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Proofs of Integration Formulas

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Integration Formula Proof

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. 


What is Integration? What is the Basic Formula of Integration?

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. 


How can one Differentiate Between Definite and Indefinite Integral?

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.


Check out Some Integration Formulas Below.

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

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

Proofs of Integration Formulas Below:

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.



Differentiation Formulas

Integration Formulas

1. \[\frac{d}{dx} (x)\] = 1

1. \[\int 1dx\] = x + C

2. \[\frac{d}{dx} (ax)\] = a

2. \[\int adx\] = ax + C

3. \[\frac{d}{dx} (x^{n}) = nx^{n-1}\]

3. \[\int x^{n}dx = \frac{x^{n+1}}{n+1}\] + C, n ≠ -1

4. \[\frac{d}{dx} (cosx)\] = -sinx

4. \[\int sinx dx\] = -cosx + C

5. \[\frac{d}{dx} (sinx)\] = cosx

5. \[\int cosx dx\] = sinx + C

6. \[\frac{d}{dx} (tanx) = sec^{2}x\]

6. \[\int sec^{2}xdx\] = tanx + C

7. \[\frac{d}{dx} (cotx) = -csc^{2}\]

7. \[\int csc^{2}xdx\] = - cotx + C

8. \[\frac{d}{dx} (secx)\] = secx tanx

8. \[\int secx(tanx)dx\] = secx + C

9. \[\frac{d}{dx} (cscx)\] = -cscx(cotx)

9. \[\int cscx(cotx)dx\] = - cscx + C

10. \[\frac{d}{dx} (ln x) = \frac{1}{x}\]

10. \[\int \frac{1}{x} dx\] = ln |x| + C 

11. \[\frac{d}{dx} (e^{x}) = e^{x}\]

11. \[\int e^{x} dx = e^{x}\] + C

12. \[\frac{d}{dx} (a^{x}) = (ln a)a^{x}\]

12. \[\int a^{x} dx = \frac{a^{x}}{ln a}\] + C, a > 0, a≠ 1

13. \[\frac{d}{dx} (sin^{-1}x) = \frac{1}{\sqrt{1-x^{2}}}\]

13. \[\int \frac{1}{\sqrt{1-x^{2}}} dx = sin^{-1}x\] + C

14. \[\frac{d}{dx} (tan^{-1}x) = \frac{1}{1+x^{2}}\]

14. \[\int \frac{1}{1+x^{2}} dx = tan^{-1}x\] + C

15. \[\frac{d}{dx} (sec^{-1}x) = \frac{1}{|x|\sqrt{x^{2} - 1}}\]

15. \[\int \frac{1}{|x|\sqrt{x^{2}-1}} dx = sec^{-1}x\] + 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.

FAQs on Proofs of Integration Formulas

1. What is the fundamental principle used to prove most standard integration formulas?

The fundamental principle is the inverse relationship between differentiation and integration, as stated in the Fundamental Theorem of Calculus. To prove that the integral of a function f(x) is F(x) + C, we simply need to differentiate the result F(x) and show that it equals the original function f(x). This confirms that F(x) is the correct antiderivative.

2. How is the proof for the integral of 1/(x² - a²) derived using partial fractions?

The derivation involves a clear, step-by-step process as per the Class 12 Maths syllabus for 2025-26:

  • Factorise the Denominator: First, the expression is written as 1/[(x - a)(x + a)].
  • Apply Partial Fractions: The single fraction is decomposed into the sum of two simpler fractions: A/(x - a) + B/(x + a).
  • Solve for Constants: By equating the numerators, we find the values A = 1/2a and B = -1/2a.
  • Integrate the Simpler Terms: The integral becomes ∫[1/2a * (1/(x - a) - 1/(x + a))] dx.
  • Combine the Results: Integrating these terms yields (1/2a) [log|x - a| - log|x + a|] + C, which simplifies to the final formula (1/2a) log|(x - a)/(x + a)| + C using logarithmic properties.

3. Why is it important to learn the proofs of integration formulas instead of just memorising them?

Understanding the proofs is crucial for a deeper conceptual grasp beyond rote learning. It helps students in several ways:

  • Builds Problem-Solving Skills: The techniques used in proofs, such as completing the square or partial fractions, are powerful tools for solving other complex, non-standard integrals.
  • Enhances Flexibility: If you encounter a problem that doesn't exactly match a standard formula, knowledge of the derivation method allows you to adapt your approach to find a solution.
  • Prevents Misapplication: Knowing how a formula is derived helps you understand its limitations and the conditions under which it is valid, preventing common errors.

4. What is the key difference in the derivation for ∫dx/(x² - a²) compared to ∫dx/(a² - x²)?

The core method for both is partial fractions, but the key difference lies in the initial factorisation and the resulting signs. For ∫dx/(a² - x²), the denominator is factored as (a - x)(a + x). When this is integrated, the term ∫1/(a - x) dx yields -log|a - x| due to the chain rule. This negative sign fundamentally alters the final logarithmic expression, leading to the formula (1/2a) log|(a + x)/(a - x)| + C, which is different from the result for ∫dx/(x² - a²).

5. How does understanding these proofs help in solving more advanced, non-standard integrals?

Understanding the proofs equips you with a versatile toolkit of integration strategies. When faced with a complex integral that doesn't fit a known formula, you can recognise if a similar technique can be applied. For example, seeing a quadratic in the denominator might prompt you to try completing the square to transform it into a form like x² ± a² or a² - x², a method central to many proofs. This ability to deconstruct a problem and apply a fundamental technique is a direct benefit of studying the proofs.

6. What is the role of the constant of integration 'C' in the proof of any indefinite integral?

The constant of integration, 'C', represents the fact that the antiderivative of a function is not unique but is a whole family of functions. Since the derivative of any constant is zero, d/dx [F(x)] = f(x) is the same as d/dx [F(x) + C] = f(x). Therefore, when proving an indefinite integral formula, including '+ C' is essential to represent the most general antiderivative, acknowledging all possible constant values.

7. Besides partial fractions, what are other key methods used to prove standard integration formulas in the Class 12 syllabus?

While partial fractions are vital, several other methods are equally important for deriving standard integration results. These include:

  • Integration by Substitution: Used to simplify the integrand by changing the variable. For example, proving ∫tan(x) dx by substituting u = cos(x).
  • Integration by Parts: Essential for integrals involving the product of two functions, such as proving the formulas for ∫log(x) dx or ∫x*sin(x) dx.
  • Trigonometric Identities: Used extensively to transform complex trigonometric integrals into simpler, integrable forms before applying a standard formula.