INTEGRATION
Definition: The process of finding a function from its derivative is called anti- differentiation or indefinite integration.
Integrand
The integral relationship
between f(x) and g(x) is expressed as .Within
the context of indefinite integral ,
the integral symbol is ?, the function g(x) is called the integrand,
c is the constant of integration and dx is the
differential that indicates x indicates x the of variable of
integration f(x) is the integral of g(x). Using
the indefinite integral of g(x) = 4x3
The constant of integration can only be determined if additional information is given
INTEGRATION OF SIMPLE FUNCTION
Generally, when given that
EXAMPLE 1
Integrate 4x3 with respect to x
EXAMPLE 2
Determine the following
SOLUTION
SOME STANDARD INTEGRALS
RULES OF INTEGRATION
a) Integral of sum integrable functions. If u1, u2, u3 – – – un are function of x then
b) If a is a constant, then
EXAMPLE 3
Determine the following integral
EXAMPLE 4
Integrate the following functions with respect to x
SOLUTION
More Examples on the integral of the form
Example 5
Evaluate
SOLUTION
THE INTEGRAL
EXAMPLE 5
SOLUTION
Integral of the form
The integral of a fraction whose numerator is the derivative of its denominator is the logarithm of the denominator.
EXAMPLE 6
EXAMPLE 7
EXAMPLE 8
SOLUTION
Integration By Algebraic Substitution
This is technique, in which the independent variable say x is change to u, where the relationship between x and u is known. This will make the integral to be reduced to a Standard Integral Form.
EXAMPLE 9
EXAMPLE 10
INTEGRATION OF RATIONAL FUNCTIONS USING PARTIAL FRACTION
At times, we may have to find an integral, which is not a standard type like . In this case the derivative will not yield the numerator even if we try to manipulate it. In such case to the integral of , we can express the rational function in terms of its partial fraction.
Therefore
is called a partial decomposition of the first integrand. It will be nice to remind ourselves about rules to follow when resolving or decomposing into partial fraction. You can also consult my book titled “Mathematics for Engineers and Scientist” where I discuss in detail how to decompose into partial fraction.
Rules
1. The numerator of the given rational function must be of lower degree than that of denominator. If it is not we apply long division to make it a lower degree.
2. If the denominator can be factorised into linear factor e.g. , the partial fraction will be like this .
3. If the denominator has one linear and irreducible quadratic factor e.g. , the partial fraction will be of the form
4. If the denominator corresponds to a linear factor of the form (ax + b)n. The partial fraction will be of the form
EXAMPLE 11
SOLUTION
INTEGRATION BY PART
Integration by part is a technique based on the product for differentiation. Let u and v be function of x, such that they are differentiable, then
Steps to follows when using integration by part formula
- Choose function u and v so that u can be easily be differentiated and dv is easy to integrate.
Organise the computation of du and dv
and substitute into the integration
- Finish up the integration by finding ∫vdu then ∫udv= uv – ∫vdu and add c at the of the computation.
EXAMPLE 12
EXAMPLE 13
This example shows that it is quite possible for you to everything right and still not to get answer given at the back of the textbook.
EXAMPLE 14
SOLUTION
REPEATED APPLICATION OF INTEGRATION BY PART
Frequently, an integration by part results in an integral that cannot be evaluate directly, but instead, one that we can evaluate only by repeating integration by parts one or more times. This is illustrated with the following examples
EXAMPLE 15
SOLUTION
INTEGRATION OF TRIGONOMETRIC FUNCTION
POWER OF SINE AND COSINE
Certain trigonometric function can be integrated after they have been expressed in terms of standard trigonometric function like the one stated above.
EXAMPLE 16
EXAMPLE 17
Example 18
It is important at this point that you should not try to memorise the integral obtained so far but try deriving the integrals on your own.
Before we move on, let us see how to derive the integrals of sec x and cosec x
EXAMPLE 19
The Integral Of The Form ∫sinmxcosnxdx Where m and n are Positive
Case 1: m or n is an odd as positive integer. If m is odd, use the substitution u = cosx, if n is odd use the substitution u = sinx
EXAMPLE 20
Case 2: Both m and n are odd (m = 2p +1, n = 2q +1, p, q ). We introduce an auxiliary function, so that, we express the integrand polynomial in terms of the auxiliary function and then perform the integration term by term.
Example 21
Case 3: If both m and n are even ( ).The integrand should be expressed in the simplest linear form in term of the cosine function and the integration can be performed term by term.
EXAMPLE 22
The Integral Of The Form ∫tanmx secnxdx Where m and n are Positive integers
Case 1: First, try to bring out sec x tan x (you will need this for du). Then, change any factors of tan2x with sec2x – 1 and use the substitution u= sec x
EXAMPLE 23
Case 2: First, bring out the factor sec2x (As you will need this as du). Then, replace the factors sec2x with 1 + tan2x and make use of the substitution u= tan x.
Integration of the form
It is good that we remind ourselves of the these four identities
EXAMPLE 24
EXAMPLE 25
EXAMPLE 26
INTEGRATION BY TRIGONOMETRIC SUBSTITUTION
Some integral form require trigonometric substitution .If an integral contain terms of the form . You can evaluate the integral by making a substitution using a trigonometric function.
If the integrand involves
EXAMPLE 27: An integral involving
The side opposite the angle θ is then 3sin θ by the Pythagorean Theorem, we get the adjacent side as as indicated, so we have
It now follows that
You must remember to change the variable back to x back. So, just like we did in the first example, we apply the same technique again.
We have substituted .
You can use integration by part to find the integral of
EXAMPLE 28: An integral involving
EXAMPLE 29: An integral involving
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