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Introduction

Let us consider Integration - Calculus | Mathematics for NDA We observe that the cos x is the derivative function of sin x or we can say that sin x is an anti-derivative (or an integral) of cos x.

Some Important Integrals
Integration - Calculus | Mathematics for NDA
Integration - Calculus | Mathematics for NDA


Some Properties of Indefinite Integrals

  1. ∫ [f(x) + g(x)] dx = ∫f(x)dx + ∫g(x)dx
  2. ∫kf(x)dx = k∫f(x)dx where k is constant
  3. ∫f'(ax + b)dx =Integration - Calculus | Mathematics for NDA
  4. ∫sinh x dx = cosh x + c
  5. ∫cosh x dx = sinh x + c
  6. ∫tanh x dx = log |cosh x| + c
  7. ∫coth x dx = log |sinh x| + c
  8. ∫sech x dx = tan-1 |sinh x | + c
  9. ∫cosec2h x dx = –coth x + c
  10. ∫cosech x  coth x = –cosech x + c
  11. ∫sech x  tanh x dx = –sech x + c

Methods of Integration 

  • Integration by substitution
  • Integration using Trigonometric Identities
  • Integration using partial fractions
  • Integration by parts

1. Integration by Substitution
The given integral ∫f(x)dx can be transformed into another form by changing the independent variable x to t by substituting x = g(t)   Consider, I = ∫f(x)dx
Put, x = g(t) so that dx / dt = g'(t)
dx = g'(t)dt  
I = ∫f(x)dx = ∫f(g(t)) g'(t)dt

Example 1:
Integrate the following function, Integration - Calculus | Mathematics for NDA
Solution:
Derivative of √x is = 1 / 2√x
We use the substitution √x  = t
So, that Integration - Calculus | Mathematics for NDA= dt ⇒ dx = 2t dt
Thus, Integration - Calculus | Mathematics for NDA
= 2∫tan4 t sec2 t dt
Again we make another substitution tan t = u
sec2 t dt = dx
= 2∫ x4 dx = 2u5/ 5 + c
= Integration - Calculus | Mathematics for NDA
[since u = tan t]
= Integration - Calculus | Mathematics for NDA

Note:

Integration - Calculus | Mathematics for NDA

2. Integration using Trigonometric Identities
The following example can be solved by using trigonometric Identities  

Example 2:
∫cos2x dx
Solution:
We know,  cos 2x = 2 cos2 x – 1
Integration - Calculus | Mathematics for NDA
= Integration - Calculus | Mathematics for NDA
= Integration - Calculus | Mathematics for NDA

Integrals of Some Particular Functions

  1.  Integration - Calculus | Mathematics for NDA
  2.  Integration - Calculus | Mathematics for NDA
  3. Integration - Calculus | Mathematics for NDA
  4. Integration - Calculus | Mathematics for NDA
  5.  Integration - Calculus | Mathematics for NDA

To find the integral of type Integration - Calculus | Mathematics for NDA, where p,q,a,b,c are constants we need to find real numbers A, B such that
Integration - Calculus | Mathematics for NDA
To determine A and B we equate from both sides the coefficients of x and the constant terms. 

3. Integration by Partial Fractions
The following example can be solved by using partial fraction method  

Example 3:
Find Integration - Calculus | Mathematics for NDA
Solution:  
Integration - Calculus | Mathematics for NDA
Integration - Calculus | Mathematics for NDA
=Integration - Calculus | Mathematics for NDA

4. Integration by Parts
This can be done by using the given formula
f(x)g(x)dx = f(x)∫g(x)dx – ∫[∫g(x)dx] f'(x)dx
In integration by parts method proper choice of first and second function:
The first function is the word ILATE
I → Inverse Trigonometric
L → Logarithmic
A → Algebraic
T → Trigonometric
E → Exponential

Example 4:
Find the value of  ∫x cos x dx
Solution:
Put f(x) = x (first function) and g(x) = cos x (second function) Integration - Calculus | Mathematics for NDA
= x sin x – ∫sin x dx
= x sin x + cos x + c

Integrals of Some Particular Functions

  1. Integration - Calculus | Mathematics for NDA
  2. Integration - Calculus | Mathematics for NDA
  3. Integration - Calculus | Mathematics for NDA
  4. Integration - Calculus | Mathematics for NDA
  5. Integration - Calculus | Mathematics for NDA
  6. Integration - Calculus | Mathematics for NDA
  7. Integration - Calculus | Mathematics for NDA
  8. Integration - Calculus | Mathematics for NDA
  9. Integration - Calculus | Mathematics for NDA
  10. Integration - Calculus | Mathematics for NDA
  11. Integration - Calculus | Mathematics for NDA

Definite Integrals

A real valued function f(x) is continuous on [a, b] thenIntegration - Calculus | Mathematics for NDA is called definite integral. Geometrically it gives area of finite regions.  

Properties of Definite Integrals

  1. Integration - Calculus | Mathematics for NDA
  2. Integration - Calculus | Mathematics for NDA
  3. Integration - Calculus | Mathematics for NDA
  4. Integration - Calculus | Mathematics for NDA
  5. Integration - Calculus | Mathematics for NDA
  6. Integration - Calculus | Mathematics for NDA
  7. Integration - Calculus | Mathematics for NDA
    = Integration - Calculus | Mathematics for NDA where k = π/2 ; if m and n are even
    = 1 ; otherwise

Example 5:
Find the value of Integration - Calculus | Mathematics for NDA
Solution:
Let, I = Integration - Calculus | Mathematics for NDA
Also, I = Integration - Calculus | Mathematics for NDA= Integration - Calculus | Mathematics for NDA= Integration - Calculus | Mathematics for NDA
Integration - Calculus | Mathematics for NDA

Solving Indefinite Integrals Using Functions 
There are two functions which can be used to solve improper indefinite integrals. They are given as

A. Gamma Function
Gamma Function is defined by: Integration - Calculus | Mathematics for NDA

Properties

  1. Γ(1) = 1 
  2. Γ(2) = 1     
  3. Γ(3) = 2
  4. Γ(n+1) = nΓ(n) = n!
  5. Integration - Calculus | Mathematics for NDA
  6. Integration - Calculus | Mathematics for NDA
  7. Integration - Calculus | Mathematics for NDA

B. β – Function (Beta Function)
It is defined as β(m, n) = Integration - Calculus | Mathematics for NDA

Properties

  1. β(m, n) = β(n, m) 
  2. β(m, n) =Integration - Calculus | Mathematics for NDA
  3. β(m, n) =Integration - Calculus | Mathematics for NDA
  4. Integration - Calculus | Mathematics for NDA

Example 6:
Find the value of Integration - Calculus | Mathematics for NDA
Solution:
Integration - Calculus | Mathematics for NDA
2m - 1 = 1 / 2
⇒ m = 3 / 4
⇒ 2n - 1 = -1 / 2
⇒ n = 1 / 4
Integration - Calculus | Mathematics for NDA

The document Integration - Calculus | Mathematics for NDA is a part of the NDA Course Mathematics for NDA.
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FAQs on Integration - Calculus - Mathematics for NDA

1. What is integration in calculus?
Ans. Integration in calculus is a mathematical operation that calculates the area under a curve. It involves finding the antiderivative of a function, which represents the original function's rate of change. Integration is used to solve a variety of problems, such as finding the area between two curves, calculating the total accumulated value over time, or determining the average value of a function.
2. How is integration different from differentiation in calculus?
Ans. Integration and differentiation are two fundamental operations in calculus, but they have different purposes. Differentiation focuses on finding the rate of change of a function, while integration deals with finding the accumulation of a function over a given interval. In simpler terms, differentiation helps determine how a function is changing, while integration helps find the total value or area under the function.
3. What are the different methods of integration in calculus?
Ans. There are various methods of integration in calculus, including: 1. Integration by substitution: This method involves substituting a new variable to simplify the integrand and then applying the chain rule to find the antiderivative. 2. Integration by parts: With this method, you split the integrand into two parts and apply a specific formula to integrate the product of the two parts. 3. Partial fraction decomposition: This method is used to break down a rational function into simpler fractions, making it easier to integrate. 4. Trigonometric substitution: This technique is employed when the integrand involves trigonometric functions. By substituting trigonometric identities, the integral can be simplified. 5. Numerical integration: If an integral cannot be evaluated analytically, numerical methods like the trapezoidal rule or Simpson's rule can be used to approximate the value.
4. What are the applications of integration in calculus?
Ans. Integration has numerous applications in various fields, including: 1. Physics: Integration is used to calculate quantities such as displacement, velocity, and acceleration from motion equations. 2. Economics: Integration helps determine total revenue, total cost, and profit functions in economics. 3. Engineering: Integration is used to analyze electrical circuits, calculate fluid flow rates, and determine the moment of inertia for mechanical systems. 4. Probability and statistics: Integration is utilized to calculate probabilities, expected values, and cumulative distribution functions in probability theory. 5. Geometry: Integration is used to find the area, volume, and centroid of irregular shapes and solids.
5. How can integration be used to find the area between two curves?
Ans. To find the area between two curves using integration, follow these steps: 1. Identify the two curves and determine their points of intersection. 2. Set up the integral by subtracting the equation of the lower curve from the equation of the upper curve. 3. Determine the limits of integration by finding the x-values where the curves intersect. 4. Integrate the difference of the two curves with respect to x over the given interval. 5. Evaluate the integral and calculate the area by taking the absolute value of the result. By following these steps, integration allows us to accurately find the area between two curves.
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