 𝑔(𝑥) are differentiable functions of 𝑥.
For example,
∫ 𝑑(𝑥 4 ) = 𝑥 4 + 𝐶
∫ 𝑑(tan 𝑥) = tan 𝑥 + 𝐶
∫ 𝑑(𝑦) = 𝑦 + 𝐶
∫ 𝑑(𝑒 𝑥 ) = 𝑒 𝑥 + 𝐶
1) ∫ 0𝑑𝑥 = 𝐶
2) ∫ 𝑑𝑥 = 𝑥 + 𝐶
3) ∫ 𝑎𝑑𝑥 = 𝑎𝑥 + 𝐶
𝑥 𝑛+1
4) ∫ 𝑥 𝑛 𝑑𝑥 = 𝑛+1 + 𝐶
Next, we discuss when 𝑑 precedes ∫ , By 𝐸𝑞𝑛 (1.1)
∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶 if 𝑑𝐹(𝑥) = 𝑓(𝑥)𝑑𝑥
Differentiating both sides of 𝐸𝑞𝑛 (1.1), we get
𝑑[∫ 𝑓(𝑥)𝑑𝑥] = 𝑑[𝐹(𝑥) + 𝐶] = 𝑑𝐹(𝑥) … (3)
5) ∫ 𝑎𝑓(𝑥)𝑑𝑥 = 𝑎 ∫ 𝑓(𝑥)𝑑𝑥
Substituting equation (1), 𝑑𝐹(𝑥) = 𝑓(𝑥)𝑑𝑥, in equation (3), we get
𝑑[∫ 𝑓(𝑥)𝑑𝑥] = 𝑓(𝑥)𝑑𝑥 … 𝐸𝑞𝑛 (1.3)

Note: When 𝒅 precedes∫ , they annul each other.
6) ∫ [𝑓(𝑥) ± 𝑔(𝑥)]𝑑𝑥 =
∫ 𝑓(𝑥)𝑑𝑥 + ∫ (𝑔(𝑥)𝑑𝑥
For example,
𝑑 ∫ 4𝑥3𝑑𝑥 = 4𝑥3𝑑𝑥
𝑑 ∫ 𝑠𝑖𝑛𝑥𝑑𝑥 = 𝑠𝑖𝑛𝑥𝑑𝑥
Proof: 𝑑(𝐶) = 0 ∙ 𝑑𝑥 = 0
Proof: 𝑑(𝑥 + 𝐶) = 𝑑𝑥 + 0 = 𝑑𝑥
Proof:
𝑑(𝑎𝑥 + 𝐶) = 𝑑(𝑎𝑑𝑥) + 𝑑𝐶 = 𝑎𝑑𝑥
for any rational number 𝑛 ≠ −1
Proof:
𝑥 𝑛+1
1
(𝑛 + 1)𝑥 𝑛+1−1 𝑑𝑥
𝑑 (
)=
𝑛+1
𝑛+1
= 𝑥 𝑛 𝑑𝑥
Proof:
𝑑(𝑎 ∫ 𝑓(𝑥)𝑑𝑥) = 𝑎𝑑 ∫ 𝑓(𝑥)𝑑𝑥 =
𝑎 𝑓(𝑥)𝑑𝑥
Proof:
𝑑[∫ 𝑓(𝑥)𝑑𝑥 ± ∫ 𝑔(𝑥)𝑑𝑥
= 𝑑∫ 𝑓(𝑥)𝑑𝑥
± 𝑑 ∫ 𝑔(𝑥)𝑑𝑥]
= 𝑓(𝑥)𝑑𝑥 ± 𝑔(𝑥)𝑑𝑥
EXAMPLE 1:
1.2 Basic Integration Formulas
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𝑋 4+1
a) ∫ 5𝑥 4 𝑑𝑥 = 5 ∫ 𝑥 4 𝑑𝑥 = 5 ( 4+1 ) + 𝐶 = 𝑋 5 + 𝐶
Formula (4)
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b) ∫ 4𝑑𝑥 = 4∫ 𝑑𝑥 = 4𝑥 + 𝐶
Let 𝑢 = 𝑥 + 1, then 𝑑𝑢 = 𝑑𝑥.
Formula (2)
c) ∫ 6(𝑥 + 1)𝑑𝑥 = 6 ∫ (𝑥 + 1)𝑑𝑥 = 6[∫ 𝑥𝑑𝑥 + ∫ 𝑑𝑥]
= 6[
𝑥2
+ 𝑥] + 𝐶 = 3𝑥 2 + 𝐶
2
Formula (6)
∫ (𝑥 + 1)4 𝑑𝑥 =
4+1
+ 𝐶 = (𝑥 + 1)2 + 𝐶 = 1⁄5 (𝑥 + 1)5 + 𝐶
Formulas (4) & (2)
d) ∫ (3𝑥 5 − 4𝑥)𝑑𝑥 = ∫ 3𝑥 5 𝑑𝑥 − ∫ 4𝑥𝑑𝑥
Formula (6)
EXAMPLE 2.
Evaluate∫ (2 − 3𝑥)5 𝑑𝑥.
= 3∫ 𝑥 5 𝑑𝑥 − 4 ∫ 𝑥𝑑𝑥 Formula (5)
𝑥6
6
(𝑥 + 1)4+1
𝑥2
2
= 3 ( ) − 4 ( ) + 𝐶 Formula (4)
= 1⁄2 𝑥 6 − 2𝑥 2 + 𝐶
Formula (3)-(6) enables us to evaluate the integral of any polynomial. For instance,
see example 2.
SOLUTION:
Let 𝑢 = 2 − 3𝑥, then 𝑑𝑢 = −3𝑑𝑥.
Note that the given integral does not have -3 preceding 𝑑𝑥. But then -3 can be
written preceding dx if we neutralize it by a factor that we write before the integral
sign. Thus, our new integral will appear in the following manner.
6
(2 − 3𝑥)
∫ (2 − 3𝑥)5 𝑑𝑥 = − 1⁄3 ∫ (2 − 3𝑥)5 (−3𝑑𝑥) = − 1⁄3
+𝐶 =
6
= − 1⁄18 (2 − 3𝑥)6 + 𝐶
EXAMPLE 2:
∫ 8𝑥 3 + 12𝑥 2 − 4𝑥 + 5)𝑑𝑥 = 8∫ 8𝑥 3 + 12∫ 𝑥 2 − 4∫ 𝑥 + 5∫ 𝑑𝑥
= 2𝑥 4 + 4𝑥 3 − 2𝑥 2 + 5𝑥 + 𝐶

1.3 The Power Formula
To insert the needed constant legitimately, we must compensate for it by
putting its reciprocal as a factor outside the integral.
EXAMPLE 3.
 7. ∫ 𝑢
Proof: 𝑑 [
𝑛
𝑢𝑛+1
𝑑𝑢 =
+ 𝐶,
𝑛+1
where 𝑛 ≠ −1
𝑢𝑛+1
1
(𝑛 + 1)𝑢𝑛+1 𝑑𝑢 = 𝑢𝑛 𝑑𝑢
+ 𝐶] =
𝑛+1
𝑛+1
Note: The differential 𝑑𝑢 in formula 7 is an exact or complete differential of the
base 𝑢. The exponent 𝑛 is a constant different from -1.
EXAMPLE 1.
Evaluate ∫ (𝑥 + 1) 𝑑𝑥.
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SOLUTION:
Let 𝑢 = 𝑥 2 − 2𝑥 + 3, then 𝑑𝑢 = 2𝑥𝑑𝑥 − 2𝑑𝑥 = 2(𝑥 − 1)𝑑𝑥
∫ (𝑥 2 − 2𝑥 + 3)4 (𝑥 − 1)𝑑𝑥 = − 1⁄2 ∫ (𝑥 2 − 2𝑥 + 3)4 (2)(𝑥 − 1)𝑑𝑥
= 1⁄10 (𝑥 2 − 2𝑥 + 3)5 + 𝐶
EXAMPLE 4.
4
SOLUTION:
Evaluate ∫ (𝑥 2 − 2𝑥 + 3)4 (𝑥 − 1)𝑑𝑥
1
1
Evaluate ∫ tan2 3 𝑥 𝑠𝑒𝑐𝑥 𝑑𝑥. sec 2 3 𝑥𝑑𝑥.
SOLUTION:
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1
1
1
Let 𝑢 = 𝑡𝑎𝑛 3 𝑥, then 𝑑𝑢 = (sec 2 3 𝑥) (3 𝑑𝑥) .
1
1
∫ tan 𝑥 sec 2 𝑥 𝑑𝑥
3
3
𝑑𝑢
 8. ∫ 𝑢−1 𝑑𝑢 = ∫ 𝑢 = 𝐼𝑛𝑢 + 𝐶, 𝑢 ≠ 0. u
2
Proof: 𝑑(𝑙𝑛𝑢 + 𝐶) =
31
tan 3 𝑥
1
1 1
= 3∫ (tan2 𝑥) ([sec 2 𝑥] 𝑑𝑥) = 3
+𝐶
3
3 3
3
1
= tan3 𝑥 + 𝐶
3
EXAMPLE 5.

𝑑𝑥
Evaluate ∫ (𝑥+1)
SOLUTION:
Let 𝑢 = ln 𝑥, then 𝑑𝑢 =
ln 𝑥 𝑑𝑥
=
𝑥
Formula 8 states that the integral of any quotient whose numerator is the
differential of the denominator is the logarithm of the denominator.
EXAMPLE 1.
ln 𝑥
Evaluate ∫ 𝑥 𝑑𝑥
∫
(ln 𝑥 ) 2
2
𝑑𝑥
.
𝑥
1
2
+ 𝐶 = ln2 𝑥 + 𝐶
SOLUTION.
Let 𝑢 = 𝑥 + 1, then 𝑑𝑢 = 𝑑𝑥. Thus the given integral takes the form of
𝑑𝑢
∫ 𝑢.
∫
EXAMPLE 6.
Evaluate ∫ (1 + 4𝑒 −𝑥 )(𝑒 −𝑥 )𝑑𝑥
SOLUTION.
𝑑𝑥
= ln(𝑥 + 1) + 𝐶
𝑥+1
EXAMPLE 2.
𝑥 𝑑𝑥
Let 𝑢 = 1 + 4𝑒 −𝑥 , then 𝑑𝑢 = −4𝑒 −𝑥
∫ (1 + 4𝑒
𝑑𝑢
𝑢
−𝑥 )(𝑒 −𝑥
1
1 (1 + 4𝑒 −𝑥 )5
𝑑𝑥) = − ∫ (1 + 4𝑒 −𝑥 )4 (−4𝑒 −𝑥 𝑑𝑥) = −
+𝐶
4
4
5
1
= − (1 + 4𝑒 −𝑥 )5 + 𝐶
20
Evaluate ∫ 𝑥 2 −3
SOLUTION.
Let 𝑢 = 𝑥 2 − 3, then 𝑑𝑢 = 2𝑥𝑑𝑥.
𝑥 𝑑𝑥
1
2𝑥 𝑑𝑥
1
∫ 𝑥 2 −3 = 2 ∫ 𝑥 2 −3 = 2 ln(𝑥 2 − 3) + 𝐶
1.4 Extension of Power Formula
𝑢𝑛+1
The power formula ∫ 𝑢𝑛 𝑑𝑢 = 𝑛+1 + 𝐶 is applicable for any rational number 𝑛
except when 𝑛 = −1. This is so because the power formula is undefined when 𝑛 =
−1. We now extend the power formula for the special case 𝑛 = −1.
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EXAMPLE 3.
𝑒 𝑥 𝑑𝑥
Evaluate ∫ 1+𝑒 𝑥
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SOLUTION.
= ∫(𝑥 + 1)𝑑𝑥 + ∫
Let 𝑢 = 1 + 𝑒 𝑥 , then 𝑑𝑢 = 𝑒 𝑥 𝑑𝑥.
∫
𝑒 𝑥 𝑑𝑥
= ln(1 + 𝑒 𝑥 ) + 𝐶
1 + 𝑒𝑥
EXAMPLE 2.
Evaluate ∫
EXAMPLE 4.
(𝑥 + 1)2
𝑑𝑥
=
+ ln(𝑥 + 1) + 𝐶
𝑥+1
2
𝑥 𝑑𝑥
√1−𝑥
SOLUTION.
sec2 3𝑥𝑑𝑥
Evaluate ∫ 2+tan 3𝑥
Rewrite 𝑥 as −[(1 − 𝑥) − 1].
SOLUTION.
∫
Let 𝑢 = 2 + tan 3𝑥, then 𝑑𝑢 = (sec 2 3𝑥)(3𝑑𝑥) .
∫
sec 2 3𝑥 𝑑𝑥 1 sec 2 3𝑥 3𝑑𝑥 1
= ∫
= ln(2 + tan 3𝑥) + 𝐶
2 + tan 3𝑥 3 2 + tan 3𝑥
3
𝑥 𝑑𝑥
√1 − 𝑥
= −∫
(1 − 𝑥) − 1
√1 − 𝑥
3
1
1
𝑑𝑥 = − ∫[(1 − 𝑥)2 − (1 − 𝑥)−2 ]𝑑𝑥
1
3
1
(1 − 𝑥)2 (1 − 𝑥)2
2
=−
+
+ 𝐶 = − (1 − 𝑥)2 + 2(1 − 𝑥)2 + 𝐶
3
1
3
2
2
1.5 Algebraic Manipulation
In Differential Calculus, we are asked to differentiate functions or algebraic
equations, and this can be done easily by simply applying the differentiation
formulas. However, in integration problems, it is necessary in some cases to perform
first certain algebraic manipulation to the integrand before an integration formula
will apply. Common algebraic manipulation makes use of expansion or performing
division in rational fractions.
EXAMPLE 3.
Evaluate ∫
𝑑𝑥
1−𝑒 −𝑥
SOLUTION.
Multiply both numerator and denominator by 𝑒 𝑥 .
∫
𝑑𝑥
𝑒 𝑥 𝑑𝑥
𝑒 𝑥 𝑑𝑥
=
∫
=
∫
= ln(𝑒 𝑥 − 1) + 𝐶
1 − 𝑒 −𝑥
𝑒 𝑥 (1 − 𝑒 −𝑥 )
𝑒𝑥 − 1
EXAMPLE 1.
EXAMPLE 4.
𝑥 2 𝑑𝑥
Evaluate ∫ 𝑥+1
(𝑥+4)𝑑𝑥
Evaluate ∫ (𝑥+2)2
SOLUTION.
𝑥2
1
Performing division gives 𝑥+1 = 𝑥 + 1 + 𝑥+1.
∫
𝑥 2 𝑑𝑥
1
= ∫ (𝑥 + 1 +
) 𝑑𝑥
𝑥+1
𝑥+1
SOLUTION.
𝑥+4
(𝑥+2)+2
Rewrite (𝑥+2)2 = (𝑥+2)2 =
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1
𝑥+2
2
+ (𝑥+2)2
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(𝑥 + 4)𝑑𝑥
𝑑𝑥
2
−2
∫
=
∫
+
∫
2(𝑥
+
2)
𝑑𝑥
=
ln(𝑥
+
2)
−
+𝐶
(𝑥 + 2)2
𝑥+2
𝑥−2
SOLUTION.
1
Let 𝑢 = 𝑐𝑜𝑠𝑥, then 𝑑𝑢 = −𝑠𝑖𝑛𝑥𝑑𝑥 or − csc 𝑥 𝑑𝑥
∫
1.6 Integration of Exponential Functions
There are two basic formulas for the integration of exponential functions:

𝑎𝑢
9. ∫ 𝑎 𝑑𝑢 =
+ 𝐶, where 𝑎
ln 𝑎
𝑢
𝑢
𝑢
> 0
 10. ∫ 𝑒 𝑑𝑢 = 𝑒 + 𝐶
𝑒 cos 𝑥 𝑑𝑥
𝑒 cos 𝑥 𝑑𝑥
= −∫
= −𝑒 𝑐𝑜𝑠𝑥 + 𝐶
csc 𝑥
−csc 𝑥
EXAMPLE 4.
2+𝑒 −𝑥
Evaluate ∫ (
𝑒 −𝑥
) 𝑑𝑥
SOLUTION.
Formula 10 is a special case of formula 9, that is when 𝑎 = 𝑒.
Algebraic manipulation:
2+𝑒 −𝑥
𝑒 −𝑥
= 2𝑒 𝑥 + 1
2 + 𝑒 −𝑥
∫ ( −𝑥 ) 𝑑𝑥 = ∫(2𝑒 𝑥 + 1)𝑑𝑥 = ∫ 2𝑒 𝑥 𝑑𝑥 + ∫ 𝑑𝑥 = 2𝑒 𝑥 + 𝑥 + 𝐶
𝑒
EXAMPLE 1.
Evaluate ∫ 2𝑥 𝑑𝑥.
SOLUTION.
1.7 Integrals of Hyperbolic Functions
Let 𝑢 = 𝑥, then 𝑑𝑢 = 𝑑𝑥.
∫ 2𝑥 𝑑𝑥 =
2𝑥
+𝐶
ln 2
EXAMPLE 2.
Evaluate ∫ 𝑒 −3𝑥 𝑑𝑥
SOLUTION.
The six hyperbolic functions are:
𝑒 𝑢 − 𝑒 −𝑢
sinh 𝑢 =
2
𝑒 𝑢 + 𝑒 −𝑢
cosh 𝑢 =
2
𝑒 𝑢 − 𝑒 −𝑢
tanh 𝑢 = 𝑢
𝑒 + 𝑒 −𝑢
csch 𝑢 =
sech 𝑢 =
coth 𝑢 =
2
𝑒 𝑢 − 𝑒 −𝑢
2
𝑒 𝑢 + 𝑒 −𝑢
𝑢
−𝑢
𝑒 +𝑒
𝑒 𝑢 − 𝑒 −𝑢
Let 𝑢 = −3𝑥, then 𝑑𝑢 = −3𝑑𝑥.
1
1
∫ 𝑒 −3𝑥 𝑑𝑥 = − ∫ 𝑒 −3𝑥 (−3𝑑𝑥) = − 𝑒 −3𝑥 + 𝐶
3
3
EXAMPLE 3.
𝑒 𝑐0𝑠𝑥 𝑑𝑥
Evaluate ∫ csc 𝑥
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The derivatives of these functions are as follows:
𝑑
𝑑
(sinh 𝑢) = cosh 𝑢 𝑑𝑢
(cosh 𝑢) = sinh 𝑢 𝑑𝑢
𝑑𝑥
𝑑𝑥
𝑑
𝑑
(tanh 𝑢) = sech2 𝑢 𝑑𝑢
(coth 𝑢) = −csch2 𝑢 𝑑𝑢
𝑑𝑥
𝑑𝑥
𝑑
(csch 𝑢) = −csch 𝑢 𝑐𝑜𝑡ℎ 𝑢 𝑑𝑢
𝑑𝑥
𝑑
(sech 𝑢) = −sech 𝑢 𝑡𝑎𝑛ℎ 𝑢 𝑑𝑢
𝑑𝑥
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San Vicente West, Urdaneta City, Pangasinan
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URDANETA CITY UNIVERSITY
College of Engineering
and Architecture
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Integrating these derivatives, we get the integration formulas for the hyperbolic
functions.
 11. ∫ 𝑐𝑜𝑠ℎ 𝑢 𝑑𝑢 = 𝑠𝑖𝑛ℎ 𝑢 + 𝐶
 12. ∫ 𝑠𝑖𝑛ℎ𝑢 𝑑𝑢 = 𝑐𝑜𝑠ℎ 𝑢 + 𝐶
 13. ∫ sech2 𝑢 𝑑𝑢 = 𝑡𝑎𝑛ℎ 𝑢 + 𝐶
 14. ∫ csch2 𝑢 𝑑𝑢 = −𝑐𝑜𝑡ℎ 𝑢 + 𝐶
 15. ∫ csch 𝑢 𝑐𝑜𝑡ℎ 𝑢 𝑑𝑢 = −𝑐𝑠𝑐ℎ 𝑢 + 𝐶
 16. ∫ sech 𝑢 𝑡𝑎𝑛ℎ 𝑢 𝑑𝑢 = −𝑠𝑒𝑐ℎ 𝑢 + 𝐶
EXAMPLES:
1
1
1) ∫ sinh 2𝑥 𝑑𝑥 = 2 ∫(sinh 2𝑥)2𝑑𝑥 = 2 cosh 2𝑥 + 𝐶
𝑥
𝑥 1
𝑥
2) ∫ cosh 𝑑𝑥 = 3 ∫(cosh ) 𝑑𝑥 = 3 sinh + 𝐶
2
3 3
3
𝑒 𝑥 −𝑒 −𝑥
3) ∫ tanh 𝑥 𝑑𝑥 = ∫ (𝑒 𝑥 +𝑒 −𝑥 ) 𝑑𝑥 = ln(𝑒 𝑥 + 𝑒 −𝑥 ) + 𝐶
1
1
4) ∫ sech 3𝑥 tanh 3𝑥 𝑑𝑥 = ∫ sech 3𝑥 tanh 3𝑥 3𝑑𝑥 = − sech 3𝑥 + 𝐶
3
3
𝑑𝑥
5) ∫ csch2 (ln 𝑥) = − coth(ln 𝑥) + 𝐶
𝑥
𝑑𝑥
𝑑𝑥
6) ∫ sech2 √𝑥 = 2 ∫ sech2 √𝑥 ( ) = 2 tanh √𝑥 + 𝐶
2√𝑥
√𝑥
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(075) 600 - 1507 | ucu.edu.ph
San Vicente West, Urdaneta City, Pangasinan
AdrianTomeldan@ucu.edu.ph | +63 917 9579 500