Integration or Integral Calculus is often lightly passed upon the students as the reverse operation of Differential Calculus, which is true but doesn’t quite capture the essence of it. While this definitely makes understanding Integral calculus easier, Integration, as a tool of mathematics, came before differentiation. Integral calculus dates back to 5th century BC when the area of a shape was obtained by imagining small polygons forming the shape which was called the method of exhaustion. It is widely debated who proposed the modern theory of integration. While most believe it was Leibniz who introduced a proper theory of integration, many believe Isaac Newton was behind the advancements in this very important tool of mathematics. Integration as a tool was needed mainly to calculate area under a curve.

## Integral Calculus Symbol

The symbol ∫ represents integration. The symbol dx, called the differential of the variable x, indicates a very small instant which is the step size of the variable of integration, x. ∫ is taken from a letter which stands for summa or sum or total. This means an integral is a sum of area of infinitesimally small rectangles under the curve, length- f(x) & width dx, being considered for integration over the variable x.

Fourier was the first person known to use limits on the top & bottom of the integral symbol, to mark the start & the end point of integration. The value on the bottom representing the start & that on the top represents the end point. So, the integral symbol with a & b as the marked limits basically represents area under the curve f(x) between these two values of the x. This form of integral is known as Definite Integral & is the more applied form of Integration.

## Definite Integral Calculus

Definite integral represents the real world application of integration, since it is mostly finite area which we are looking to calculate. This is also one of the most asked topics in JEE. One needs an eye to identify the category to which a problem of definite integral belongs. There is one and only one way to develop oneself in any form of integration i.e. practice as many problems as possible. Important formulae of definite integral are same as those of the indefinite integral. It is the properties of definite integrals which play a key role in solving problems efficiently. Here are few of the most important properties of definite Integrals.

There are functions which can be integrated over definite limits, using the properties listed above, but their indefinite integrations do not exist.

## Indefinite Integral Calculus

When there are no limits marked on the integral symbol, the result of integration is known as indefinite integral & is a generalised result of the area under the curve f(x) for the variable x, wherever it is continuous. If limits are applied to the result of indefinite integration, we obtain its definite value. Indefinite Integral is known to have a wide range of applications as well. Most of the theory in Physics & Chemistry is a result of Indefinite Integration. Let’s have a look at some of the important formulae in indefinite integration.

## Integration Formulas

The integration of a function f(x) is given by F(x) and is represented as

∫f(x) dx = F (x) + C

Where, the Right Hand Side of the equation means integral of f(x) with respect to x.

F(x) is called anti-derivative or primitive.

f(x) is called the integrand.

dx is called the integrating agent.

C is an arbitrary constant known as the constant of integration.

x is the variable of integration.

We are aware of the Anti-derivatives of the basic functions. The integrals of these functions can be obtained by us.

Let’s discuss a few important formulae and their applications in determining the integral value of other functions.

1.

Proof: We can express the integrand as:

Multiplying the numerator and the denominator by 2a and simplifying the obtained expression we get:

Hence, upon integrating the obtained expression with respect to x, we get

According to the properties of integration, the integral of sum of two functions is equal to the sum of integrals of the given functions, i.e.

Therefore, equation 1 can be represented as:

Integrating with respect to x we get:

Proof: The integrand can be expressed as:

Multiplying the numerator and the denominator by 2a and simplifying the obtained expression we get:

Hence, upon integrating the obtained expression with respect to x we get:

According to the properties of integration, the integral of sum of two functions is equal to the sum integrals of the given functions, i.e.

Therefore, the equation 2 can be rewritten as

Integrating with respect to x, we get:

Proof: Let x = a tan Ɵ. Differentiating both sides of this equation with respect to x we have;

dx = a sec2Ɵ dƟ

Hence, using this, the integral can be expressed as:

Integrating with respect to x we get:

4.

Proof: Let x = a tan Ɵ. Differentiating both sides of this equation with respect to x we have;

dx = a secƟ dƟ

Therefore, using this, the integral can be expressed as:

Using the trigonometric identity sec2Ɵ = 1 + tan2Ɵ, the above equation can be written as:

Integrating with respect to x, we have:

Substituting the value of Ɵ in the above equation we get;

Here, C = C1 – log |a|

6.

Proof: Let x = a sin Ɵ. Differentiating both sides of this equation with respect to x we have;

dx = a cosƟ dƟ

Hence, using this, the integral can be expressed as:

Using the trigonometric identity 1 – sin2Ɵ =cos2Ɵ, the above equation can be written as:

Integrating with respect to x, we have:

Substituting the value of Ɵ in the above equation we get;

7. Integration by parts:

Integration by parts is the integration technique most commonly used when you have to integrate the product of two functions.

The generalised formula for integration by parts can be given as:

Let’s understand with an example. There are many a instances when a given function is a product of two simpler functions.

∫x sinx dx  is a product of two different simpler functions x. To integrate such product of functions, the integration by parts rule is used. In this example, v= x and du/dx =sin x.

Derivation of formula:

This final formula is known as integration by parts.

Now, let us solve the above example using this formula:

I=x∫sin⁡xdx−∫(ddx(x)∫sin⁡xdx)dx

I=−xcosx−∫(1⋅∫sinxdx)dx

⟹I=−xcos⁡x+∫cos⁡xdx

⟹I=−xcosx+sinx+C

In case the final product is still a product a two integrals, we have to use the above formula again to simplify the integral.

The final formula of Integration by parts would therefore be:

The LILATE rule: This rule was was proposed by Herbert Kasube of Bradley University.

LIATE stands for:

L: Logarithmic functions

I: Inverse trigonometric functions

A: Algebraic functions: x squared, etc.

T: Trigonometric functions

E: Exponential functions

This defines the order in which integration needs to be performed on particular functions.

For example, consider:

∫xsinx dx

Here we have the product of two functions: One algebraic (x) and one trigonometric (SinX).

According to LIATE, A (algebraic) comes before T (trigonometric). So, the algebraic function (x in this case) should be the function f(x).

8. Integration by substitution:

Also known as the Reverse Chain Rule, this is commonly used to solve relatively complex integrals.

Here, g’(x) is the derivative of g(x). In case the integral is of the above form, it can be simplified by substituting g(x) as u. hence the function now becomes ∫abf(u)du
The above form is a simple integration, where f(u) can be integrated and u can finally be substituted back as g(x).

Let us understand this with an easy example.

∫sin(a x + b) dx

Let us substitute (ax+b) with u.

So, differentiating this, we get du/dx=a.

Or, dx=1/a du.

Now substituting these above, we get:

I = (1/a) ∫sin(u) du

= (1/a) cos(u)

= (1/a) cos(a x + b) + K

## How to Approach Integration

Integration, like any other topic of mathematics, should be practiced as much as possible. One thing special about integration is that there is almost always more than one way of solving a given problem. But, there is only one quickest way to solve it. In a time-bound test, it is very important that one’s solution is as quick as possible. This is the ‘having an eye for integral problems’ talked about earlier. Here are some standard ways to prepare yourself for JEE integral calculus.

### Learn to identify the Form

Instead of jumping right away on a problem, it is best to judge the form it is in. In the beginning, this just means you have to stare at the problem and identify the form it is in. This will keep you from going astray in the problem.

### Rearrange the Integral

There are times when an integration problem seems difficult to solve but a simple rearrangement can make solving it very convenient. Generally, a seemingly complex problem in JEE can easily be solved by rearrangement. Basic rearrangement involves simple steps which don’t change the problem. Some of the simplest rearrangements are as follows

• Multiply & Dividing
• Breaking the Numerator

### Memorise the Conventional Formulae

It might sound mildly offensive to some aspirants when asked to ‘memorise‘ the formulae, but memorising a formula cuts a lot on time. You should of course try and derive the formula before memorising it. This will offer you a definite edge over the others.

### Know the Standard Integration Techniques

All aspirants must be aware of the standard integration techniques such as Substitution, By Parts, Method of Partial Fractions etc. This can only be achieved by practicing problems from each of these techniques.

### Diversified Practice

Other than practicing from NCERT, one should make sure they practice from at least two other different books. This is to ensure exposure to a different type of problems. NCERT is, of course, must because there have been times when JEE asked a direct problem from NCERT Mathematics in integration. ‘Problems in Calculus of One Variable’ by I.A. Maron is one of the highly recommended books on calculus.

Find out an advanced guide to calculus in this article!

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