Integrals of Particular Functions: Definition, Proofs

Integration is used to sum the functions on a large scale. Integration can be said as the process of finding anti-derivatives. It is driven by the problem of defining and calculating the area of the region bounded by the graph of the functions. Integrals are applied in many fields in real life. From calculating the area between curves to calculating the kinetic energy, or work done in physics and mechanics, integration is used. Depending on the type of the problem, various integral functions can be used to obtain the desired answer. 

Key terms: function, integral, derivative, calculus, particular functions, anti-derivative, integral calculus, solve, problems.

Also Read: Differentiation and Integration Formula

The development of integral calculus arises out of the efforts of solving the problems of the following types: 

(a) the problem of finding a function whenever its derivative is given

(b) the problem of calculating the area bounded by the graph of a function under certain conditions. 

These two problems lead to the two forms of the integrals, e.g., indefinite and definite integrals, which together constitute the Integral Calculus.

Integrals Detailed Video Explanation

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Integrals of Particular Functions

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1. \(\int \frac{dx}{x^2 - a^2}= \frac{1}{2a} log \mid \frac{x-a}{x+a} \mid +C\)
2. \(\int \frac{dx}{a^2 - x^2}= \frac{1}{2a} log \mid \frac{a+x}{a-x} \mid +C\)
3. \(\int \frac{dx}{x^2 + a^2} = \frac{1}{a} tan^{-1} \frac{x}{a}+ C\)
4. \(\int \frac{dx}{ \sqrt{x^2-a^2}} = log \mid x+ \sqrt{x^2-a^2} \mid + C\)
5. \(\int \frac{dx}{ \sqrt{a^2-x^2}} =sin^{-1} \frac{x}{a} +C\)
6. \(\int \frac{dx}{ \sqrt{x^2+a^2}} = log \mid x+ \sqrt{x^2+a^2} \mid + C\)

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Proofs of Integral Functions

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These standard formulae can be used to obtain new formulae and can be applied directly to evaluate other integrals. 

Show that: 

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

Hence proved. 

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

From the above equation (1), we have

Hence proved. 

3. \(\int \frac{dx}{x^2 + a^2} = \frac{1}{a} tan^{-1} \frac{x}{a}+ C\)

Put x = a tan \(\theta\) . Then dx = a sec²\(\theta\) d\(\theta\)

Therefore,

Hence proved. 

4. \(\int \frac{dx}{ \sqrt{x^2-a^2}} = log \mid x+ \sqrt{x^2-a^2} \mid + C\)

Let x = sec\(\theta\) . Then dx = a sec\(\theta\) tan\(\theta\) d\(\theta\)

Therefore, 

Hence proved. 

5. \(\int \frac{dx}{ \sqrt{a^2-x^2}} =sin^{-1} \frac{x}{a} +C\)

Let x = a sin\(\theta\). Then dx = a cos\(\theta\) d\(\theta\)

Therefore, 

Hence proved. 

6. \(\int \frac{dx}{ \sqrt{x^2+a^2}} = log \mid x+ \sqrt{x^2+a^2} \mid + C\)

Let x = a tan\(\theta\). Then dx = a sec² \(\theta\)d\(\theta\)

Therefore, 

Hence proved. 

Applying these standard formulae, other integrals needed for solving of the given problem can be obtained directly. 

7. To find the integral of \(\int \frac{dx}{ax^2+bx+c}\), we write 

\(ax^2+bx+c =a[x^2+ \frac{b}{a}x+ \frac{c}{a}] =a [(x+ \frac{b}{a})^2+( \frac{c}{a} -\frac{b^2}{4a^2})]\)

Now, put \(x+\frac{b}{a}\)= t so that dx=dt and writing \(\frac{c}{a} -\frac{b^2}{4a^2}\)= ± k²

We find the integral reduced to the form \(\frac{1}{a} \int \frac{dt}{t^2 \pm k^2}\) depending upon the sign of \(( \frac{c}{a} -\frac{b^2}{4a^2})\) and hence can be evaluated. 

8. To find the integral of the type \(\int \frac{dx}{\sqrt{ax^2+bx+c}}\), proceeding as in (7), 

We obtain the integral using the standard formula. 

9. To find the integral of the type \(\int \frac{px + q}{ax^2+bx+c}dx\), where p, q, a, b, c, are the constants. 

We are to find the real numbers A, B such that

px + q = A \(\frac{d}{dx}\)(ax² + bx + c) + B = A(2ax + b) + B

To determine A and B, we equate from both sides the coefficients of x and the constant terms. A and B are thus obtained and hence the integral is reduced to one of the known forms.

Read More: Conditional Probability Formula

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Things to remember 

1. Each type of integral is solved using a specific particular function. 
2. An integral problem can be solved by decuding it to a form of the particular function. 
3. Anti-derivative means the integration of a function
4. Integral of ex=ex+ C, where C is the integration constant. 
5. Complicated integral problems can be solved by the method of integration by substitution. 

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