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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Physics » Methods of Integration

Methods of Integration

Introduction, the definition of integration, rules of integration, integration methods, an example of integration methods, conclusion, and the FAQs

Table of Content
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We all know that when the values are very small, they are difficult to solve. At such times integration comes into the picture. It is the method to unit the small parts or values to find out the value of the whole. In physics, it is normally used to find out the areas of the bounded regions. This is done by finding the function with the given differential. So, let us learn more about integration in this article and get an overview of integration methods and some of the integration rules.

Definition of integration

Finding functions whose derivative is given is known as anti-differentiation or integration. When we cannot perform general addition operations, we use integration to feature values on an outsized scale.

Rules for Integration:

  1. A constant is always included in the expression for the indefinite integral, i.e.,

 If g′(x)=f(x), then∫f(x)dx=g(x)+Cifg′(x)=f(x), then∫f(x)dx=g(x)+C, since, the derivative of a constant is 0.

  1. The integral of a derivative gives back the same function itself (with a constant):

∫f′(x)dx=f(x)+C∫f′(x)dx=f(x)+C

If we do the derivation of an equation we get the same function:

ddx(∫f(x)dx)=f(x)

  1. 3. ∫{f(x)±g(x)}dx=∫f(x)dx+∫g(x)dx∫{f(x)±g(x)}dx=∫f(x)dx+∫g(x)dx
  2. 4. ∫kf(x)dx=k∫f(x)dx∫kf(x)dx=k∫f(x)dx
  3.  ∫f(x)dx=g(x)+C, ∫f(x)dx=g(x)+C, then

∫f(ax+b)dx= 1/a*g(ax+b)+C

Methods of Integration

Substitution Method

Integration can be solved by introducing a new independent variable when it is difficult to seek out the integration of a function. By changing the independent variable x to t, in a given form of integral function, say (∫f(x)). (∫f(x)), we can transform the integral.

Let’s substitute the value of independent x = g(t) in the integral function ∫f(x), 

We get, dx / dt = g’(t)

Or, dx = g’(t) • dt

Thus, from the above substitution, we get,

I=∫f(x).dx=f(g(t).g′(t)).dt

I=∫f(x).dx=f(g(t).g′(t)).dt

Parts Integration method

This Integration rule is employed to seek out the integral of two functions. 

By product rule of derivatives, we have

pmi1

So, Equation (2) becomes

∫f(x)g(x)dx=f(x)∫g(x)dx−∫[f′(x)∫g(x)dx]dx

Using Trigonometric Identities –

To simplify any integral function that consists of trigonometric functions, trigonometric identities are used. 

pmi2

Particular Functions Integration

Integration of various particular functions have some important formulae of integration that will be applied to form other integration into the quality sort of integrand. Integration of those standard integrands are often easily found employing a direct sort of integration methods. Here are the 6 important formulas listed below – 

  • ∫ dx/ (x²– a²) = ½  a log | (x – a) / (x + a) | + c
  • ∫ dx /√ (a² – x²) = sin–¹ (xa) + c
  • ∫ dx /√ (x² + a²) = log | x + √(x² + a²) | + c
  • ∫ dx/ (a² – x²) = ½ a log | (a + x) / (a – x) | + c
  • ∫ dx /√ (x² – a²) = log| x+√(x² – a²) | + c
  • ∫ dx / (x² + a²) = 1/a tan–1 (x/a) + c

Where, c = constant

Using Partial Factors

pm3

Let us find out the value of A and B

We get, 1=A(x+2)+B(x+1), when we compare the equation.

From this, we get a group of two linear equations.

A+B=0 and 2A+B =1

If we solve the equation we get, A=1 and B=-1.

pmi4

Integration methods examples

Calculate ∫ cos2 x dx

Integrate the function f(x)=2x sin(x2+1) with respect to x.

Solution:

Let x2+1=z

Then, 2x dx = dz

∫f(x)dx=∫2xsin(x2+1)dx

=∫sinz dz=−cosz+C

=−cos(x2+1)+C

∴∫2xsin(x2+1)dx = −cos(x2+1) + C

Conclusion

Thus, to conclude, there are 5 integration methods-

The substitution method is when integration can be solved by introducing a new independent variable when it is difficult to seek out the integration of a function. Parts integration method is employed to seek out the integral of two functions. Using trigonometric identities to simplify an integral function that consists of trigonometric functions. 

Partial Fraction, Particular Integration functions have some important formulae of integration which will be applied to form other integration into the integrand in standard form.

By using integration by substitution.

faq

Frequently asked questions

Get answers to the most common queries related to the IIT JEE Examination Preparation.

What are the rules of integration?

Ans. A constant is always included in the expression for the indefinite integral, i.e...Read full

What are the uses of integration in real life?

Ans. In real life, integration is used in engineering, where engineers find the shape of the building by doi...Read full

Give details about the fundamental theorem of calculus?

Ans. The fundamental theorem of calculus gives the concept of linking differentiating functions and integrat...Read full

What are the formulae of integration?

Ans. The formulae of integration are as follows. ...Read full

What are the five methods of solving integration problems?

Ans. The integration is solved by using five methods which are as follows: ...Read full

Ans.

  • A constant is always included in the expression for the indefinite integral, i.e.,

 If g′(x)=f(x), then∫f(x)dx=g(x)+Cifg′(x)=f(x), then∫f(x)dx=g(x)+C, since, the derivative of a constant is 0.

  • The integral of a derivative gives back the same function itself (with a constant):

∫f′(x)dx=f(x)+C∫f′(x)dx=f(x)+C

If we do the derivation of an equation we get the same function:

ddx(∫f(x)dx)=f(x)

  • c. ∫{f(x)±g(x)}dx=∫f(x)dx+∫g(x)dx∫{f(x)±g(x)}dx=∫f(x)dx+∫g(x)dx
  • d. ∫kf(x)dx=k∫f(x)dx∫kf(x)dx=k∫f(x)dx
  •  ∫f(x)dx=g(x)+C,∫f(x)dx=g(x)+C, then

∫f(ax+b)dx=1/a*g(ax+b)+C

Ans. In real life, integration is used in engineering, where engineers find the shape of the building by doing integration. In physics, it is used to find the centre of gravity, etc. The three-dimensional models demonstrated in graphical representation are used to find speed, acceleration, and velocity. It is used in economics to determine the total cost function and revenue cost function from the given marginal cost.

Ans. The fundamental theorem of calculus gives the concept of linking differentiating functions and integrating functions. There are two main fundamental theorems of calculus.

Ans. The formulae of integration are as follows.

Ans. The integration is solved by using five methods which are as follows:

  • Substitution Method
  • Parts Integration Method 
  • Using Trigonometric Identities 
  • Partial Fraction
  • Particular Functions Integration

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