Table of Contents
List of all Integration Formulas: Integral is considered the calculation of an integral. Integrals are generally employed in arithmetic calculations to solve a number of quantities like volume, area, displacement, etc. The integrals can be definite or indefinite. Integration is one of two major calculus branches of Mathematics apart from Differentiation. Integration involves determining how a function changes with its variables. While it’s part of the 11th and 12th-grade syllabus, it holds special importance when getting ready for the IIT-JEE exams. In this article, we provide the basics of Integration and the list of all Integration Formulas.
Integration Formulas
The integration Formula can be called the reverse step of differentiation formula. Sometimes it is called Inverse Differentiation. In calculus study, Integration is defined as the process of calculating a function with its derivative. The basic integration formulas for the different functions are explained here. Additionally, some sample questions are also mentioned here in this article. In this study, the integration by parts means integrating the product of two functions such as y = uv.
Basic Integration Formulas
The basic formulas in integration for the different functions are tabulated here. These given Integral formulas are important for solving the problems of integral calculus.
Basic Integration Formulas | |
S.no | Formula |
1. | ∫ 1 dx = x + C |
2. | ∫ a dx = ax+ C |
3 | ∫ xn dx = ((xn+1)/ (n+1))+ C ; n ≠1 |
4. | ∫ sin x dx = – cos x + C |
5. | ∫ cos x dx = sin x + C |
6. | ∫ sec²x dx = tan x + C |
7. | ∫ cosec²x dx = – cot x + C |
8. | ∫ sec x (tan x) dx = sec x + C |
9. | ∫ cosec x ( cot x) dx = – cosec x + C |
10. | ∫ (1/x) dx = ln |x| + C |
11. | ∫ ex dx = ex+ C |
12. | ∫ ax dx = (ax / ln a) + C ; a>0, a≠1 |
13. | ∫ tanx.dx =log|secx| + C |
14. | ∫ cotx.dx = log|sinx| + C |
15. | ∫ secx.dx = log|secx + tanx| + C |
16. | ∫ cosecx.dx = log|cosecx – cotx| + C |
Integration Formulas of Inverse Trigonometric Functions
The integral formulas related to the Inverse Trigonometric Functions with Limits are given below.
Integration Formula of Inverse Trigonometric Functions | |
S.no |
Formula |
1. | ∫ 1/(1 +x²).dx = -cot-1x + C |
2. | ∫ 1/x√(x² – 1).dx = sec-1x + C |
3. | ∫ 1/x√(x² – 1).dx = -cosec-1 x + C |
4. | ∫1/√(1 – x²).dx = sin-1x + C |
5. | ∫ /1(1 – x²).dx = -cos-1x + C |
6. | ∫1/(1 + x²).dx = tan-1x + C |
Difficult Integration Formulas
The Difficult Integral Formulas related to the Trigonometric Functions and Logarithms are given below.
Difficult Integration Formula | |
S.no |
Formula |
1. | ∫ √(x² + a² ).dx =1/2.x.√(x² + a² )+ a²/2 . log|x + √(x² + a² )| + C |
2. | ∫1/(x² + a²).dx = 1/a.tan-1 x/a + C |
3. | ∫1/√(x² – a²)dx = log|x +√(x² – a²)| + C |
4. | ∫ √(x² – a²).dx =1/2.x.√(x² – a²)-a²/2 log|x + √(x² – a²)| + C |
5. | ∫1/√(a² – x²).dx = sin-1 x/a + C |
6. | ∫1/(x² – a²).dx = 1/2a.log|(x – a)(x + a| + C |
7. | ∫ 1/(a² – x²).dx =1/2a.log|(a + x)(a – x)| + C |
8. | ∫1/√(x² + a² ).dx = log|x + √(x² + a²)| + C |
9. | ∫√(a² – x²).dx = 1/2.x.√(a² – x²).dx + a²/2.sin-1 x/a + C |
Application of Integration Formulas
The Integrals can be either Definite Integrals or Indefinite Integrals. The application of Integration depends on the types of integrals used in calculus.
Definite Integral Formulas
These are integration calculations with the pre-existing values of their limits. These result in a determined end value of the Integral.
b∫ag (x). dx = G(b) – G(a)
Indefinite Integral Formulas
These are integration calculations without the pre-existing values of their limits. These result in the indefinite final value of the integral. The Integration Constant ‘C’ is applied in this case.
g'(x) = g(x) + C
Integration Formulas: Questions with Solutions
Question: Find the value of ∫ 16 x³ dx.
Solution: ∫ 16 x³ dx = 16 ∫ x³ dx
= 16 x^4 / 4 + C
= 4 x^4 + C
Question: Determine the value of ∫ Cos x + x dx.
Solution: ∫Cosx + xdx = ∫Cosxdx + ∫xdx
∫Cosx + xdx = sinx + x² / 2 + C
Question: What is the value of ∫2x cos (x² – 5) ?
Solution: I = ∫2x cos (x² – 5). dx
Let x2–5 = t …..(1st)
2x. dx = dt
After putting above values, we get
I = ∫ cos (t). dt
I = sin t + C …..(2nd)
After putting the values of 1st equation in 2nd then we get
∫2x cos (x² – 5) = sin(x² – 5) + C