JEE Advanced 2025 Revision Notes for 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

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