Integration Formulas

Integration Formulas are the basic formulas 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 beneficial for finding the integration of various functions. 

The various integral calculus formulas are

• d/dx {φ(x)} = f(x) ⇔ ∫f(x) dx = φ(x) + C
• ∫ xⁿ dx = \frac{x^{n+1}}{n+1} + C, n ≠ -1
• ∫(1/x) dx = log_e|x| + C
• ∫e^x dx = e^x + C
• ∫a^x dx = (a^x / log_ea) + C

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

Check- Integration in Maths

Basic Integration Formulas

Some of the basic formulas of integration, which are used to solve integration problems are discussed below. They are derived from the fundamental theorem of integration. The list of basic integral formulas is given below:

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

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 that are of high importance for solving integrals are discussed below:

• ∫1/(x² - a²) dx = 1/2a ln|(x - a)(x + a| + C
• ∫ 1/(a² - x²) dx =1/2a ln|(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
• Integration by Substitution
• Integration by Partial Fractions

Integration by Parts Formula

The 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 ∫ xe^x dx

Solution:

∫ xe^x dx is of the form ∫ f(x) g(x) dx

let f(x) = x and g(x) = e^x

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

∫ xe^x dx = x ∫e^x dx - ∫( 1 ∫e^x dx) dx+ c

= xe^x - e^x + c

Integration by Substitution Formula

The 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 /4 
= u⁴ /16
= (4x +3)⁴/16

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
• 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.

Application of Integrals

Integral formulas are highly useful formulas in mathematics that are used for a variety of tasks. Various applications of integrals include:

• 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 center of mass or center of gravity

Articles Related to Integration Formulas

• Calculus
• Calculus Formulas
• Integral Calculus
• Definite Integrals
• Indefinite Integrals
• Definite Integral Properties

Solved Examples of Integral Formulas

Example 1: Evaluate

• ∫ x⁶ dx  
• ∫1/x⁴ dx  
• ∫³√x dx  
• ∫3^x dx  
• ∫4e^x 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 

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

= (x ^(1/3)+1/((1/3)+ 1) + C 

= x^4/3 / (4/3) + C 

=  (3/4)(x^4/3) + C

(iv) ∫3^x dx 

= (3^x / log_e 3) + C        [ ∫a^x dx = (a^x / log_ea) + C]                                            

(v) ∫4e^x dx  

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

= 4 e^x + C                   [∫e^x dx = e^x + 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, ∫[1/√(a² - x²)] dx = sin^-1(x/a) + C]

= sin^-1(x/2) + C

(ix) ∫[1/{3√(x²  - 9)}] dx 

= ∫[1/{3√(x² - 3²)}] dx   [we know that, \int\frac{1}{x\sqrt{x^2-a^2}}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 ∫{e^9log_e^x + e^8log_e^x}/{e^6log_e^x + e^5log_e^x} dx

Solution:

 Since, e^alog_e^x = x^a

∫{e^9log_e^x + e^8log_e^x}/{e^6log_e^x + e^5log_e^x} 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 ∫4^x+2 dx

Solution:

 ∫4^x+2 dx = ∫4^x. 4² dx 
= ∫16. 4^x dx              [ we known that ∫k.f(x) dx = k∫f(x) dx , where k is constant]
= 16∫ 4^x dx               [∫a^x dx = (a^x / log_ea) + C]
= 16 (4^x/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 = 2cos²x 

∫[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 

Practice Problems on Integration Formulas

Question 1: ∫ x² dx.

Question 2: ∫ e^x dx.

Question 3: ∫ 1/x dx.

Question 4: ∫ sin (x) dx.

Question 5: ∫ (2x³ + 3x² + x + 1)dx.

Answer Sheet

1. x³/3 + C
2. e^x + C
3. ln |x| + C
4. -cos (x) + C
5. \frac{x(x+2)(x^2+1)}{2}