MODULE

Integration Concepts
1 and Formula

What is this all about?

In mathematics, we usually need to find the derivative of some

mathematical functions. It gives the rate of change of one variable with
respect to others. Integration is the opposite process of differentiation. The
fundamental use of integration is to get back the function whose derivatives
are known. So, it is like an anti-derivative procedure. Thus, integrals are
computed by viewing an integration as an inverse operation to differentiation.
In this module, the student will learn the Integration concepts as well
as some integration formula with examples.

What should you learn?

At the end of the lesson, the students should be able to:

a. Understand the concept of anti-differentiation.
b. Understand the integration of logarithmic functions.
c. Understand how to integrate exponential functions in various
contexts.
d. Learn how to integrate trigonometric functions using suitable
strategies.
 e. Learn how to integrate morse complex power functions using
appropriate method.
f. Apply the power formula to integrate functions with simple powers.
g. Apply the integration techniques for trigonometric functions.
h. Apply the integration techniques for hyperbolic functions.
i. Evaluate indefinite integrals using various techniques.

Get Started

Instructions: Choose the best answer for each question.

1. What is the antiderivative (integral) of the function f (x) = 2 x ?
a. x^2 + 1 c. 2x^3
b. x^2 d. 1/x^2
2. The integral of sin(x) dx represents:
a. The derivative of cos(x)
b. An indefinite integral with no constant of integration
c. The area under the curve of sin(x)
d. Cannot be determined without a specific interval
3. What is the integral of x^3 dx?
a. x^4 + C c. 3x^2 + C
b. x^2 + C d. 1/x^4
4. What is the integral of cos(x) dx?
a. sin(x) + C c. -sin(x) + C
b. cos(x^2) + C d. x*cos(x)
5. The integral of ln(x) dx is:
a. x*ln(x) + C c. x^ln(x) + C
b. 1/ln(x) + C d. [ln(x)]^2 + C
6. The integral of e^x dx is:
a. x*e^x + C c. e^x + C)
b. e^(x^2) + C d. 1/e^x + C
7. What is the integral of arctan(x) dx?
a. tan(x) + C c. x*arctan(x) + C
b. 1/arctan(x) + C d. ln(1 + x^2) + C
8. Which of the following is NOT generally true about hyperbolic functions (
sinh , cosh )?
a. They have similar properties to trigonometric functions (sine, cosine).
b. Their integrals can involve logarithmic terms.
c. They can be expressed in terms of exponential functions.
d. Their derivatives are not hyperbolic functions themselves.
 9. The integral of x n ⅆx is possible for what values of n (except for special
cases)?
a. Only positive integers
b. Only negative integers
c. Any real number except for -1
d. Only rational numbers
10. The Fundamental Theorem of Calculus relates:
a. Integration and differentiation of exponential functions
b. Integration and area under a curve
c. Differentiation and limits of sequences
d. Integration and trigonometric identities

Read and Learn

1.1 Review on Anti-Differentiation

The notation used to represent all anti-derivatives of a function f(x) is the

indefinite integral symbol (∫ ¿ ,1where ∫ f ( x ) dx=F ( x )+ c . The function f(x) is
called integrand, and c is referred to as the constant of the integration. The
expression F(x) + c is called the indefinite integral of F with respect to the
independent variable x.
Example 1.1.1: Find the anti-derivative of f(x)= cos x
Since the derivative of F(x)= sin x is F'(x)= cos x, write
∫ cos x dx=sin x + c
Example 1.1.2: Find the anti-derivative of f(x)= -8x
Since the derivative of F(x)= -8x is F'(x)= -8, write;
∫−8 dx=−8 x +c
Note that "c" is an arbitrary constant that is put in every solution of an
indefinite integration.

1.1.1 Basic Techniques for Anti-Differentiation

Many anti-differentiation formulas can be derived directly from there
corresponding derivative formulas, while other anti-differentiation problems
require more work.
Here are some of the basic anti-differentiation techniques;

Constant Multiple Rule

The constant rule of integration tells you how to find an integral for a
constant quantity. The rule is defined as:
∫ a dx, where "a is an element of any integers."
 Example 1.1.1: Evaluate the indefinite integral ∫ 4 dx .
∫ 4 dx
= 4 ∫ dx Take the constant 4 out of the integration
and integrate what is left.
= 4 (x) + C Remove the integral sign then add "+ C"
on the result and simplify if possible.
= 4x + C

Power Rule
n+1
x
The power rule says that ∫ x dx=
n
n+1
+C , To apply this rule, we simply
add "1" to the exponent and we divide the result by the same exponent of the
result. Finally, add C to the final result.
Example 1.1.2: Evaluate the indefinite integral ∫ x dx
3

∫ x 3 dx
3 +1
x
= +C Apply the concept of power rule.
3+1
Then simplify.
4
x
= +C
4

Example1.1.3: Evaluate the indefinite integral ∫ √ x dx

= ∫ √ x dx Transform radical into a fraction exponent.
1
= ∫ x 2 dx
1
+1
x2
= 1 +C Apply the concept of power rule.
+1
2
Then simplify.
3
2
x
= 3 +C
2
3
2
= x × 2 +C
1 3
3

= 2 x +C , or = √ +C
2 2 x3
3 3
 Sum and Difference Rule
The sum and difference rule tells us how we should integrate functions
that are the sum or difference of several terms. It basically tells us that we must
integrate each term in the sum or difference separately, and then just add the
results together. The order in which the terms appear in the result is not
important. We can state this formally as follows:
∫ f ( x ) ± g ( x ) dx=∫ f ( x ) dx ±∫ g ( x ) dx
You may be wondering at this point why the rule is written in the way that it is. It
is quite important to realize here that, in a function that is the sum or difference of
two (or more) terms, each term can be considered to be a function in its own right
- even a constant term.
Example 1.1.4: Evaluate the indefinite integral ∫ 6 x +8 x−10 dx
2

= ∫ 6 x +8 x−10 dx
2

= ∫ 6 x dx +∫ 8 x dx−∫ 10 dx
2

= 6 ∫ x dx+ 8∫ x dx−10 ∫ dx
2

( ) ( )
2+1 1+1
x x
=6 +8 −10 ( x ) +C
2+1 1+1

= 6 ( )+8 ( )−10 x +C
3 2
x x
3 2

= 6 ( )+8 ( )−10 x +C
3 2
x x
3 2
= 2 x3 + 4 x 2−10 x+ C

1.2 Review on Indefinite Integrals

Indefinite integrals represent the family of functions that have the same
derivative. They are written in the form ∫ f (x )dx =F (x )+ C , where F (x) is an
antiderivative of f (x), and C is the constant of integration. Indefinite integrals do
not have specified limits of integration. (See examples 1.1.1 – 1.1.4)

1.3 Simple Power Formula

n+1
x
The power rule for integration states that ∫ x n dx= + C , where n ≠−1.
n+1
This formula is used to find the integral of functions involving powers of x . (See
example 1.1.2 and 1.1.3)

1.4 Simple Trigonometric Function

Integration of Trigonometric Functions

Integration is the process of summing up small values of a function under the
region of limits. It is just opposite to the differentiation. Integration is also known
as anti-derivative.
Integration of Trigonometric functions involves basic simplification techniques.
These techniques use different trigonometric identities which can be written in an
alternative form that are more amenable to integration.
The integration of a function f(x) is given by F(x) and it is represented by:
∫ f ( x ) dx=F ( x )+ c
f (x) is called anti-derivative or primitive.
F (x) is called the integrand.
dx is called the integrating agent.
C is called the constant of integration or arbitrary constant.
X is the variable of integration.
Trigonometric formulas:
• ∫ sin x dx = -cos x + C
• ∫ cos x dx = sin x + C
• ∫ sec2 x dx = tan x + C
• ∫ cosec2 x dx = -cot x + C
• ∫ sec x tan x dx = sec x + C
• ∫ cosec x cot x dx = -cosec x + C
• ∫ tan x dx = ln | sec x | + C
• ∫ cot x dx = ln | sin x | + C
• ∫ sec x dx = ln | sec x + tan x | + C
• ∫ cosec x dx = ln | cosec x – cot x | + C
Example 1.4.1:

Simplify ∫ 3x2 sin (x3) dx.

Let I = ∫ 3 x 2sin (x 3)dx .

In order to evaluate the given integral lets substitute any variable by a new
variable as:

Let x 3be t for the given integral.

Then, dt=3 x 2dx

Therefore,

I = ∫ 3 x 2sin (x 3)dx= ∫ sin (x 3)(3 x 2 dx )

Now, substitute t for x3 and dt for 3x2 dx in the above integral.

I = ∫ sin (t)(dt )
 As ∫ sin x dx=−cos x+C , thus

I =−cos t+C

Again, substitute back x3 for t in the expression as:

I = ∫ 3x2 sin (x3) dx = -cos x3 + C

Which is the required integral.

Hence, the General Form of integration by substitution is:

∫ f[g(x)].g'(x).dx = f(t).dx

Where t = g(x)
Usually, the method of integration by substitution is extremely useful when
we make a substitution for a function whose derivative is also present in the
integrand. By doing so, the function simplifies and then the basic formulas of
integration can be used to integrate the function.

In calculus, the integration by substitution method is also known as the

“Reverse Chain Rule” or “U-Substitution Method”. We can use this method to find
an integral value when it is set up in the special form.

Example 1.4.1:

Determine the integral of the following function: f(x) = cos 3 x.

Solution:

Let us consider the integral of the given function as,

I = ∫ cos3 x dx

It can be rewritten as:

I = ∫ (cos x) (cos2x) dx

Using trigonometry identity; cos2x = 1 – sin2x, we get

I = ∫ (cos x) (1 – sin2x) dx

⇒ I = ∫ cos x – cos x sin2x dx

⇒ I = ∫ cosx dx – ∫ cosx sin2x dx

As ∫ cos x dx = sin x + C,

Thus, I = sin x – ∫ sin2x cos x dx . . . (1)

Let, sin x = t

⇒ cos x dx = dt.

Substitute t for sin x and dt for cos x dx in second term of the above integral.

I = sin x – ∫ t2 dt

⇒ I = sin x – t3/3 + C

Again, substitute back sin x for t in the expression.

Hence, ∫ cos3x dx = sin x – sin3 x / 3 + C.

Example 1.4.2:

If f(x) = sin2 (x) cos3 (x) then determine ∫ sin2(x) cos3(x) dx.

Solution:

Let us consider the integral of the given function as,

I = ∫ sin2(x) cos3(x) dx

Using trigonometry identity; cos2 x = 1 – sin2 x, we get

I = ∫ sin2 x (1 – sin2 x) cos x dx

Let sin x = t then,

⇒ dt = cos x dx

Substitute these in the above integral as,

I = ∫ t2 (1 – t2) dt

⇒ I = ∫ t2 – t4 dt

⇒ I = t 3 / 3 – t5 / 5 + C

Substitute back the value of t in the above integral as,

Hence, I = sin3 x / 3 – sin5 x / 5 + C.

Example 1.4.3:

Let f(x) = sin4(x) then find ∫ f(x)dx. i.e. ∫ sin4(x) dx.

Solution:

Let us consider the integral of the given function as,

I = ∫ sin4(x) dx

⇒ I = ∫ (sin2(x))2 dx

Use\ing trigonometry identity; sin2(x) = (1 – cos (2x)) / 2, we get

I = ∫ {(1 – cos (2x)) / 2}2 dx

⇒ I = (1/4) × ∫ (1+cos2(2x)- 2 cos2x) dx

⇒ I = (1/4) × ∫ 1 dx + ∫ cos2(2x) dx – 2 ∫ cos2x dx

⇒ I = (1/4) × [ x + ∫ (1 + cos 4x) / 2 dx – 2 ∫ cos2x dx ]

⇒ I = (1/4) × [ 3x / 2 + sin 4x / 8 – sin 2x ] + C

⇒ I = 3x / 8 + sin 4x / 32 – sin 2x / 4 + C

Hence, ∫ sin4(x) dx = 3x / 8 + sin 4x / 32 – sin 2x / 4 + C

1.5 Logarithmic Function

In mathematics, the logarithmic function is an inverse function to

exponentiation. The logarithmic function is defined as

For x > 0 , a > 0, and a ≠1,

y= loga x if and only if x = ay

Then the function is given by

f(x) = loga x

The base of the logarithm is a. This can be read it as log base a of x. The
most 2 common bases used in logarithmic functions are base 10 and base e.
Common Logarithmic Function
The logarithmic function with base 10 is called the common logarithmic function
and it is denoted by log10 or simply log.

f(x) = log10 x

Natural Logarithmic Function

The logarithmic function to the base e is called the natural logarithmic function
and it is denoted by loge.

f(x) = loge x

Logarithmic Functions Properties

Logarithmic Functions have some of the properties that allow you to
simplify the logarithms when the input is in the form of product, quotient or the
value taken to the power. Some of the properties are listed below.

Product Rule

logb MN = logb M + logb N

Multiply two numbers with the same base, then add the exponents.

log 30 + log 2 = log 60

Quotient Rule

logb M/N = logb M – logb N

Divide two numbers with the same base, subtract the exponents.

log8 56 – log8 7 = log8(56/7)=log88 = 1

Power Rule

Raise an exponential expression to power and multiply the exponents.

Logb Mp = P logb M

log 1003 = 3. Log 100 = 3 x 2 = 6

Zero Exponent Rule

loga 1 = 0.

Change of Base Rule

logb (x) = ln x / ln b or logb (x) = log10 x / log10 b

Here are some example of logarithmic function:

Example 1.5.1:
Use the properties of logarithms to write as a single logarithm for the given
equation: 5 log9 x + 7 log9 y – 3 log9 z
Solution:
By using the power rule , Logb Mp = P logb M, we can write the given equation as
5 log9 x + 7 log9 y – 3 log9 z = log9 x5 + log9 y7 – log9 z3
From product rule, logb MN = logb M + logb N
5 log9 x + 7 log9 y – 3 log9 z = log9 x5y7 – log9 z3
From Quotient rule, logb M/N = logb M – logb N
5 log9 x + 7 log9 y – 3 log9 z = log9 (x5y7 / z3 )
Therefore, the single logarithm is 5 log9 x + 7 log9 y – 3 log9 z = log9 (x5y7 / z3 )

Example 1.5.2:
Use the properties of logarithms to write as a single logarithm for the given
equation: 1/2 log2 x – 8 log2 y – 5 log2 z
Solution:
By using the power rule , Logb Mp = P logb M, we can write the given equation as
1/2 log2 x – 8 log2 y – 5 log2 z = log2 x1/2 – log2 y8 – log2 z5
From product rule, logb MN = logb M + logb N
Take minus ‘- ‘ as common
1/2 log2 x – 8 log2 y – 5 log2 z = log2 x1/2 – log2 y8z5
From Quotient rule, logb M/N = logb M – logb N
1/2 log2 x – 8 log2 y – 5 log2 z = log2 (x1/2 / y8z5 )

lo g2 x – 8 lo g2 y – 5 lo g2 z =log 2 ⁡( √8 5 )
1 x
The solution is
2 y z
1.6 Exponential Function

An exponential function is a Mathematical function in the form f (x) = ax,

where “x” is a variable and “a” is a constant which is called the base of the
function and it should be greater than 0. The most commonly used exponential
function base is the transcendental number e, which is approximately equal to
2.71828.
Exponential Function Formula
An exponential function is defined by the formula f(x) = ax, where the input
variable x occurs as an exponent. The exponential curve depends on the
exponential function and it depends on the value of the x.
The exponential function is an important mathematical function which is of the
form
x
f ( x )=a
Where a>0 and a is not equal to 1.
X is any real number.
If the variable is negative, the function is undefined for -1 < x < 1.
Here,
“x” is a variable
“a” is a constant, which is the base of the function.
An exponential curve grows, or decay depends on the exponential function. Any
quantity that grows or decays by a fixed per cent at regular intervals should
possess either exponential growth or exponential decay.

Exponential Growth
In Exponential Growth, the quantity increases very slowly at first, and then
rapidly. The rate of change increases over time. The rate of growth becomes
faster as time passes. The rapid growth is meant to be an “exponential increase”.
The formula to define the exponential growth is:
x
y=a(1+r )
Where r is the growth percentage.

Exponential Decay
In Exponential Decay, the quantity decreases very rapidly at first, and then
slowly. The rate of change decreases over time. The rate of change becomes
slower as time passes. The rapid growth meant to be an “exponential decrease”.
The formula to define the exponential growth is:
x
y=a(1−r)
Where r is the decay percentage.

Exponential Function Graph

The following figure represents the graph of exponents of x. It can be seen
that as the exponent increases, the curves get steeper and the rate of growth
increases respectively. Thus, for x > 1, the value of y=f n (x )increases for
increasing values of (n).
 From the above, it can be seen that the nature of polynomial functions is
dependent on their degree. The higher the degree of any polynomial function, the
higher its growth. A function which grows faster than a polynomial function is
x
y=f ( x )=a , where a>1. Thus, for any of the positive integer’s n the function f (x)
is said to grow faster than that of f n ( x).
Thus, the exponential function having base greater than 1, i.e., a > 1 is defined
as y=f ( x )=a x . The domain of exponential function will be the set of entire real
numbers R and the range are said to be the set of all the positive real numbers.

It must be noted that the exponential function is increasing and the point
(0, 1) always lies on the graph of an exponential function. Also, it is very close to
zero if the value of x is mostly negative.

Exponential function having base 10 is known as a common exponential function.

Consider the following series:

Exponential Function Rules

1. Law of Zero Exponent:
a0 = 1
2. Law of Negative Exponent:
a-m = 1/am
Example 1.6.1:
 f(x) = 2x

Example 1.6.2:
g(x) = (1/2)x.

1.7 Inverse Trigonometric Functions

Integrating functions with denominators of the forms

√ a +u , a +u ,∧u √ u +a , will result in inverse trigonometric functions. Integrals
2 2 2 2 2 2

resulting in inverse trigonometric functions are normally challenging to integrate

without the formulas derived from the derivative of inverse function.

Integral involving inverse trigonometric functions.

Let u be a differentiable function in terms of x and a > 0.

du −du
∫ 2 2 =sin−1 +c ∫ 2 2 =cos−1 + c
√ a −u √ a −u
 du 1 u −du 1 u
∫ a 2+u 2 = a tan−1 a +c ∫ a 2+u 2 = a cot−1 a + c
du 1 −1 u −du 1 −1 u
∫ = sec +c ∫ = csc +c
u √ u −a a
2 2 a u √ u −a a
2 2 a
 The table above shows the integral rules to keep in mind. Take note of the
denominator forms closely since it immediately tell you the integral rule we need
to apply.
Using the problem given at the start of the discussion, let us try to
evaluate the given problem.
Example 1.7.1:
dx
Evaluate the indefinite integral ∫
√ 1−25 x 2

Solution:
Having a direct look at the denominator, we have √ 12−(5 x )2, so the best formula
du −1 u
to use for our function is ∫ 2 2 =sin a +C , where a=1 and u=5 x .
√ a −u
For us to apply the formula, we’ll need to use the substitution method and rewrite
the integrand as follows;

u=5 x

du=5 dx

1
du=dx
5

1
du
dx 5
∫ =∫
√ 1−25 x 2 √ 12−u2
1 du
¿ ∫
5 √ 1−u2

Apply the appropriate formula that will return a sine inverse function.

du u
∫ −1
=sin +C
√ a −u
2 2 a

1 du 1 −1 u
∫
5 √ 1−u 5
2
= sin
1
+C

1 −1
¿ sin u+C
5

Since we earlier assigned uto be 5 x , we substitute this expression back so we

have an anti-derivative that is in terms of the original variable, x .

dx 1 −1
∫ = sin 5 x +C
√ 1−25 x 5
2

Example 1.7.2:
dx
Evaluate the indefinite integral ∫ 2
4 x +9
Solution:
When you see a denominator that is the sum of two perfect squares, this is a
great indicator that we’re expecting an inverse tangent function as its
antiderivative.
du
Since the function we’re working with has a form of 2 2 , use the formula that
a +u
du 1 −1 u
results to an inverse tangent function: ∫ 2 2 = tan +C , where a=3 and
a +u a a
u=2 x .
Let’s apply the substitution method to rewrite the integrand.
u=2 x
du=2 dx
1
du=dx
2
1
du
dx 2
∫ 4 x2 +9 =∫ u2 +9
1 du
¿ ∫ 2
2 u +9

Apply the appropriate integral properties and formulas to evaluate our new
expression.
1 du 1 du
∫ = ∫
2 u 2+ 9 2 32 +u2
¿
2 3[
1 1 −1 u
tan
3
+C
]
1 −1 u
¿ tan +C
6 3
Make sure to replace u with 2 x back to return an integral in terms of x .
dx 1 u
∫ 4 x2 +9 =¿ 6 tan−1 3 + C ¿

dx
Example 1.7.3: Evaluate the indefinite integral ∫
x √ 16 x 2−25
Solution:
 du
The integrand has the form, , so apply the formula that returns an inverse
u √u2 −a2
secant function:
du 1 u
∫ 2 2 = a sec−1 a + c , where a=5 and u=4 x . What makes this form unique is
u √ u −a
that aside from the radical expression, we see a second factor in the
denominator. If the second factor remains after simplifying the integrand, then
expect an inverse secant function for its antiderivative.

Since we still have a coefficient before the variable inside the radical, use the
substitution method and use u=4 x∧u 2=16 x 2.
u=4 x
1
u=x
4
du=4 dx
1
du=dx
4
1
du
dx 4
∫ =∫
x √16 x −25 1
2
u √u2 −25
4
1
4 du
¿ ∫
1 u √ u2−25
4
1 4 du
¿ ∙ ∫
4 1 u √ u 2−25
du
¿∫
u √ u2−25

Now that we’ve rewritten the integrand into a form where the inverse secant
function formula applies, let’s now integrate the expression as shown below.

du du
∫ =∫
u √ u −25
2
u √ u2 −52
1 −1 u
¿ sec +C
5 5

Substitute u=4 x back into the resulting expression.

dx 1 −1 4 x
∫ = sec +C
x √ 16 x −25 5
2 5

1.8 Hyperbolic Function

 Hyperbolic functions are mathematical functions that are analogs to the
trigonometric functions. It is essential in integral calculus as they provide
alternative methods to solve complex integrals involving exponential functions.
Hyperbolic functions are functions that parametrize a parabola. In integral
calculus, hyperbolic functions play a significant role in simplifying integrals
involving exponential functions. The primary hyperbolic functions include:
 Hyperbolic sine (sinh )
 Hyperbolic cosine (cosh )
 Hyperbolic tangent ( tanh)
 Hyperbolic cosecant (csch )
 Hyperbolic secant ( sech)
 Hyperbolic cotangent (coth )

We can use either the derivative rules or the exponential form of the rest
of the hyperbolic functions. But no worries, here are the summary of all six
hyperbolic functions’ integration rules as shown below.

Derivative Rule Integration Rule

d
sinh x=cosh x ∫ cosh x dx=sinh x +C
dx
d
cosh x=sinh x ∫ sinh x dx=cosh x +C
dx
d 2
tanh x=sech x ∫ sech2 x dx=tanh x +C
dx
d
coth x=−csch x ∫ csch2 x dx=−coth+C
dx
d
sech x=−sech x tanh x ∫−sech x tanh x dx=¿−sech x +C ¿
dx
d
csch x=−csch x coth x ∫−csch x coth x dx=−csch +C
dx

Here some guidelines on how to use these integral rules to integrate

hyperbolic expressions completely:

 Identify the hyperbolic expressions found in the function and take note of
their corresponding antiderivative formula.
 If the hyperbolic function contains an algebraic expression within it, apply
the substitution method first.
 If the function that needs to be integrated is a product of two simpler
functions, use integration by parts only when the substitution method does
not apply.

When you’re ready, go ahead and study some the examples provided. Learn
how to integrate different types of functions that contain hyperbolic expressions.
Example 1.8.1
cosh x
Evaluate the indefinite integral ∫ dx .
3+ 4 sinh x

Solution:
If we take a look at the derivative of the denominator, we have
d
3+ 4 sinh x=4 cosh x , so we use the substitution method to cancel the
dx
numerator out.
u=3+4 sinh x
du=4 cosh x dx
1
du=dx
4 cosh x
If we let u=3+4 sinh x we can cancel out cosh x once we replace dx with
1
du .
4 cosh x
cosh x cosh x 1
∫ 3+ 4 sinh x dx=∫ u
∙
4 cosh x
du

cosh x 1 1 1
¿∫ ∙ du¿ ∫ ∙ du
u 4 cosh x u 4
1 1
¿ ∫ du
4 u
1
Use the antiderivative formula, ∫ x dx=ln|x|+ C , Rewrite the antiderivative back
in terms of x by substituting u=3+4 sinh x back.
1 1 1
∫ du= 4 ln|u|+C
4 u
1
¿ ln ¿ 3+ 4 sinh x∨+C
4
cosh x 1
This means that ∫ dx= ln ¿ 3+ 4 sinh x∨+C .
3+ 4 sinh x 4

Example 1.8.2
Evaluate the integral, ∫ e cosh x dx
x
Solution:
We’re integrating the expression, e x cosh x , which is the product of two
expressions: e x and cosh x . We can’t apply the substitution method for this
expression. Instead, what we’ll do is rewrite cosh x using its exponential form
x −x
e +e
cosh x= .
2

( )
x −x
∫ e x cosh x dx =∫ e x e +2e dx
x x −x x
e ∙ e +e ∙ e
¿∫ dx
2
2x 0
e +e
¿∫ dx
2
1 x
¿ ∫ e +1 dx
2
We can then let ube 2 x and apply the substitution method as shown below.
u=2 x
du=2 dx
1
du=dx
2
1 1 1
∫ 2 ( e x +1 ) dx = 2 ∫ (e u +1 ¿ )∙ 2 du ¿
1 1
¿ ∙ ∫ e +1 du
u
2 2
1
¿ ∫ e + 1du
u
4
Evaluate the new integral expression by applying the sum rule and exponential
rule;
1 1
∫ e + 1du= (∫ e du+∫ 1 du )
u u
4 4
1 u
¿ ( e +u ) +C
4
Substitute u=2 x back into the expression so that we have our antiderivative in
terms of x .
1 u
( e +u ) +C= 1 ( e 2 x +2 x )+ C
4 4
2x 2x
e 2x e +2 x
¿ + + C∨¿ +C
4 4 4
2x
e +2 x
This means that ∫ e x cosh x dx =¿ +C ¿ .
4

Example 1.8.3
Evaluate the indefinite integral ∫ sinh x dx
2

Solution:
Rewrite sinh2 x using the hyperbolic identities, 2 2
cosh x−sinh x=1 and
2 2
cosh 2 x=sinh x +cosh x
2 2
−sinh x=1−cosh x
2 2
sinh x=cosh x−1
2 2 2
2 sinh x=sinh x+ cosh x−1
2
2 sinh x=cosh 2 x−1
2
2sinh x cosh 2 x−1
=
2 2
2 cosh 2 x−1
sinh x=
2

Substitute this expression back into our indefinite integral, ∫ sinh x dx

2

cosh 2 x−1
∫ sinh 2 x dx=∫ 2 dx
1
¿ ∫ cosh 2 x−1 dx
2

1
Apply the substitution method and use, u=2 x → du=2 dx → du=dx . Integrate
2
cosh u using the integral rule cosh u, ∫ cosh u dx=sinh x +C .
1 1 1
2
∫ cosh 2 x−1 dx= ∫ (cosh u−1¿) du¿
2 2
1 1
¿ ∙ ∫ cosh u−1 du
2 2
1
¿ ∫ cosh u du−∫ 1 du
4
1
¿ ( sinh u−u )+ C
4
1 1
¿ sinh u− u+C
4 4

Substitute u=2 x back into the expression.

1 1 1 1
sinh u− u+C= sinh 2 x− 2 x+C
4 4 4 4
1 1
¿ sinh 2 x − x +C
4 2

1 1
Hence, we have ∫ sinh x dx=¿ sinh 2 x− x+C ¿
2
4 2

1.9 General Power Formula

 The general power rule for integration extends the simple power formula
to handle a broader range of power functions. It states that
n (n+1)
∫ x dx=(x )/(n+1)+C for all real numbers n except when n=−1.

Example 1.9.1
1
∫ dx
x √ x2
25

Look at the denominator, it’s quite intimidating to integrate. But it will

2
be simple to integrate if we transform √5 x 2=x 5 ,
1 1
∫ dx=∫ dx
x ∙ √x
2 5 2 2
2 5
x ∙x
2 10+2 12
Multiply x 2 ∙ x =x
5 5
=x 5

1 1
∫ 2
dx=∫ 12
2 5 5
x ∙x x
−12
¿∫ x 5
dx
−12
+1
5
x
¿ +C
−12
+1
5
−7
x5
¿ +C
−7
5
−7
5
5x
¿−+C
7
5 −5
¿− 7 + C∨ 5 7 +C
7√x
7 x5
 Try this!

A. Evaluate the following integration.

1
1. ∫ x 5 dx 5.∫ 4
dx
3x
1
2. ∫ x 4 dx 6. ∫ ( 3 x 4 −6 x 2+ 12 ) dx
3
4 x −5
3. ∫ √3 x 2 dx 7. ∫ 2
dx
x
1
8. ∫ dx
( x +2 )2
6 3
6−5 x −2 x
∫−6 x dx 9. ∫
5
4. 3
dx
x
B. Integrate the following simple trigonometric function.
1. ∫ cos 5 x dx
2. ∫ sin 2 xdx
3. ∫ csc2 x dx
4. ∫ x tan x 2 dx
5. ∫ sec2 θ dθ
C. Integrate the following exponential and logarithmic function.
1. ∫ e
2 x−1
dx
∫ e x dx
2

2.
7
3. ∫ x dx
1
4. ∫ x+ 5 dx
1
5. ∫ e x dx

D. Integrate the following functions yielding inverse trigonometric

functions.
dx
1. ∫
√ 81−x 2
 dx
2. ∫ x 2+16
dx
3. ∫ x 2−6 x +18

E. Integrate the following hyperbolic functions.

1. ∫ x sinh x dx
2 3

2 sinh x
2. ∫ 5+6 cosh x dx
3. ∫ cosh 2 x dx
4. ∫ 4 e x sinh x dx
x
5. ∫ coth 6 dx

Do this!

A. Evaluate the following indefinite integral.

1
1. ∫ x π dx
1
2. ∫ 2 + x32 +1 dx
2 x

3. ∫ ( 2 x +1 )4 dx
3
4. ∫ 5 √ y− √ y dy
4 7
5. ∫ 3 t3 + 2t dt
B. Integrate the following:
2
1. ∫ 7−6 sin θ
x
sin θ
dθ

2. ∫ 4 x cos ( x 2 ) dx
3. ∫ 4 x 3−9+2 sinx+7 e x dx
23 y 9
4. ∫ (¿ y 2+1 +6 csc y + y )dy ¿
5. ∫ 12+ cscθ [ sinθ+ cscθ ] dθ
C. Evaluate the following indefinite integral yielding exponential and
logarithmic functions.
1. ∫ 2 x e dx
4
3 x

2. ∫ log 2 x dx
∫ x e−x dx
2

3.
dx
4. ∫ eln x x
D. Evaluate the following indefinite integral yielding inverse trigonometric
functions.
x−3
1. ∫ 2 dx
x +1
t +5
2. ∫ dt
√ 9− ( t−3 )
2

2
x −2
3. ∫ dx
x √ x 2−4
x +3
4. ∫ dx
( x +2 )2+ 4
3 dy
5. ∫
2 √ y ( 1+ y )

E. Evaluate the following indefinite integral yielding hyperbolic functions.

1. ∫ cosh 3 x sinh 3 x dx
2. ∫ cosh 2 x dx
sinh x dx
3. ∫ ( 1+2 cosh x )5
4. ∫ sinh 2 x dx
cosh x
5. ∫ 3+ 4 sinh x dx
 REFERENCES

enginerdmath. (2022). Integration of hyperbolic functions. Retrieved from

https://www.youtube.com/live/DT7EOhN0px4?si=_AjWvYZP507Ir9fV

Integration of trigonometric functions - formulas, solved examples. Retrieved from

https://byjus.com/maths/integration-trigonometric-functions/#:~:text=Below%20are
%20the%20list%20of,ln%7Csec%20x%7C%20%2B%20C

Integrals of inverse trig functions - definition, formulas, and examples. (2023). Retrieved
from https://www.storyofmathematics.com/integrals-of-inverse-trig-functions/

Answer Key:
Try this!!!
A.
1. ∫ x dx
5

5 +1
x
= +c
5+1
6
x
= +c
6
1
2. ∫ 4 dx
x
=∫ x dx
−4

− 4+1
x
= +c
−4 +1
−3
x
= +c
−3
−1
= 3 +c
3x

3.∫ √3 x 2 dx
2
= ∫ x 3 dx
2
+1
3
x
= 2 +c
+1
3
 5
3
x
= 5 +c
3
5
3 3
= x +c
5
4. ∫ −6 x dx
5

= −6 ∫ x dx
5

5+1
x
= −6 +c
5+1
6
x
= −6 +c
6
6
= −x +c

1
5. ∫ 3 x 4 dx
1 1
= ∫ dx
3 x4
1
=
3
∫ x−4 dx
− 4+1
1 x
= +c
3 −4 +1
−3
1x
= +c
3 −3
−1 −3
= x +c
9

−1
= 3
+c
9x

6. ∫ (3 x −6 x +12)dx
4 2

=3∫ x −6 ∫ x +12∫ dx
4 2

4+1 2+1
x x
=3 −6 + 12 x +c
4+1 2+1
5 3
x x
= 3 −6 +12 x+ c
5 3
5
x
= 3 −2 x 3+12 x +c
5
3
4 x −5
7. ∫ 2
dx
x
3
4x 5
= ∫ 2 −∫ 2 dx
x x
 2
= 2x − ( −5x )+ c
2 5
= 2 x + +c
x

1
8. ∫ ¿¿ ¿
Let u= x +2
du= dx
1
= ∫ 2 du
u
= ∫ u du
−2

−2+1
u
= +c
−2+1
−1
u
= +c
−1
−1
= +c
u
−1
= +c
x+2
6 2
6−5 x −2 x
9. ∫ 3
dx
x
6 3
6 5x 2x
= ∫ 3 − 3 − 3 dx
x x x
6
= ∫ 3 −∫ 5 x −∫ 2 dx
3

x
−3 +1 3+ 1
x x
=6 −5 −2 x +c
−3+1 3+1
−2 4
x x
=6 −5 −2 x+ c
−2 4
4
−3 x
= 2 −5 −2 x +c
x 4

B.
1. ∫ cos 5 x dx
let u=5 x
du 5 dx
=
5 5
du
=dx
5
du
= ∫ cos u
5
 1
=
5
∫ cos u du
1
= sin u+c
5
1
= sin(5 x )+ c
5

2.∫ sin 2 x dx
1
= ∫ [ 1−cos(2 x) ] dx
2
1
=
2
∫ [ 1−cos (2 x)] dx
=
1
2 [
x−
sin(2 x)
2 ]
+c
1 1
= x− sin(2 x)+ c
2 4

3.∫ csc2 x dx
1
= ∫ 2 dx
sin x
1
2
cos x
=∫ dx
2
sin x
2
( 1
2
cos x )
sec x
=∫ 2
dx
tan x
let u=tanx
2
du=sec x dx
1
=∫ 2 du
u
= ∫ u du
−2

= −u−1+ c
= −cot x +c

4. x tan x 2 dx
2
let u=x
du 2 x dx
=
2 2
du
=x dx
2
du
=∫ tan(u)
2
 1
=
2
∫ tan (u)du
1 sin(u)
= ∫ du
2 cos(u)
let u=cos u
du=−sin( u) du
−du=sin(u)du
1 1
= ∫ (−du)
2 u
−1 1
=
2 u
∫ du
−1
= ln |u|+ c
2
−1
= ln |cos (u)|+c
2
−1
ln |cos (x )|+ c
2
=
2

5. ∫ sec2 θ dθ
= tanθ+ c

C.
1. ∫ e
2 x−1
dx
let u=2 x−1
du 2 dx
=
2 2
du
=dx
2
u du
= ∫e
2
1
= ∫ e du
u
2
1 u
= e +c
2
1 2 x−1
= e +c
2

2. ∫ 2 x e x dx
2

Use U-substitution
u du
¿∫ 2 x e
2x
¿ ∫ e du
u

u
¿ e +C
 2

¿ e x +C

7
3. ∫ x dx
1
= 7 ∫ dx
x
= 7 ln |x|+ c

1
4. ∫ x+ 5 dx
let u=x +5
du=dx
1
= ∫ du
u
= ln |u|+c
= ln |x +5|+ c

1
5.∫ e x dx
= ∫ e dx
−x

let u=−x
du=−1 dx
−du=dx
= ∫ e (−du)
u

= −∫ e du
u

= −e u +c
= −e− x + c

D.
dx
1. ∫
√ 81−x 2
1
= ∫ dx
√ 81−x 2
∫ du
√ a −u
2 2
=arcsin ( ua )+c
2 2 2
a =81 u =x
a=9 u=x
= arcsin
x
9
+c ()
 dx
2. ∫ x 2+16
1
=∫ 2
dx
x +16
∫ 2du 2 = 1a arctan ua +c
a +u ()
2 2 2
a =16 u =x
a=4 u=x
1
= arctan
4
x
4 ()
+c

1
3. ∫ x 2−6 x +8 dx
=∫ 1 ¿
¿¿
1
∗1
9
=∫ 2
dx
(x−3)
+1
9
1
∗1
= 9
∫ ¿¿ ¿
2 x−6
let u=
6
2
du= dx
6
dx=3 du
1
∗3∗1
= 9
∫ u 2+1 du
1 1
= ∫ ∗¿ 2 du ¿
3 u +1
1
= arctan u
3
1
= arctan
3
2 x−6
6 ( +c )
E.
du
1. ∫ x sinh x dx=∫ x sinh u
2 3 2
2
3x
du
¿ ∫ sinh u
3
1
¿
3
∫ sinh u du
 1
¿ cosh u+C
3
1 3
¿ cosh x +C
3

2 sinh x 2 sinh x du
2. ∫ dx=¿ ∫ (use u−substitution∈the denominator)¿
5+6 cosh x u 6 sinh x
1 1
¿ ∫ ∙ du
u 3
1 1
¿ ∫ du
3 u
1
¿ ln u+C
3
1
¿ ln 5+6 cosh x +C
3

1
3. ∫ cosh x dx=∫ cosh 2 x +1 dx
2
2
1
¿ ∫ cosh 2 x +1 dx
2
1
¿ ∫ cosh 2 x dx+ ¿∫ 1dx ¿
2
1 du
¿ ∫ cosh u +¿ ∫ 1 dx ¿
2 2
1 1
¿ ∙ sinh u+ x +C
2 2
1
= sinh 2 x+ x +C
4

4. ∫ 4 e sinh x dx =¿ 4 ∫ e sinh x dx ¿
x x

( )
x −x
e −e
¿ 4∫ e
x
dx
2
2x 0
e −e
¿ 4∫ dx
2
4
¿ ∫ e −1 dx
2x
2
¿ 2∫ e dx −¿∫ 1 dx ¿
2x

( )
2x
e
¿2 −x +C
2
2x
¿ e −2 x+ C
 x
5.∫ coth dx=∫ coth u 6 du
6
¿ 6 ∫ coth u du
¿ 6 ln sinh u+C
x
¿ 6 ln sinh +C
6
Do this!
A.
1
1 +1
xπ
1. ∫
π
x dx= C
1
+1
π
1
+1
π
x
= 1+ π C
π
1
+1
π
=π x +C
1+ π
1 1
3 3
2. ∫ 2 x 2 + 2
+1 dx = ∫ 2 x 2 dx +¿ ∫ 2 dx+ ¿∫ 1dx ¿ ¿
x x
1
1
¿ 2∫ x 2 dx +¿ 3∫ 2 dx +1∫ dx ¿
x
3
x 2
x−1
¿2 +3 + x +C
3 −1
2
3
2∙ 2 x 3 x−1 2
¿ + + x +C
3 −1
3
4 x2 3
¿ − + x+C
3 x

3. ∫ ( 2 x +1 ) dx=∫ 16 x +32 x + 24 x + 8 x +1 dx
4 4 3 2

¿ ∫ 16 x dx + ∫ 32 x dx + ∫ 24 x dx + ∫ 8 xdx + ∫ 1 dx
4 3 2

¿ 16 ∫ x +32 ∫ x + 24 ∫ x +8 ∫ x +1 ∫ dx
4 3 2

4+1 3 +1 2+1 1+1

x x x x
¿ 16 +32 +24 +8 + x+C
4+ 1 3+1 2+1 1+1
5 4 3 2
x x x x
¿ 16 +32 +24 +8 + x+C
5 4 3 2
5
16 x 4 3 2
¿ +8 x +8 x + 4 x + x+ C
5
 3 3
4. ∫ 5 √ y− dy=∫ 5 √ y dy−¿ ∫ dy ¿
√y √y
1
1
¿ 5∫ y 2 dy −3∫ 1 dy
y2
1 −1
+1 +1
2 2
y y
¿5 −3 +C
1 −1
+1 +1
2 2
3 1
5 y2 3 y2
¿ − +C
3 1
2 2
3
10 √ y 3
2 1
10 y
¿ 2
−6 y +C∨ −6 √ y +C
3 3

4 7 4 7
5.∫ + dt=∫ 2 dt +¿ ∫ dt ¿
3 t 2t 2t
2
3t
4 1 7 1
¿ ∫ 2 dt +¿ ∫ dt ¿
3 t 2 t
−2 +1
4 t 7
¿ + ln t +C
3 −2+ 1 2
−1
4t 7
¿ + ln t+C
−3 2
4 7
¿− + ln t+C
3t 2

B.
2
7−6 sin θ
1.∫ x
dθ
sin θ
2
7 dθ 6 sin θ
=∫ 2 −∫ 2
dθ
sin θ sin θ
2
dθ sin θ
= 7 ∫ 2 −6∫ 2 d θ
sin θ sin θ
2
dθ sin θ
= 7 ∫ 2 −6∫ 2 d θ
sin θ sin θ
=7 ∫ csc θ−6∫ dθ
2

=−7 cotθ−6 θ+c

2.∫ 4 x cos ( x 2 ) dx
 du
Let u=x2 =4∫ x cos u( )
du 2 x dx 2x
= 4
2x 2x =
2
∫ cos u du
du
dx= =2 sin u+c
2x
=2 sin x2 + c

3.∫ 4 x −9+2 sinx+7 e dx

3 x

=4 ∫ x dx−9∫ dx+ 2∫ sinx dx +7∫ e dx

3 x

=4¿)−9 x +2 (−cosx ) +7 e x +c

( )
4
x x
=4 −9 x−2 cosx +7 e +c
4
= x −9 x−2 cosx +7 e x +c
4

23 y 9
4. ∫ (¿ +6 csc y + )dy ¿
2
y +1 y
23 ( 2 ) y dy
Let u= y + 1
2
=
2
∫ 2
y +1
dy +6 ∫ cscy coty dy +9∫
y
du
=2 y 23 du dy
dy = ∫ +6∫ cscy coty dy+∫
du=2 ydy 2 u y
23
= ln|u|−6 cscy+9 ln | y|+C
2
23
= ln| y +1|−6 cscy+ 9 ln | y|+C
2
2

5. ∫ 12+ cscθ [ sinθ+ cscθ ] dθ

=∫ 12+ sinθ cscθ+ csc θ dθ
2

=12∫ dθ+∫ sinθ cscθ dθ+ (−cotθ )+C

=12θ+∫ sinθ
1
sinθ ( )
dθ+ (−cotθ ) +C
=12θ+θ−cotθ +C
=13θ−cotθ+C

C.
du
∫ 2 x 3 e x dx=∫ 2 x 3 eu 4 x 3
4

1.
2
¿
4
∫ u
e du
1 u
¿ e +C
2
 1 x 4

¿ e +C
2
2. ∫ log 2 x dx
u=log 2 x v =x
1
du= dx dv =dx
ln 2 x
1
¿ log 2 x ∙ x−∫ x dx
ln 2 x
1
¿ x ∙ log 2 x−∫ dx
ln 2
1
¿ x ∙ log 2 x− ∙ x+C
ln 2
x
¿ x ∙ log 2 x− +C
ln 2

du
∫ x e−x dx=∫ x eu −2 x
2

3.
1
¿
2
∫ e u du
1 u
¿ e +C
2
1 −x 2

¿ e +C
2

dx
4. ∫ eln x x
=∫ e x du
u

u
¿ x e +C
ln x
¿ x e +C

D.
x−3 x 3
1. ∫ x 2+1 dx=∫ x 2+1 − x 2+1 dx
x 3
¿∫ 2
dx−¿∫ 2 dx ¿
x +1 x +1
¿¿
1 2 −1
¿ ln x + 1−3 tan x +C
2

1 u
2. ∫ dt =sin
−1
+C
√ 9− (t−3 ) 2 a
−1 t−3
¿ sin +C
3
 2 2
x −2 x 2
3. ∫ dx=∫ dx−∫ dx
x √ x −4
2
x √ x −4
2
x √ x 2−4
x 2
¿∫ 2 dx−∫ dx
√ x −4 x √ x2 −4
−1
1
¿ ∫ x ∙ ( x −4 ) dx−2 ∫
2 2
dx
x √ x −4
2

=u−substitution−ITF :(arctan)
−1 x
¿ √ x −4−sec
2
+C
2
1
( 2 x + 4 ) +1
4. x +2 2
∫ ( x +2 )2+ 4 dx=∫ ( 2 dx
x + 4 x +4 ) + 4
1
( 2 x +4 )
2 1
¿∫ 2 dx +¿ ∫ 2 dx ¿
x + 4 x+ 8 x + 4 x +8
1 ( 2 x + 4) 1
¿ ∫ 2 dx +¿ ∫ 2 dx ¿
2 x +4 x+ 8 x + 4 x+ 8
¿ u −for the denominator+ ITF :( arctan)
1 1 −1 x+2
¿ ln |x +4 x +8|+ tan
2
+C
2 2 2

3 dy 3 dy
5. ∫ 2 √ y ( 1+ y ) = 2 ∫ √ y ( 1+ y )
3 dy
¿ ∫
2 √ y+√ y ∙ y
3 dy
¿ ∫ 1 3
2
y 2+ y 2
3 2
¿ ∫ du
2 1+u 2
1
¿ 3∫ 2
du
1+u
¿ 3 arctan u+C
¿ 3 arctan √ y +C

E.
1. ∫ cosh 3 x sinh 3 x dx=u−: u=sinh 3 x
du
¿ ∫ cosh 3 x ∙u
3 cosh 3 x
du
¿∫ u
3
1
¿
3
∫ u du
 1+1
1u
¿ +C
3 1+ 1
2
1u
¿ +C
3 2
2
u
¿ +C
6
2
sinh(3 x ¿)
¿ ¿
6

1
2. ∫ cosh 2 x dx=∫ 2 cosh 2 x +1 dx
1
¿
2
∫ cosh 2 x +1 dx
1
¿ ∫ cosh 2 x dx+ ¿∫ 1dx ¿
2
1 du
¿ ∫ cosh u +¿ ∫ 1 dx ¿
2 2
1 1
¿ ∙ sinh u+ x +C
2 2
1
= sinh 2 x+ x +C
4

sinh x sinh x du
3. ∫ ( 1+2 cosh x )5 dx =∫ ( u )5 2 sinh x
1 1
¿∫ ∙ du
u 2
5

1
¿
2
∫ −5
u du

( )
−4
1 u
¿ +C
2 −4
1
¿− 4 +C
8u
1
¿− 4
+C
8 ( 1+2 cosh x )
4. ∫ sinh x dx
2

2 2
−sinh x=1−cosh x
2 2
sinh x=cosh x−1
2 2 2
2 sinh x=sinh x+ cosh x−1
2
2 sinh x=cosh 2 x−1
2
2sinh x cosh 2 x−1
=
2 2
2 cosh 2 x−1
sinh x=
2
 cosh 2 x−1
∫ sinh 2 x dx=∫ 2
dx
1
¿
2
∫ cosh 2 x−1 dx
1 1 1
2
∫ cosh 2 x−1 dx= ∫ (cosh u−1¿) du¿
2 2
1 1
¿ ∙ ∫ cosh u−1 du
2 2
1
¿ ∫ cosh u du−∫ 1 du
4
1
¿ ( sinh u−u )+ C
4
1 1
¿ sinh u− u+C
4 4
1 1 1 1
sinh u− u+C= sinh 2 x− 2 x+C
4 4 4 4
1 1
¿ sinh 2 x − x +C
4 2

5.
cosh x
∫ 3+ 4 sinh x dx
d
3+ 4 sinh x=4 cosh x
dx
u=3+4 sinh x
du=4 cosh x dx
1
du=dx
4 cosh x
cosh x cosh x 1
∫ 3+ 4 sinh x dx=∫ u
∙
4 cosh x
du

cosh x 1 1 1
¿∫ ∙ du¿ ∫ ∙ du
u 4 cosh x u 4
1 1
¿ ∫ du
4 u
1 1 1
∫
4 u
du= ln |u|+C
4
1
¿ ln ¿ 3+ 4 sinh x∨+C
4