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Integration formulas are the basic formulas that are used to solve various integral problems. They are used to find the integration of algebraic expressions, trigonometric ratios, inverse trigonometric functions, and logarithmic and exponential functions. These integration formulas are very useful to find the integration of various functions. 

Integration is the inverse process of differentiation, i.e. if d/dx (y) = z, then ∫zdx = y. Integration of any curve gives the area under the curve. We find the integration by two methods Indefinite Integration and Definite Integration. In indefinite integration, there is no limit of the integration whereas in definite integration there is a limit under which the function is integrated.

Let us learn about these integral formulas in detail in this article.

Integral Calculus

Integral calculus is a branch of calculus that deals with the theory and applications of integrals. The process of finding integrals is called integration. Integral calculus helps in finding the anti-derivatives of a function. The anti-derivatives are also called the integrals of a function. It is denoted by ∫f(x)dx. Integral calculus deals with the total value, such as lengths, areas, and volumes. The integral can be used to find approximate solutions to certain equations of given data. Integral calculus involves two types of integration: 

  • Indefinite Integrals
  • Definite Integrals

What are Integration Formulas?

The integration formulas have been broadly presented as the following sets of formulas. The formulas include basic integration formulas, integration of trigonometric ratios, inverse trigonometric functions, the product of functions, and some advanced sets of integration formulas. Basically, integration is a way of uniting the part to find a whole. It is the inverse operation of differentiation. Thus the basic integration formula is 

∫ f'(x) dx = f(x) + C. 

Using this, the following integration formulas are derived.

The various integral calculus formulas are

  1. d/dx {φ(x)} = f(x)  <=> ∫f(x) dx = φ(x) + C
  2. ∫ xn dx = [Tex] \frac{x^{n+1}}{n+1}       [/Tex] + C, n ≠ -1
  3. ∫(1/x) dx = loge|x| + C
  4. ∫ex dx = ex + C
  5. ∫ax dx = (ax / logea) + C

More, integral formulas are discussed below in the article,

Note:

  • d/dx [∫f(x) dx] = f(x)
  • ∫k . f(x) dx = k ∫f(x) dx , where k is constant
  • ∫{f(x) ± g(x)} dx = ∫f(x) dx  ± ∫g(x) dx

Basic Integration Formulas

Some of the basic formulas of integration which are used to solve integration problems are discussed below. They are derived by the fundamental theorem of integration.

  • ∫ 1 dx = x + C
  • ∫ xn dx = x(n + 1)/(n + 1)+ C
  • ∫ 1/x dx = log |x| + C
  • ∫ ex dx = ex + C
  • ∫ ax dx = ax /log a+ C
  • ∫ ex [f(x) + f'(x)] dx = ex f(x) + C   {where, f'(x) = d/dx[f(x)]}

Classification of Integral Formulas

Integral Formulas are classified into various categories based on the following function.

  • Rational functions
  • Irrational functions
  • Hyperbolic functions
  • Inverse hyperbolic functions
  • Trigonometric functions
  • Inverse trigonometric functions
  • Exponential functions
  • Logarithmic functions

Integration Formulas of Trigonometric Functions

Integration Formulas of Trigonometric functions are used to solve the integral equations involving Trigonometric functions. A list of integral formulas involving trigonometric and inverse trigonometric functions is given below,

  • ∫ cos x dx = sin x + C
  • ∫ sin x dx = -cos x + C
  • ∫ sec2x dx = tan x + C
  • ∫ cosec2x dx = -cot x + C
  • ∫ sec x tan x dx = sec x + C
  • ∫ cosec x cot x dx = -cosec x + C
  • ∫ tan x dx = log |sec x| + C
  • ∫ cot x dx = log |sin x| + C
  • ∫ sec x dx = log |sec x + tan x| + C
  • ∫ cosec x dx = log |cosec x – cot x| + C

Integration Formulas of Inverse Trigonometric Functions

Various Integration Formulas of Inverse Trigonometric Functions which are used to solve integral questions are given below,

  • ∫1/√(1 – x2) dx = sin-1x + C
  • ∫ -1/√(1 – x2) dx = cos-1x + C
  • ∫1/(1 + x2) dx = tan-1x + C
  • ∫ -1/(1 + x2) dx = cot-1x + C
  • ∫ 1/x√(x2 – 1) dx = sec-1x + C
  • ∫ -1/x√(x2 – 1) dx = cosec-1 x + C

Advanced Integration Formulas

Some other advanced integration formulas which are of high importance for solving integrals are discussed below,

  • ∫1/(x2 – a2) dx = 1/2a log|(x – a)(x + a| + C
  • ∫ 1/(a2 – x2) dx =1/2a log|(a + x)(a – x)| + C
  • ∫1/(x2 + a2) dx = 1/a tan-1x/a + C
  • ∫1/√(x2 – a2)dx = log |x +√(x2 – a2)| + C
  • ∫ √(x2 – a2) dx = x/2 √(x2 – a2) -a2/2 log |x + √(x2 – a2)| + C
  • ∫1/√(a2 – x2) dx = sin-1 x/a + C
  • ∫√(a2 – x2) dx = x/2 √(a2 – x2) dx + a2/2 sin-1 x/a + C
  • ∫1/√(x2 + a2 ) dx = log |x + √(x2 + a2)| + C
  • ∫ √(x2 + a2 ) dx = x/2 √(x2 + a2 )+ a2/2 log |x + √(x2 + a2)| + C

Different Integration Formulas

Various types of integration methods are used to solve different types of integral questions. Each method is a standard result and can be considered a formula. Some of the important methods are discussed below in this article. Let’s check the three important integration methods.

  • Integration by Parts Formula
  • Integration by Substitution Formula
  • Integration by Partial Fractions Formula

Integration by Parts Formula

Integration by Parts Formula is applied when the given function is easily described as the product of two functions. The integration by Parts formula used in mathematics is given below,

∫ f(x) g(x) dx = f(x) ∫g(x) dx – ∫ (∫f'(x) g(x) dx) dx + C

Example: Calculate ∫ xex dx

Solution:

∫ xex dx is of the form ∫ f(x) g(x) dx

let f(x) = x and g(x) = ex

we know that, ∫ f(x) g(x) dx = f(x) ∫g(x) dx – ∫ (∫f'(x) g(x) dx) dx + C

∫ xex dx = x ∫ex dx – ∫( 1 ∫ex dx) dx+ c

              = xex – ex + c

For more detail, Check this Integration by Parts

Integration by Substitution Formula

Integration by Substitution Formula is applied when a function is a function of another function. i.e. let I = ∫ f(x) dx, where x = g(t) such that dx/dt = g'(t), then dx = g'(t)dt

Now, I = ∫ f(x) dx = ∫ f(g(t)) g'(t) dt

Example: Evaluate ∫ (4x +3)3 dx

Solution:

Let u = (4x+3) ⇒ du = 4 dx

∫ (4x +3)3 dx 

= 1/4 ∫(u)3 du

= 1/4. u4 /5 

= u4 /20

= 4x +3)4/20

For more detail, Check this Integration by Substitution Formula

Integration by Partial Fractions Formula

Integration by Partial Fractions Formula is used when the integral of P(x)/Q(x) is required and P(x)/Q(x) is an improper fraction, such that the degree of P(x) is less than the (<) the  degree of Q(x), then the fraction P(x)/Q(x) is written as

P(x)/Q(x) = R(x) + P1(x)/ Q(x)

where 
R(x) is a polynomial in x and 
P1(x)/ Q(x) is a proper rational function. 

Now the integration of R(x) + P1(x)/ Q(x) is easily calculated using the formulas discussed above.

For more detail, Check this Integration by Partial Fractions

Application of Integral Formulas

Integral formulas are highly useful formulas in mathematics that are used for a variety of tasks. They are used for 

  • Finding the length of the curve
  • Finding the area under the curve
  • Finding approximate values of the function
  • Determining the path of an object and others
  • To find the area under the curve
  • To find the surface area and volume of irregular shapes
  • To find the centre of mass or centre of gravity

These formulas are basically categorized into two categories,

  • Definite Integration Formulas
  • Indefinite Integration Formulas

Definite Integration Formulas

Definite integral formulas are used when the limit of the integration is given. In definite integration, the solution to the question is a constant value. Generally, the definite integration is solved as,

ab f(x) dx = F(b) – F(a)

Learn more about Definite Integral Properties

Indefinite Integration Formulas

Indefinite Integration Formulas are used to solve the indefinite integration when the limit of integration is not given. In indefinite integration, we use the constant of the integration which is generally denoted by C

∫f(x) = F(x) + C

Also, Check

Solved Examples on Integral Formulas

Example 1: Evaluate

  • ∫ x6 dx  
  • ∫1/x4 dx  
  • 3√x dx  
  • ∫3x dx  
  • ∫4ex dx  
  • ∫(sin x/cos2x) dx  
  • ∫(1/sin2x) dx 
  • ∫[1/√(4 – x2)] dx  
  • ∫[1/3√(x2 – 9)] dx  
  • ∫(1 /cos x tan x) dx 

Solution:

 (i)∫x6 dx 

= (x6+1)/(6 + 1) + C       [∫xn dx = {xn+1/(n+1)} + C    n ≠ -1]

= (x7/7) + C 

(ii) ∫1/x4 dx 

= ∫x-4 dx                       [∫xn dx = {xn+1/(n+1)} + C    n ≠ -1]

= (x-4+1)/(-4 + 1) + C 

= -(x-3/ 3) + C

= -(1/3x3) + C

(iii) 3√x dx 

= ∫x1/3 dx                     [∫xn dx = {xn+1/(n+1)}+ C    n ≠ -1]

= (x (1/3)+1/((1/3)+ 1) + C 

= x4/3 / (4/3) + C 

=  (3/4)(x4/3) + C

(iv) ∫3x dx 

= (3x / loge 3) + C        [ ∫ax dx = (ax / logea) + C]                                            

(v) ∫4ex dx  

= 4∫ex dx                     [∫k . f(x) dx = k f(x) dx , where k is constant]

= 4 ex + C                   [∫ex dx = ex + C]

(vi) ∫(sin x/cos2x) dx 

= ∫[(sin x/cos x) .(1/cos x)] dx

= ∫tan x . sec x dx        [ ∫tan x .sec x dx = sec x + C ]       

= sec x + C

(vii) ∫(1/sin2x) dx 

= ∫cosec2x dx               [∫cosec2x dx  = -cot x + C ]

= -cot x  + C

(viii) ∫[1/√(4 – x2)] dx 

= ∫[1/√(22 – x2)] dx      [we know that, dx = sin-1(x/a) + C]

= sin-1(x/2) + C

(ix) ∫[1/{3√(x – 9)}] dx 

= ∫[1/{3√(x2 – 32)}] dx   [we know that, [Tex]\int\frac{1}{x\sqrt{x^2-a^2}}       [/Tex]dx = (1/a)sec-1(x/a) + C]

= (1/3)sec-1(x/3) + C

(x) ∫(1 /cos x tan x) dx 

= ∫[cos x /(cos x sin x)] dx

= ∫(1/ sin x) dx

= ∫cosec x dx                [we know that, ∫cosec x dx =  log |cosec x – cot x| + C]

= log |cosec x – cot x| + C

Example 2: Evaluate ∫{e9logex + e8logex}/{e6logex + e5logex} dx

Solution:

 Since, ealogex = xa

∫{e9logex + e8logex}/{e6logex + e5logex} dx 

= ∫{x9 + x8}/{x6 + x5} dx

= ∫[x8(x + 1)]/[x5(x + 1)] dx

=∫ x8/x5 dx

= ∫x3 dx                    [we know that, ∫xn dx = {xn+1/(n+1)} + C    n ≠ -1]

= (x4/4) + C 

Example 3: Evaluate ∫ sin x + cos x  dx 

Solution:

  ∫(sin x + cos x) dx 

= ∫sin x dx + ∫cos x dx             [we know that, ∫{f(x) ± g(x)} dx = ∫f(x) dx ± ∫g(x) dx]         

= -cos x + sin x + C                  [we know that, ∫sin x dx = -cos x + C, ∫cos x dx = sin x + C ]

Example 4: Evaluate ∫4x+2 dx

Solution:

 ∫4x+2 dx = ∫4x. 42 dx 

= ∫16. 4x dx              [ we known that∫k.f(x) dx = k∫f(x) dx , where k is constant]

= 16∫ 4x dx               [∫ax dx = (ax / logea) + C]

= 16 (4x/log 4) + C

Example 5: Evaluate ∫(x2 + 3x + 1) dx

Solution:

 ∫(x2 + 3x + 1) dx 

= ∫x2 dx+ 3∫x dx + 1∫ x0dx         [We know that, ∫xn dx = {xn+1/(n+1)}+ C  n ≠ -1]

= [x2+1/2+1] + 3[[x1+1/1+1]] + [x0+1/0+1] + C

= [x3/3] + 3[x2/2] + x + C

Example 6: Evaluate ∫[4/(1 + cos 2x)] dx 

Solution:

1 + cos 2x = 2cos2

∫[4/(1 + cos 2x)] dx 

= ∫[4/(2cos2x)] dx

= ∫(2/cos2x) dx

= ∫2 sec2xdx

= 2∫sec2x dx     [We know that, ∫sec2x dx = tan x + C ]

= 2 tan x + C 

Example 7: Evaluate ∫(3cos x – 4sin x + 5 sec2x) dx

Solution:

∫(3cos x – 4sin x + 5 sec2x) dx 

= ∫3cos x dx – ∫4sin x dx + ∫5sec2x dx   [∫k.f(x) dx = k ∫f(x) dx, where k is constant]

=  3∫cos x dx – 4∫sin x dx + 5∫sec2x dx

= 3sin x – 4(-cos x) + 5 tan x + C

=  3sin x + 4cos x + 5 tan x + C 

FAQs on Integration Formulas

Q1: What are all Integration Formulas?

Answer:

Integration formulas are the formulas which are used to solve various integration problems,

  • ∫ 1 dx = x + C
  • ∫ xn dx = x(n + 1)/(n + 1)+ C
  • ∫ 1/x dx = log |x| + C
  • ∫ ex dx = ex + C
  • ∫ ax dx = ax /log a+ C
  • ∫ ex [f(x) + f'(x)] dx = ex f(x) + C   {where, f'(x) = d/dx[f(x)]}

Q2: What is the integration formulas of uv?

Answer:

The integration formula of uv is,

∫uvdx = u∫vdx – ∫[d/dx(u) × ∫vdx] dx

Q3: What does integration in mathematics mean?

Answer:

If the derivative of the function g(x) is f(x) then the integration of f(x) is g(x) i.e. ∫f(x)dx = g(x). Integration is represented by the symbol “

Q4: How do we Integrate using Integration Formulas?

Answer:

Integration can be achieved using the formulas,

  • Define a small part of an object in certain dimensions which by adding infinitely times makes the complete object.
  • Using integration formulas over that small part along the varying dimensions get us the complete object.

Q5: What is the Integral Formula by Part?

Answer:

Integral formula by part is used to solve the integral where improper fraction is given.

Q6: What is the Use of Integration Formulas?

Answer:

Integration formulas are used to solve various integral problems. Various problems which we encounter in our daily life can be easily solved with the help of integration, such as finding center of mass of any object, finding the trajectory of missile, rockets, planes and others.



Last Updated : 18 Feb, 2024
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