If a student aspires to become a successful engineer, then he or she should have a strong knowledge of integration. Integration is important in every step of engineering, from measuring cable lengths to planning long projects, integration is an integral part of Mathematics and very much important in this field of career.

There are many significant integration formulas that are used for integrating many of the standard integrals. All the different Integration methods have their own merits and play a vital role in yielding proper end results. In this section, we will look at the different integrals of the typical functions.

Integration is generally the mixing of items that got separated earlier. If we consider the figure ∫ f(x)dx = F(x) + C, if F′(x)=f(x), ∫ is the integral symbol there. F(x) is the integrand, x is the variable, and C remains the constant of integration.

When there are integrals with a specified upper limit and lower limit, then it is referred to as a definite integral. It is also called definite because it provides a precise answer at the end of every problem that is stated. As the definite integral exists on the curve and x-axis over a specified interval, it is obtained as positive, otherwise negative when below or above the x-axis.

While when we talk about the indefinite integral, it is specified in the general terms. It is in contrast to the referred integration form. The indefinite integral is considered as functions antiderivative, as its interpretation is not possible by the state of nature. The area of an interval cannot be judged easily depending on the general nature of integration.

\[\int \frac{dx}{(x^{2} - a^{2})}\] = \[\frac{1}{2a}\] log |\[\frac{(x – a)}{(x + a)}\]| + C

\[\int \frac{dx}{(a^{2} - x^{2})}\] = \[\frac{1}{2a}\] log |\[\frac{(a - x)}{(a + x)}\]| + C

1: \[\int \frac{1}{(x^{2} - a^{2})}\] = \[\frac{1}{2a}\] log |\[\frac{(x - a)}{(x + a)}\]| + C

We know that,

\[\frac{1}{(x^{2} - a^{2})} = \frac{1}{(x - a)(x + a)} = \frac{1}{2a} [\frac{(x + a) - (x - a)}{(x - a)(x + a)}]\]

Therefore, \[\frac{1}{2a}\] [1/(x – a) – 1/(x + a)]

Therefore it goes like,

\[\int \frac{dx}{(x^{2} – a^{2})} = \frac{1}{2a}\] [ \[\int \frac{dx}{(x – a)} – \int \frac{dx}{(x + a)}\]]

= \[\frac{1}{2a}\] [log |(x – a) – log |(x + a)] + C

= \[\frac{1}{2a}\] log |(x – a) / (x + a)| + C

2: \[\int \frac{1}{(a^{2} - x^{2})}\] = \[\frac{1}{2a}\] log |\[\frac{(a - x)}{(a + x)}\]| + C

We know,

\[\frac{1}{(a^{2} - x^{2})} = \frac{1}{(a - x)(a + x)} = \frac{1}{2a} [\frac{(a + x) + (a - x)}{(a - x)(a + x)}]\]

= \[\frac{1}{2a}\] [1/(a – x) + 1/(a + x)]

Hence,

\[\int \frac{dx}{(a^{2} - x^{2})} = \frac{1}{2a}\] [ \[\int \frac{dx}{(a - x)} - \int \frac{dx}{(a + x)}\]]

= \[\frac{1}{2a}\] [– log |(a – x) + log |(a + x)] + C

= 1/2a log |(a + x) / (a – x)| + C.

After looking at the integration formulas & proof we will solve an example now.

Example 1:

Find out the integral of \[\frac{(x + 3)}{\sqrt{(5 – 4x + x^{2})}}\] with respect to x.

Solution:

We say,

W x + 3 = \[A\frac{d}{dx} \sqrt{(5 - 4x + x^{2})}\] + B = A (– 4 – 2x) + B

After equating the coefficient we get,

A = – ½ and B = 1

So, \[\frac{(x + 3)}{\sqrt{(5 – 4x + x^{2})}}\]dx = – ½ \[\int (- 4 - 2x){\sqrt{(5 - 4x + x^{2})}}\] dx + \[\int \frac{dx}{\sqrt{(5 - 4x + x^{2})}}\]

= – ½ I1 + I2 … (a)

Solving I1

We can substitute,

(5 – 4x + x2) = t,

So, (– 4 – 2x) dx = dt.

Hence

I1 = \[\int \frac{(– 4 – 2x)}{\sqrt{(5 – 4x + x^{2})}}\] dx = \[\int \frac{dt}{\sqrt{t}} = 2 \sqrt{t}\] + C1

= 2 \[\sqrt{(5 – 4x + x^{2})}\] + C1 … (b)

Solving I2

I2 = \[\frac{dx}{\sqrt{(5 – 4x + x^{2})}}\] = \[\int \frac{dx}{[9 – (x + 2)^{2}]}\]

After substituting,

(x + 2) = t,

So, dx = dt.

Hence,

I2 = \[\int \frac{dt}{\sqrt{(3^{2} – t^{2})}}\] = sin–1 (t/3) + C2

= sin–1 [(x + 2) / 3] + C2 … (c)

After substituting (b) and (c) we can get,

\[\frac{(x + 3)}{\sqrt{(5 – 4x + x^{2})}}\]dx = – ½ I1 + I2

= – \[\sqrt{(5 – 4x + x^{2})}\] + sin–1 [(x + 2) / 3] + C …

Where, C = C2 = C1/2.

After concentrating on the example let us look at some of the commonly asked questions on integration formulas.

FAQ (Frequently Asked Questions)

1. What are the Applications and the Rules of the Integration?

Answer: Integration formulas proof plays the main role in the field of research and analysis, engineering (electrical, computer, etc.), physics, science, and graphical designing for application in real life. Integration is used for finding the area, curves, volumes, central point, length, and many other things.

There are rules such as power rule, sum role, difference rule, and coefficient rule for integration. The indefinite integral of the variable is raised to power in the rule of power. The coefficient tells us about the indefinite integral of coefficient c. The difference between two or more terms deals with the rule of difference. Then the sum of several terms’ functions about the integration of sum rule.

2. How can the Integration Formulas be Proved?

Answer: For integration formula proof there are Fundamental Theorems of Calculus available. In some cases, there should be L.H.S. equal to the R.H.S. for correctly proving the formula or statement. The theorem is derived continuously with the variable to get the final result similar to the argument given in the statement. The precise result should be obtained for proving the result through the theorem.

There are specific functions in integration formula proof for proving the average value, mean value, range, and many more. To prove the big integration formula, there is a specific method called Integration by Parts which is mostly used when there are more than two functions that are multiplied together.