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# JEE Advanced 2023 Revision Notes for Integral Calculus

Last updated date: 24th Mar 2023
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Integral Calculus is the advanced concept of Calculus covered in the JEE Advanced syllabus. This chapter introduces the different rules and formulae to determine anti-derivatives of integrals. It explains the various properties of integration, method of integration, integration of substitution, trigonometric identities, etc. To get a good hold of these mathematical principles, refer to the Integral Calculus JEE Advanced notes on Vedantu.

These revision notes are prepared by our subject experts keeping the mathematical principles in mind. Their simpler explanations will help students to learn the concepts faster and to recall them before an exam during revision sessions.

Competitive Exams after 12th Science

## Access JEE Advanced Revision Notes Mathematics Integral Calculus

### Integrations:

Let $f(x)$ be a function then the collection of all its primitives is called the indefinite integral of $f(x)$ and is denoted by $\int {f(x)dx}$ .

Integration as inverse operation of differentiation. If $\dfrac{d}{{dx}}\left\{ {\varphi \left( x \right)} \right\} = f\left( x \right),\int {f(x)dx = \varphi (x) + C}$,

Where $C$ is called the constant of integration or arbitrary constant.

### Symbols:

$f(x)$: Integrand

$f(x)dx$: Element of integration

$\int {}$: Sign of integral

$\varphi \left( x \right)$: Anti-derivative or primitive or integral of function $f(x)$.

### Integration of Function:

The integral or primitive of a function$f(x)$ with respect to $x$ is that function $\varphi \left( x \right)$ whose derivative with respect to $x$ is the given function$f(x)$ .it is expressed symbolically as

$\int {f(x)dx = \varphi (x)}$

Thus $\int {f(x)dx = \varphi (x)} \Leftrightarrow \dfrac{{dy}}{{dx}}\left[ {\varphi \left( x \right)} \right] = f(x)$

The process of finding the integrals of a function is called integration, and the given function is called the integrand. It is obvious to note that the operation of integration is the inverse operation of differentiation. Hence integral of a function is also named as anti-derivative of that function.

Further observe that

$\dfrac{{dy}}{{dx}}\left[ {\varphi \left( x \right) + c} \right] = f(x)$

$\therefore \int {f(x)dx = \varphi (x) + c}$

These constant numbers are generally denoted by $c$, called the constant of integration. Due to the presence of this constant, such an integral is called an indefinite integral.

### Basic Theorems of Integration:

If $f(x),~g(x)$ are two function of a variable $x$ and $k$ is constant, then

$\int {Kf(x)dx = K\int {f(x)dx} }$

$\int {[f(x) \pm g(x)]dx = \int {f(x)dx \pm \int {g(x)dx} } }$

$\dfrac{d}{{dx}}\left( {\int {f\left( x \right)dx} } \right) = f(x)$

$\int {\left( {\dfrac{d}{{dx}}f(x)} \right)} dx = f(x)$

### Standard Integrals:

•  $\int {0dx = c}$

•  $\int {1.dx = x + c}$

•  $\int {k.dx = kx + c(k \in R)}$

•  $\int {{x^n}} .dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c,~(n \ne - 1)$

•  $\int {\dfrac{1}{x}dx = {{\log }_e}} x + c$

•  $\int {{e^x}} dx = {e^x} + c$

•  $\int {{a^x}} dx = \dfrac{{{a^x}}}{{{{\log }_e}a}} + c = {a^x}{\log _a}e + c$

•  $\int {\sin xdx = - \cos x + c}$

•  $\int {\cos xdx = \sin x + c}$

•  $\int {\tan xdx = \log \sec x + c} = - \log \cos x + c$

•  $\int {\cot xdx = \log \sin x + c}$

•  $\sec xdx = \log \left( {\sec x + \tan x} \right) = - \log \left( {\sec x - \tan x} \right) = \log \tan \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right) + c$

• $\int {co\sec xdx = - \log \left( {co\sec x + \cot x} \right) = \log \left( {co\sec x - \cot x} \right) = \log \tan \left( {\dfrac{x}{2}} \right) + c}$

•  $\int {\sec x\tan xdx = \sec x + c}$

•  $\text{cosec}~ x \cot xdx = - \text{cosec}~ x + c$

•  $\int {{{\sec }^2}xdx = \tan x + c}$

•  $\int {co{{\sec }^2}xdx = - \cot x + c}$

•  $\int {\sinh xdx = \cosh x + c}$

•  $\int {\cosh xdx = \sinh x + c}$

•  $\int {{{{\mathop{\rm sech}\nolimits} }^2}xdx = \tanh x + c}$

•  $\int {co{\mathop{\rm sech}\nolimits} x\coth xdx = - \cos echx + c}$

• $\int {\dfrac{1}{{{x^2} + {a^2}}}} dx = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) + c$

• $\int {\dfrac{1}{{{x^2} - {a^2}}}} dx = \dfrac{1}{{2a}}\log \left( {\dfrac{{x - a}}{{x + a}}} \right) + c$

• $\int {\dfrac{1}{{{a^2} - {x^2}}}} dx = \dfrac{1}{{2a}}\log \left( {\dfrac{{a + x}}{{a - x}}} \right) + c$

• $\int {\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}} dx = {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right) + c = - {\cos ^{ - 1}}\left( {\dfrac{x}{a}} \right) + c$

• $\int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}} dx = {\sinh ^{ - 1}}\left( {\dfrac{x}{a}} \right) + c = \log \left( {x + \sqrt {{x^2} + {a^2}} } \right) + c$

• $\int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}} dx = {\cosh ^{ - 1}}\left( {\dfrac{x}{a}} \right) + c = \log \left( {x + \sqrt {{x^2} - {a^2}} } \right) + c$

• $\int {\sqrt {{a^2} - {x^2}} } dx = \dfrac{x}{2}\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}.{\sin ^{ - 1}}\dfrac{x}{a} + c$

• $\int {\sqrt {{x^2} + {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} + {a^2}} + \dfrac{{{a^2}}}{2}.\log \left( {x + \sqrt {{x^2} + {a^2}} } \right)$

• $\int {\sqrt {{x^2} - {a^2}} } dx = \dfrac{x}{2}\sqrt {{a^2} - {x^2}} - \dfrac{{{a^2}}}{2}.\log \left( {x + \sqrt {{x^2} - {a^2}} } \right)$

• $\int {\dfrac{1}{{x\sqrt {{x^2} + {a^2}} }}} dx = \dfrac{1}{a}{\sec ^{ - 1}}\dfrac{x}{a} + c$

•  $\int {{e^{ax}}} \sin bxdx = \dfrac{{{e^{ax}}}}{{{a^2} + {b^2}}}\left( {a\sin bx - b\cos bx} \right) + c$

$= \dfrac{{{e^{ax}}}}{{\sqrt {{a^2} + {b^2}} }}\sin \left\{ {bx - {{\tan }^{ - 1}}\left( {\dfrac{b}{a}} \right)} \right\} + c$

•  $\int {{e^{ax}}} \cos bxdx = \dfrac{{{e^{ax}}}}{{{a^2} + {b^2}}}\left( {a\cos bx + b\sin bx} \right) + c$

$= \dfrac{{{e^{ax}}}}{{\sqrt {{a^2} + {b^2}} }}\cos \left\{ {bx - {{\tan }^{ - 1}}\left( {\dfrac{b}{a}} \right)} \right\} + c$

### Methods of Integration:

When integration cannot be reduced into some standard form then integration is performed using following methods.

• Integration by substitution

• Integration by parts

• Integration using by partial fractions

• Integration by reduction formula

### Integration by Substitution:

• Integration if the form $\int {f(ax + b)dx:}$

• If $\int {f\left( x \right)dx = \varphi \left( x \right)}$ then $\int {f(ax + b)dx = \dfrac{1}{a}} \varphi (ax + b)$

•  $\int {{{\left( {ax + b} \right)}^n}} dx = \dfrac{1}{a}.\dfrac{{{{\left( {ax + b} \right)}^{r - 1}}}}{{n + 1}} + c,n \ne - 1$

•  $\int {\dfrac{1}{{ax + b}}} dx = \dfrac{1}{a}\log |ax + b| + C$

•  $\int {{e^{ax + b}}} dx = \dfrac{1}{a}{e^{ax + b}} + C$

•  $\int {{a^{bx + b}}} dx = \dfrac{1}{b}.\dfrac{{{a^{bx + c}}}}{{\log a}} + C,a > 0$ and$a \ne 1$

•  $\int {\sin \left( {ax + b} \right)dx = - \dfrac{1}{a}} \cos \left( {ax + b} \right) + C$

•  $\int {\cos \left( {ax + b} \right)dx = \dfrac{1}{a}} \sin \left( {ax + b} \right) + C$

•  $\int {{{\sec }^2}\left( {ax + b} \right)dx = \dfrac{1}{a}} \tan \left( {ax + b} \right) + C$

•  $\int {co{{\sec }^2}\left( {ax + b} \right)dx = - \dfrac{1}{a}} \cot \left( {ax + b} \right) + C$

•  $\int {\sec \left( {ax + b} \right)\tan \left( {ax + b} \right)dx = \dfrac{1}{a}} \sec \left( {ax + b} \right) + C$

•  $\int {\cos ec\left( {ax + b} \right)\cot \left( {ax + b} \right)dx = - \dfrac{1}{a}} co\sec \left( {ax + b} \right) + C$

•  $\int {\tan \left( {ax + b} \right)dx = - \dfrac{1}{a}} \log |\cos \left( {ax + b} \right)| + C$

•  $\int {\cot \left( {ax + b} \right)dx = \dfrac{1}{a}} \log |\sin \left( {ax + b} \right)| + C$

•  $\int {\sec \left( {ax + b} \right)dx = \dfrac{1}{a}} \log |\sec \left( {ax + b} \right) + \tan \left( {ax + b} \right)| + C$

•  $\int {co\sec \left( {ax + b} \right)dx = \dfrac{1}{a}} \log |\cos ec\left( {ax + b} \right) - \cot \left( {ax + b} \right)| + C$

Integrals of the form$\int {\dfrac{{f’\left( x \right)}}{{f\left( x \right)}}} dx:$

$\int {\dfrac{{f\left( x \right)}}{{f\left( x \right)}}} dx = \log \left\{ {f\left( x \right)} \right\}$

Integrals of the form ${\int {\left\{ {f\left( x \right)} \right\}} ^n}f\left( x \right)dx:$

${\int {\left\{ {f\left( x \right)} \right\}} ^n}f\left( x \right)dx = \dfrac{{{{\left\{ {f\left( x \right)} \right\}}^{n + 1}}}}{{n + 1}},n \ne - 1$

Integrals of the form $\int {\dfrac{{f\left( x \right)}}{{\sqrt {f\left( x \right)} }}} dx:$

$\int {\dfrac{{f\left( x \right)}}{{\sqrt {f\left( x \right)} }}} dx = 2\sqrt {f\left( x \right)} + c$

### Some Special Integrals:

In this section, we will introduce some important formula of integrals and apply them to evaluate many integrals.

Expression                                                Substitution

${a^2} + {x^2}$                                                       $x = a\tan \theta$ or $a\cot \theta$

${a^2} - {x^2}$                                                        $x = a\sin \theta$ or $a\cos \theta$

${x^2} - {a^2}$                                                        $x = a\sec \theta$ or $a\cos ec\theta$

$\sqrt {\dfrac{{a - x}}{{a + x}}}$ or$\sqrt {\dfrac{{a + x}}{{a - x}}}$                                      $x = a\cos 2\theta$

$\sqrt {\dfrac{{x - \alpha }}{{\beta - x}}}$ or $\sqrt {\left( {x - \alpha } \right)\left( {x - \beta } \right)}$             $x = \alpha {\cos ^2}\theta + \beta {\sin ^2}\theta$

### Integration of Rational and Irrational Functions:

Integrals of the type $\int {\dfrac{1}{{a{x^2} + bx + c}}dx}$ or reducible to the $\int {\dfrac{1}{{a{x^2} + bx + c}}dx}$

To evaluate this type of integrals we express $a{x^2} + bx + c$ as the sum or difference of two squares by using the following algorithm

• Make the coefficient of ${x^2}$ unity, if it is not, by multiplying and dividing by it.

• Add and subtract square of the half of coefficient of $x$ to express $a{x^2} + bx + c$ in the form

$a\left[ {{{\left( {x + \dfrac{b}{{2a}}} \right)}^2} + \dfrac{{4ac - {b^2}}}{{4{a^2}}}} \right]$

### Integration by Parts:

It u and v are two functions of $x$ , then

$\int {\left( {uv} \right)dx = \left( {u\int {vdx} } \right) - \int {\left( {\dfrac{{du}}{{dx}}} \right).\left( {\int {vdx} } \right)dx} }$

Integral of the product of two functions

= first $x$ integral of second $- \int {}$ (derivative of first) $x$ (integral of second)

### Selection of the Dirst Function:

For applying this method, we take ${x^n}$ as the first function, provided we know the integral of the second. If a logarithmic function or inverse trigonometric function is one of the products, then that should be taken as the first function. If logarithmic function a. 1 inverse trigonometric function has no second product, then I would be considered the second function.

### Integrals of Rational Algebraic Functions By Using Partial Fraction:

If $f(x)$ and $g(x)$ are two polynomials, then $\dfrac{{f(x)}}{{g(x)}}$ defines a rational algebraic function or a rational function of $x$ .

If degree of$f(x)$ < degree $g(x)$  of, Then $\dfrac{{f(x)}}{{g(x)}}$ is called a proper rational function.

If degree $f(x)$  $\ge$ degree $g(x)$ then $\dfrac{{f(x)}}{{g(x)}}$ is called improper rational function.

If $\dfrac{{f(x)}}{{g(x)}}$ is an improper rational function, we divided $f(x)$ by $g(x)$ so that he rational function $\dfrac{{f(x)}}{{g(x)}}$

Is expressed in the form that of $g(x)$ .Thus $\dfrac{{f(x)}}{{g(x)}}$ is expressible as the sum og a polynomial and a proper rational function.

### Some of the Integrates of Different Expression of ${E^X}$ :

•  $\int {\dfrac{{a{e^x}}}{{b + c{e^x}}}} dx$               [Put ${e^x} = t$ ]

•  $\int {\dfrac{1}{{1 + {e^x}}}} dx$                  [multiply and divide by ${e^{ - x}}$ and put ${e^{ - x}} = t$ ]

•  $\int {\dfrac{1}{{1 - {e^x}}}} dx$                  [multiply and divided by ${e^{ - x}}$ and put ${e^{ - x}} = t$ ]

•  $\int {\dfrac{1}{{{e^x} - {e^x}}}} dx$                [multiply and divided by ${e^x}$ ]

•  $\int {\dfrac{{{e^x} - {e^x}}}{{{e^x} + {e^x}}}} dx$                 $\left[ {\dfrac{{f\left( x \right)}}{{f\left( x \right)}}} \right]$ form

•  $\int {\dfrac{{{e^x} + 1}}{{{e^x} - 1}}} dx$                   [multiply and divide ${e^{ - x/2}}$ ]

•  $\int {\left( {\dfrac{{{e^x} - {e^x}}}{{{e^x} + {e^{ - x}}}}} \right)} dx$          [integrated = ${\tanh ^2}x$ ]

•  ${\int {\left( {\dfrac{{{e^x} - {e^x}}}{{{e^x} + {e^{ - x}}}}} \right)} ^2}dx$          [integrated= ${\coth ^2}x$ ]

•  $\int {\dfrac{1}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} dx$          [integrated= $\dfrac{1}{4}{{\mathop{\rm sech}\nolimits} ^2}x$ ]

•  $\int {\dfrac{1}{{{{\left( {{e^x} - {e^{ - x}}} \right)}^2}}}} dx$         [integrated= $\cos ec{h^2}x$ ]

•  $\int {\dfrac{1}{{\left( {1 + {e^x}} \right)\left( {1 - {e^x}} \right)dx}}}$  [multiply and divide by ${e^x}$ and put${e^x} = t$ ]

•  $\int {\dfrac{1}{{\sqrt {1 - {e^x}} }}dx}$                [multiply and divide by ${e^{ - x}}$ ]

•  $\int {\dfrac{1}{{\sqrt {1 + {e^x}} }}} dx$                 [multiply and divide by ${e^{ - \dfrac{x}{2}}}$ ]

•  $\int {\dfrac{1}{{\sqrt {{e^x} - 1} }}} dx$                 [multiply and divide by ${e^{ - \dfrac{x}{2}}}$ ]

• $\int {\dfrac{1}{{\sqrt {2{e^x} - 1} }}} dx$                [multiply and divide by $\sqrt 2 {e^{ - \dfrac{x}{2}}}$ ]

•  $\int {\sqrt {1 - {e^x}} } dx$                 [integrand =$\dfrac{{(1 - {e^x})}}{{\sqrt {1 - {e^x}} }}$ ]

•  $\int {\sqrt {1 - {e^x}} } dx$                 [integrand =$\dfrac{{(1 + {e^x})}}{{\sqrt {1 + {e^x}} }}$ ]

•  $\int {\sqrt {{e^x} - 1} } dx$                 [integrand =$\dfrac{{({e^x} - 1)}}{{\sqrt {{e^x} - 1} }}$ ]

•  $\int {\sqrt {\dfrac{{{e^x} + a}}{{{e^x} - a}}} } dx$                [integrand =$\dfrac{{({e^x} + a)}}{{\sqrt {{e^{2x}} - {a^2}} }}$ ]

### Examples:

If $\int {\dfrac{{\sin x}}{{{{\sin }^3}x + {{\cos }^3}x}}} dx =$

$\alpha {\log _e}|1 + \tan x| + \beta {\log _e}|1 - \tan x + {\tan ^2}x| + \gamma {\tan ^{ - 1}}\left( {\dfrac{{2\tan x - 1}}{{\sqrt 3 }}} \right) + C,$ when C is constant of integration, then the value of $18\left( {\alpha + \beta + {\gamma ^2}} \right)$ is

Solution:

$= \int {\dfrac{{\dfrac{{\sin x}}{{{{\cos }^3}x}}}}{{1 + {{\tan }^3}x}}} dx$

=$\int {\dfrac{{\tan x.{{\sec }^2}x}}{{\left( {\tan x + 1} \right)\left( {1 + {{\tan }^2}x - \tan x} \right)}}} dx$

Let $\tan x = t \Rightarrow {\sec ^2}x.dx = dt$

$= \int {\dfrac{t}{{\left( {{\mathop{\rm t}\nolimits} + 1} \right)\left( {{t^2} - t + 1} \right)}}} dt$

$= \int {\left( {\dfrac{A}{{t + 1}} + \dfrac{{B\left( {2t - 1} \right)}}{{{t^2} - t + 1}} + \dfrac{C}{{{t^2} - t + 1}}} \right)} dx$

$\Rightarrow A\left( {{t^2} - t + 1} \right) + B(2t - 1)\left( {{t^2} - t + 1} \right) + C(t + 1) = t$

$\Rightarrow {t^2}\left( {A + 2B} \right) + t\left( { - A + B + C} \right) + A - B + C = 1$

$\therefore A + 2B = 0$                       ….(1)

$- A + B + C = 1$                     ….(2)

$A - B + C = 0$                      ……(3)

$\Rightarrow C = \dfrac{1}{2} \Rightarrow A - B = - \dfrac{1}{2}$      …….(4)

$A + 2B = 0$

$A - B = - \dfrac{1}{2}$

$3B = \dfrac{1}{2}$

$B = \dfrac{1}{6}$

$A = - \dfrac{1}{3}$

$I = - \dfrac{1}{3}\int {\dfrac{{dt}}{{1 + t}}} + \dfrac{1}{6}\int {\dfrac{{2t - 1}}{{{t^2} - t + 1}}} dt + \dfrac{1}{2}\int {\dfrac{{dt}}{{{t^2} - t + 1}}}$

$= - \dfrac{1}{3}\ln |(1 + \tan x)| + \dfrac{1}{6}\ln |{\tan ^2}x - \tan x + 1| + \dfrac{1}{2}.\dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{\left( {\tan x - \dfrac{1}{2}} \right)}}{{\dfrac{{\sqrt 3 }}{2}}}} \right)$

$= - \dfrac{1}{3}\ln |(1 + \tan x)| + \dfrac{1}{6}\ln |{\tan ^2}x - \tan x + 1| + \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2\tan x - 1}}{{\sqrt 3 }}} \right) + C$

$\alpha = - \dfrac{1}{3}$

$\beta = \dfrac{1}{6}$

$18\left( {\alpha + \beta + {\gamma ^2}} \right)$ $\gamma = \dfrac{1}{{\sqrt 3 }}$

=$18\left( { - \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{3}} \right) = 3$

## Importance of Maths Integral Calculus

Integral Calculus is the opposite of Differential Calculus. Here, the formulae are derived to find out the anti-derivatives of functions within the limits of the variable. This chapter will explain how integration can be defined as the inverse function matching the features of differentiation.

Studying this chapter is very important as it will reveal how the formulae of differentiation can be reversed to find out the integration. There are different types of functions that can go through integration operations. Learn from this chapter the outcomes of integrating various functions of Mathematics.

This chapter will also explain indefinite and definite integration processes and their differences. You will also learn how to do integration by parts and how to determine the integral of partial functions.

With the help of Integral Calculus JEE Advanced notes PDF, you will be able to learn the fundamental concepts and theorems of integral calculus faster. This chapter holds immense importance for the JEE exam.

### Benefits of Integral Calculus JEE Advanced Revision Notes

• These revision notes are compiled by the experts for easy understanding of students. The formulae and derivations are defined and described using simple examples so that you can grab the concepts perfectly while preparing this chapter.

• Develop a good grip on this chapter by using these notes and learn how Differential Calculus and Integral Calculus are linked to each other. Understand how the experts have explained the concepts and use the process to do the same to find the accurate answers in the JEE Advanced exam.

• When you have completed this chapter, use these notes to revise what you have studied before an exam. It will take less time than using the textbook to revise the chapter. Download the consolidated presentation of the Integral Calculus JEE Advanced revision notes for free from Vedantu.

Remember the application of the integral calculus formulae due to the concise format of this chapter in the notes. It will help you to remember which formula to use during an exam.

The free PDF version of these notes can be used at your convenience. You can add this file to your study material for this chapter and revise it at your convenience. Discover how the experts have excellently defined and described the important formulae and concepts of Integral Calculus faster. You can use these notes to make your study and revision sessions shorter.

## FAQs on JEE Advanced 2023 Revision Notes for Integral Calculus

1. What is integration?

In Mathematics, integration is a process where the parts of a function are added or summed to find the whole answer. It has upper and lower limits and the value of a function can be determined by putting the values into the variable positions.

2. What is the application of integration?

Integration is used to study the rate of change of two variables and their relationship.

3. What is the integration method?

When the addition to a large scale is not possible, we use an integration method on a function of two variables.

4. What are the limits?

The upper and lower end or boundaries of an integration operation are called limits.