An integral is a mathematical operation that combines the values of a function over an interval, effectively reversing the process of differentiation. Integrals are used to find areas, volumes, and accumulated quantities.

What are definite integrals of functions?

Definite integrals calculate the accumulated value of a function over a given interval [a, b]. The result is a numerical value representing the area under the curve between the limits a and b.

What techniques are used to solve integrals of particular functions?

Common techniques include: Substitution: Simplifies the integral by changing variables. Integration by Parts: Derived from the product rule for differentiation. Partial Fractions: Decomposes rational functions into simpler fractions.

Why are integrals important in real-world applications?

Integrals are used in various fields such as physics (to calculate work and energy), engineering (for signal processing and system optimization), and economics (to model growth and compute total surplus).

Integrals of Particular Functions: Proofs with Solved Examples

Q: What is the difference between a particular solution and a particular integral?

Particular Solution: In differential equations, a particular solution is a specific solution that meets the given initial or boundary conditions. Particular Integral: In the context of integration, a particular integral refers to the antiderivative of a function, often found without considering any specific initial conditions.

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Integrals of Particular Functions

Integration is a method to sum functions on a large scale. Here, we discuss integrals of some particular functions commonly used in calculations. These integrals have significant real-life applications, such as finding areas between curves, volumes, average values of functions, kinetic energy, centers of mass, and work done, among others.

1.0Integrals of Particular Functions

$\int \frac{d x}{x ^{2} - a ^{2}} = \frac{1}{2 a} lo g \frac{x - a}{x + a} + C$

$\int \frac{d x}{a ^{2} - x ^{2}} = \frac{1}{2 a} lo g \frac{a + x}{a - x} + C$

$\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} \frac{x}{a} + C$

$\int \frac{dx}{x ^{2} - a ^{2}} = lo g x + x^{2} - a^{2} + C$

$\int \frac{d x}{a ^{2} - x ^{2}} = sin^{- 1} \frac{x}{a} + C$

$\int \frac{dx}{x ^{2} - a ^{2}} = lo g x + x^{2} + a^{2} + C$

2.0Proofs of Integral Functions

We have $\frac{1}{x ^{2} - a ^{2}} = \frac{1}{( x - a ) ( x + a )}$

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

Therefore, $\int \frac{d x}{x ^{2} - a ^{2}} = \frac{1}{2 a} [\int \frac{d x}{x - a} - \int \frac{d x}{x + a}]$

$= \frac{1}{2 a} [lo g ∣ (x - a) ∣ - lo g ∣ (x + a) ∣] + C$

$= \frac{1}{2 a} lo g \frac{x - a}{x + a} + C$

In view of (1) above, we have

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

Therefore,

$\int \frac{d x}{a ^{2} - x ^{2}} = \frac{1}{2 a} [\int \frac{d x}{a - x} + \int \frac{d x}{a + x}]$

$= \frac{1}{2 a} [- lo g ∣ a - x ∣ + lo g ∣ a + x ∣] + C$

$= \frac{1}{2 a} lo g \frac{a + x}{a - x} + C$

Put x = a tan θ. Then dx = a sec² θ dθ.

Therefore,

$\int \frac{d x}{x ^{2} + a ^{2}} = \int \frac{a ^{2} s e c ^{2} θ d θ}{a ^{2} t a n ^{2} θ + a ^{2}}$

$= \frac{1}{a} \int d θ = \frac{1}{a} θ + C = \frac{1}{a} tan^{- 1} \frac{x}{a} + C$

Let x = a sec θ. Then dx = a sec θ tan θ dθ.

Therefore,

$\int \frac{d x}{x ^{2} - a ^{2}} = \int \frac{a s e c θ t a n θ d θ}{a ^{2} s e c ^{2} θ - a ^{2}}$

$= \int sec θ d θ = lo g ∣ sec θ + tan θ ∣ + C_{1}$

$= lo g \frac{x}{a} + \frac{x ^{2}}{a ^{2}} - 1 + C_{1}$

$= lo g x + x^{2} - x^{2} - lo g ∣ a ∣ + C_{1}$

$= lo g x + x^{2} - a^{2} + C, where C = C_{1} - lo g ∣ a ∣$

Let x = a sinθ. Then dx = a cos θ dθ.

Therefore,

$\int \frac{d x}{a ^{2} - x ^{2}} = \int \frac{a c o s θ d θ}{a ^{2} - a ^{2} s i n ^{2} θ}$

$= \int d θ = θ + C = sin^{- 1} \frac{x}{a} + C$

Let x = a tan θ. Then dx = a sec² θ dθ.

$\int \frac{d x}{x ^{2} + a ^{2}} = \int \frac{a s e c ^{2} θ d θ}{a ^{2} t a n ^{2} θ + a ^{2}}$

$= \int sec θ d θ = lo g ∣ (sec θ + tan θ) ∣ + C_{1}$

$= lo g \frac{x}{a} + \frac{x ^{2}}{a ^{2}} + 1 + C_{1}$

$= lo g x + x^{2} + a^{2} - lo g ∣ a ∣ + C_{1}$

$= lo g x + x^{2} + a^{2} + C, where C = C_{1} - lo g ∣ a ∣$

Applying these standard formulae, we now obtain some more formulae which are useful from applications point of view and can be applied directly to evaluate other integrals.

3.0Some more Integrals

To find the integral $\int \frac{d x}{a x ^{2} + b x + c}$ , we write

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

Now, put $x + \frac{b}{2 a} = t$ so that dx = dt and writing $\frac{c}{a} - \frac{b ^{2}}{4 a ^{2}} = \pm k^{2}$ . We find the integral reduced to the form $\frac{1}{a} \int \frac{d t}{t ^{2} \pm k ^{2}}$ depending upon the sign of $(\frac{c}{a} - \frac{b ^{2}}{4 a ^{2}})$ and hence can be evaluated.

To find the integral of the type $\int \frac{d x}{a x ^{2} + b x + c}$ , proceeding as in (1), we obtain the integral using the standard formulae.
To find the integral of the type $\int \frac{p x + q}{a x ^{2} + b x + c} d x$ , where given p, q, a, b, c are constants. Now, we have to find real numbers, A, B such that:

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

To determine A and B, we equate the coefficients of x and the constant term from both sides. By solving these equations, we obtain A and B, thus reducing the integral to a known form.

For the evaluation of the integral of the type $\int \frac{( p x + q ) d x}{a x ^{2} + b x + c}$ , we proceed as in (3) and reduce the integral into known standard forms.

4.0Difference Between a Particular Solution and a Particular Integral

Aspect	Particular Solution	Particular Integral
Definition	A specific solution to a differential equation that satisfies given initial or boundary conditions.	The antiderivative of a function, representing a general solution with an arbitrary constant C.
Context	Used in solving differential equations.	Used in integration, particularly indefinite integrals.
Form	Includes initial conditions or boundary values.	Includes an arbitrary constant of integration C.
Example (Differential Eq.)	For $\frac{d y}{d x} = 2 x$ with y(0) = 3, the particular solution is y = x² + 3.	For $\int 2 x d x$ , the particular integral is x² + C.
Example (Physical Context)	For $\frac{d ^{2} y}{d x ^{2}}$ =-g with y(0) = h₀ and y'(0) = 0, the solution is $y = h_{0} - \frac{1}{2} g t^{2}$ .	The integral $\int - g d t$ gives the velocity v = –gt + C, and integrating again gives $y = h_{0} - \frac{1}{2} g t^{2}$ .
Purpose	Finds a unique solution based on specific criteria.	Provides a general solution applicable to a range of scenarios.
Usage	Involves applying specific conditions to determine the exact function.	Involves finding the indefinite integral, which includes a constant of integration.

5.0Solved Examples of Integrals of Particular Functions

Example 1: $\int \frac{s i n x d x}{c o s ^{2} x - 36}$

Solution: put cos x = t ⇒ –sin x dx = dt

$\int \frac{- d t}{t ^{2} - ( 6 ) ^{2}} = - \frac{1}{2.6} ln \frac{t - 6}{t + 6} + C$

$= - \frac{1}{12} ln \frac{c o s x - 6}{c o s x + 6} + C$

Example 2: $\int \frac{d x}{3 x ^{2} + 6 x + 15}$

Solution: ⇒ $\frac{1}{3} \int \frac{d x}{x ^{2} + 2 x + 5}$

⇒ $\frac{1}{3} \int \frac{d x}{( x + 1 ) ^{2} + 4} = \frac{1}{3} \int \frac{d x}{( x + 1 ) ^{2} + ( 2 ) ^{2}}$

$= \frac{1}{6} tan^{- 1} (\frac{x + 1}{2}) + C$

Example 3: $\int \frac{d x}{x ^{2} - 5 x}$

Solution: $\int \frac{d x}{( x - \frac{5}{2} ) ^{2} - ( \frac{5}{2} ) ^{2}}$

$= ln [(x - \frac{5}{2}) + (x - \frac{5}{2})^{2} - (\frac{5}{2})^{2}] + C$

Example 4: $\int \frac{2 x}{x ^{4} + 2 x ^{2} + 4} d x$

Solution: put x² = t ⇒ 2x dx = t

$\int \frac{d t}{t ^{2} + 2 t + 4} = \int \frac{d t}{( t + 1 ) ^{2} + ( 3 ) ^{2}}$

$= ln (t + 1) + (t + 1)^{2} + (3)^{2} + C$

t = x²

Example 5: Evaluate $\int \frac{d x}{( x - a ) ( b - x )}$

Solution: Put x = a cos²θ + b sin²θ, the given integral becomes

$I = \int \frac{2 ( b - a ) s i n θ c o s θ d θ}{{ ( a c o s ^{2} θ + b s i n ^{2} θ - a ) ( b - a c o s ^{2} θ - b s i n ^{2} θ } ^{\frac{1}{2}}}$

$= \int \frac{2 ( b - a ) s i n θ c o s θ d θ}{( b - a ) s i n θ c o s θ} = (\frac{b - a}{b - a}) \int 2 d θ = 2 θ + C = 2 sin^{- 1} (\frac{x - a}{b - a}) + C$ Ans.

Example 6: Evaluate $\int \frac{d x}{2 x ^{2} + x - 1}$

Solution:

$I = \int \frac{d x}{2 x ^{2} + x - 1} = \frac{1}{2} \int \frac{d x}{x ^{2} + \frac{x}{2} - \frac{1}{2}} = \frac{1}{2} \int \frac{d x}{x ^{2} + \frac{x}{2} + \frac{1}{16} - \frac{1}{16} - \frac{1}{2}}$

$= \frac{1}{2} \int \frac{d x}{( x + 1/4 ) ^{2} - 9/16} = \frac{1}{2} \int \frac{d x}{( x + 1/4 ) ^{2} - ( 3/4 ) ^{2}}$

$= \frac{1}{2} \cdot \frac{1}{2 ( 3/4 )} lo g \frac{x + 1/4 - 3/4}{x + 1/4 + 3/4} + C$

${using, \int \frac{d x}{x ^{2} - a ^{2}} = \frac{1}{2 a} lo g \frac{x - a}{x + a} + C}$

$= \frac{1}{3} lo g \frac{x - 1/2}{x + 1} + C = \frac{1}{3} lo g \frac{2 x - 1}{2 ( x + 1 )} + C$

Ans.

Example 7: Evaluate $\int \frac{3 x + 2}{4 x ^{2} + 4 x + 5} d x$

Solution: Express 3x + 2 = ℓ (d.c. of 4x² + 4x + 5) + m

or, 3x + 2 = ℓ (8x + 4) + m

Comparing the coefficients, we get

8ℓ = 3 and 4ℓ + m = 2 ⇒ ℓ = 3/8 and m = 2 – 4ℓ = 1/2

⇒ $I = \frac{3}{8} \int \frac{8 x + 4}{4 x ^{2} + 4 x + 5} d x + \frac{1}{2} \int \frac{d x}{4 x ^{2} + 4 x + 5}$

$= \frac{3}{8} lo g 4 x^{2} + 4 x + 5 + \frac{1}{8} \int \frac{d x}{x ^{2} + x + \frac{5}{4}}$

$= \frac{3}{8} lo g 4 x^{2} + 4 x + 5 + \frac{1}{8} tan^{- 1} (x + \frac{1}{2}) + C$

Ans.

6.0Practice Questions of Integrals of Particular Functions

a. $\int \frac{d x}{2 x ^{2} + 9}$

b. $\int \frac{d x}{4 - x ^{2}}$

c. $\int \frac{d x}{1 + 9 x ^{2}}$

d. $\int \frac{d x}{1 - 25 x ^{2}}$

e. $\int \frac{d x}{25 + 4 x ^{2}}$

f. $\int \frac{x ^{2}}{9 + 16 x ^{6}} d x$

g. $\int \frac{d x}{x x ^{2} + 4}$

h. $\int \frac{d x}{4 x - 3 - x ^{2}}$

i. $\int \frac{c o s α d α}{a ^{2} + s i n ^{2} α}$

j. $\int \frac{x d x}{x ^{4} + 1}$

Answers:

a. $\frac{1}{3 2} arctan \frac{2}{3} x + C$

b. $arc sin \frac{x}{2} + C$

c. $\frac{1}{3} arc tan 3 x + C$

d. $\frac{1}{5} arc sin 5 x + C$

e. $\frac{1}{10} tan^{- 1} \frac{2 x}{5} + C$

f. $\frac{1}{36} tan^{- 1} (\frac{4 x ^{3}}{3}) + C$

g. $\frac{1}{2} ln [\frac{x ^{2} + 4 - 2}{x}] + C$

h. $arc sin (x - 2) + C$

i. $\frac{1}{a} arctan \frac{s i n α}{a} + C$

j. $\frac{1}{2} arctan x^{2} + C$

7.0Sample Questions on Integrals of Particular Functions

Q. What is the integral of $\frac{1}{x ^{2} - a ^{2}}$ ?

Ans: $\int \frac{1}{x ^{2} - a ^{2}} d x = \frac{1}{2 a} lo g \frac{x - a}{x + a} + C$

Q. What is the integral of $\frac{1}{x ^{2} + a ^{2}}$ ?

Ans: $\int \frac{1}{x ^{2} + a ^{2}} d x = \frac{1}{a} tan^{- 1} (\frac{x}{a}) + C$

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Integrals of Particular Functions

1.0Integrals of Particular Functions

2.0Proofs of Integral Functions

3.0Some more Integrals

4.0Difference Between a Particular Solution and a Particular Integral

5.0Solved Examples of Integrals of Particular Functions

6.0Practice Questions of Integrals of Particular Functions

7.0Sample Questions on Integrals of Particular Functions

Table of Contents

Frequently Asked Questions

What is an integral?

What is the difference between a particular solution and a particular integral?

What are definite integrals of functions?

What techniques are used to solve integrals of particular functions?

Why are integrals important in real-world applications?

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Related Articles:-

Application of Integrals

Double Integral

Conic Sections

Area Under The Curve

Binomial Theorem

Statistics

Tangent and Normal