Integration Formula Proof

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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. 

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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,

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

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= \[\frac{1}{2a}\] [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 + \[\int \frac{dx}{\sqrt{(5 – 4x + x^{2})}}\]

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= – ½ I1 + I2 … (a)

Solving I1

We can substitute,

(5 – 4x + x2) = t, 

So, (– 4 – 2x) dx = dt. 

HenceI1 = \[\int \frac{(– 4 – 2x)}{\sqrt{(5 – 4x + x^{2})}}\] 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.

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