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 âŦzdx = 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 âŦf(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
âŦ f'(x) dx = f(x) + C.
Using this, the following integration formulas are derived.
The various integral calculus formulas are
d/dx {Ï(x)} = f(x) <=> âŦf(x) dx = Ï(x) + C
âŦ xn dx = + C, n â -1
âŦ(1/x) dx = loge |x| + C
âŦex dx = ex + C
âŦax dx = (ax / loge a) + C
More, integral formulas are discussed below in the article,
Note:
d/dx [âŦf(x) dx] = f(x)
âŦk . f(x) dx = k âŦf(x) dx , where k is constant
âŦ{f(x) Âą g(x)} dx = âŦf(x) dx Âą âŦg(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.
âŦ 1 dx = x + C
âŦ xn dx = x(n + 1) /(n + 1)+ C
âŦ 1/x dx = log |x| + C
âŦ ex dx = ex + C
âŦ ax dx = ax /log a+ C
âŦ 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,
âŦ cos x dx = sin x + C
âŦ sin x dx = -cos x + C
âŦ sec2 x dx = tan x + C
âŦ cosec2 x dx = -cot x + C
âŦ sec x tan x dx = sec x + C
âŦ cosec x cot x dx = -cosec x + C
âŦ tan x dx = log |sec x| + C
âŦ cot x dx = log |sin x| + C
âŦ sec x dx = log |sec x + tan x| + C
âŦ 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,
âŦ1/â(1 – x2 ) dx = sin-1 x + C
âŦ -1/â(1 – x2 ) dx = cos-1 x + C
âŦ1/(1 + x2 ) dx = tan-1 x + C
âŦ -1/(1 + x2 ) dx = cot-1 x + C
âŦ 1/xâ(x2 – 1) dx = sec-1 x + C
âŦ -1/xâ(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,
âŦ1/(x2 – a2 ) dx = 1/2a log|(x – a)(x + a| + C
âŦ 1/(a2 – x2 ) dx =1/2a log|(a + x)(a – x)| + C
âŦ1/(x2 + a2 ) dx = 1/a tan-1 x/a + C
âŦ1/â(x2 – a2 )dx = log |x +â(x2 – a2 )| + C
âŦ â(x2 – a2 ) dx = x/2 â(x2 – a2 ) -a2 /2 log |x + â(x2 – a2 )| + C
âŦ1/â(a2 – x2 ) dx = sin-1 x/a + C
âŦâ(a2 – x2 ) dx = x/2 â(a2 – x2 ) dx + a2 /2 sin-1 x/a + C
âŦ1/â(x2 + a2 ) dx = log |x + â(x2 + a2 )| + C
âŦ â(x2 + a2 ) dx = x/2 â(x2 + a2 )+ a2 /2 log |x + â(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,
âŦ f(x) g(x) dx = f(x) âŦg(x) dx – âŦ (âŦf'(x) g(x) dx) dx + C
Example: Calculate âŦ xex dx
Solution:
âŦ xex dx is of the form âŦ f(x) g(x) dx
let f(x) = x and g(x) = ex
we know that, âŦ f(x) g(x) dx = f(x) âŦg(x) dx – âŦ (âŦf'(x) g(x) dx) dx + C
âŦ xex dx = x âŦex dx – âŦ( 1 âŦex 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 = âŦ f(x) dx, where x = g(t) such that dx/dt = g'(t), then dx = g'(t)dt
Now, I = âŦ f(x) dx = âŦ f(g(t)) g'(t) dt
Example: Evaluate âŦ (4x +3)3 dx
Solution:
Let u = (4x+3) â du = 4 dx
âŦ (4x +3)3 dx
= 1/4 âŦ(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,
âŦa b f(x) dx = F(b) â 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
âŦf(x) = F(x) + C
Also, Check
Solved Examples on Integral Formulas
Example 1: Evaluate
âŦ x6 dx
âŦ1/x4 dx
âŦ3 âx dx
âŦ3x dx
âŦ4ex dx
âŦ(sin x/cos2 x) dx
âŦ(1/sin2 x) dx
âŦ[1/â(4 – x2 )] dx
âŦ[1/3â(x2 – 9)] dx
âŦ(1 /cos x tan x) dx
Solution:
(i)âŦx6 dx
= (x6+1 )/(6 + 1) + C [âŦxn dx = {xn+1 /(n+1)} + C n â -1]
= (x7 /7) + C
(ii) âŦ1/x4 dx
= âŦx-4 dx [âŦxn dx = {xn+1 /(n+1)} + C n â -1]
= (x-4+1 )/(-4 + 1) + C
= -(x-3 / 3) + C
= -(1/3x3 ) + C
(iii) âŦ3 âx dx
= âŦx1/3 dx [âŦxn dx = {xn+1 /(n+1)}+ C n â -1]
= (x (1/3)+1 /((1/3)+ 1) + C
= x4/3 / (4/3) + C
= (3/4)(x4/3 ) + C
(iv) âŦ3x dx
= (3x / loge 3) + C [ âŦax dx = (ax / loge a) + C]
(v) âŦ4ex dx
= 4âŦex dx [âŦk . f(x) dx = k f(x) dx , where k is constant]
= 4 ex + C [âŦex dx = ex + C]
(vi) âŦ(sin x/cos2 x) dx
= âŦ[(sin x/cos x) .(1/cos x)] dx
= âŦtan x . sec x dx [ âŦtan x .sec x dx = sec x + C ]
= sec x + C
(vii) âŦ(1/sin2 x) dx
= âŦcosec2 x dx [âŦcosec2 x dx = -cot x + C ]
= -cot x + C
(viii) âŦ[1/â(4 – x2 )] dx
= âŦ[1/â(22 – x2 )] dx [we know that, dx = sin-1 (x/a) + C]
= sin-1 (x/2) + C
(ix) âŦ[1/{3â(x2 – 9)}] dx
= âŦ[1/{3â(x2 – 32 )}] dx [we know that, dx = (1/a)sec-1 (x/a) + C]
= (1/3)sec-1 (x/3) + C
(x) âŦ(1 /cos x tan x) dx
= âŦ[cos x /(cos x sin x)] dx
= âŦ(1/ sin x) dx
= âŦcosec x dx [we know that, âŦcosec x dx = log |cosec x – cot x| + C]
= log |cosec x – cot x| + C
Example 2: Evaluate âŦ{e9log e x + e8log e x }/{e6log e x + e5log e x } dx
Solution:
Since, ealog e x = xa
âŦ{e9log e x + e8log e x }/{e6log e x + e5log e x } dx
= âŦ{x9 + x8 }/{x6 + x5 } dx
= âŦ[x8 (x + 1)]/[x5 (x + 1)] dx
=âŦ x8 /x5 dx
= âŦx3 dx [we know that, âŦxn dx = {xn+1 /(n+1)} + C n â -1]
= (x4 /4) + C
Example 3: Evaluate âŦ sin x + cos x dx
Solution:
âŦ(sin x + cos x) dx
= âŦsin x dx + âŦcos x dx [we know that, âŦ{f(x) Âą g(x)} dx = âŦf(x) dx Âą âŦg(x) dx]
= -cos x + sin x + C [we know that, âŦsin x dx = -cos x + C, âŦcos x dx = sin x + C ]
Example 4: Evaluate âŦ4x+2 dx
Solution:
âŦ4x+2 dx = âŦ4x . 42 dx
= âŦ16. 4x dx [ we known thatâŦk.f(x) dx = kâŦf(x) dx , where k is constant]
= 16âŦ 4x dx [âŦax dx = (ax / loge a) + C]
= 16 (4x /log 4) + C
Example 5: Evaluate âŦ(x2 + 3x + 1) dx
Solution:
âŦ(x2 + 3x + 1) dx
= âŦx2 dx+ 3âŦx dx + 1âŦ x0 dx [We know that, âŦxn dx = {xn+1 /(n+1)}+ C n â -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 âŦ[4/(1 + cos 2x)] dx
Solution:
1 + cos 2x = 2cos2 x
âŦ[4/(1 + cos 2x)] dx
= âŦ[4/(2cos2 x)] dx
= âŦ(2/cos2 x) dx
= âŦ2 sec2 xdx
= 2âŦsec2 x dx [We know that, âŦsec2 x dx = tan x + C ]
= 2 tan x + C
Example 7: Evaluate âŦ(3cos x – 4sin x + 5 sec2 x) dx
Solution:
âŦ(3cos x – 4sin x + 5 sec2 x) dx
= âŦ3cos x dx – âŦ4sin x dx + âŦ5sec2 x dx [âŦk.f(x) dx = k âŦf(x) dx, where k is constant]
= 3âŦcos x dx – 4âŦsin x dx + 5âŦsec2 x 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,
âŦ 1 dx = x + C
âŦ xn dx = x(n + 1) /(n + 1)+ C
âŦ 1/x dx = log |x| + C
âŦ ex dx = ex + C
âŦ ax dx = ax /log a+ C
âŦ 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,
âŦuvdx = uâŦvdx – âŦ[d/dx(u) Ã âŦvdx] 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. âŦf(x)dx = g(x). Integration is represented by the symbol ââŦ â
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.
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Last Updated :
09 May, 2023
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