Class 12 Maths Chapter 7 Integrals Notes FREE PDF Download
Integration class 12 notes FREE PDF covers fundamental and advanced integration techniques, including key formulas and applications. With clear explanations and practical examples, our notes are designed to help you master integrals for your board exams.
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Access Revision Notes for Class 12 Maths Chapter 7 Integrals
Differentiation is the inverse of integration. Integration is the process of determining a function whose differential coefficient is known.
So from the above, if the differential coefficient of ${\text{F}}({\text{x}})$ is $f({\text{x}})$
i.e. $\dfrac{{\text{d}}}{{{\text{dx}}}}[{\text{F}}({\text{x}})] = f({\text{x}})$, then one can say that the antiderivative or integral of $f(x)$ is $F(x)$, written as $\int f (x)dx = F(x)$,
Here $\int {\text{d}} {\text{x}}$ is the notation of integration $f({\text{x}})$ is the integrand, $x$ is the variable of integration and dx denotes the integration with respect to ${\text{X}}$.
Indefinite Integral:
We know that if $\dfrac{{\text{d}}}{{{\text{dx}}}}[{\text{F}}({\text{x}})] = f({\text{x}})$, then $\int f ({\text{x}}){\text{dx}} = {\text{F}}({\text{x}})$.
Also, for any arbitrary constant C,
$\dfrac{{\text{d}}}{{{\text{dx}}}}[{\text{F}}({\text{x}}) + {\text{C}}] = \dfrac{{\text{d}}}{{{\text{dx}}}}[{\text{F}}({\text{x}})] + 0 = f({\text{x}})$
$\therefore \int f ({\text{x}}){\text{dx}} = {\text{F}}({\text{x}}) + {\text{C}}$
This shows that ${\text{F}}({\text{x}})$ and ${\text{F}}({\text{x}}) + {\text{C}}$ are both integrals of the same function ${\text{f}}({\text{x}})$. Thus, for different values of ${\text{C}}$, we obtain different integrals of ${\text{f}}({\text{x}})$. This implies that the integral of ${\text{f}}({\text{x}})$ is not definite. By virtue of this property ${\text{F}}({\text{x}})$ is called the indefinite integral of ${\text{f}}({\text{x}})$.
1.1 Properties of Indefinite Integration
$\quad \dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {\int {\text{f}} ({\text{x}}){\text{dx}}} \right] = f({\text{x}})$
$\int {{f^\prime }} (x){\text{dx}} = \int {\dfrac{{\text{d}}}{{{\text{dx}}}}} [f({\text{x}})]{\text{dx}} = f({\text{x}}) + {\text{c}}$
$\quad \int {\text{k}} f({\text{x}}){\text{dx}} = {\text{k}}\int f ({\text{x}}){\text{dx}}$, where ${\text{k}}$ is any constant
If ${f_1}({\text{x}}),{f_2}({\text{x}}),{f_3}({\text{x}}), \ldots $ (finite in number) are functions of ${\text{x}}$, then
$ \int{\left[ {{f}_{1}}(x)\pm {{f}_{2}}(x)\pm {{f}_{3}}(x)\ldots \right]}dx $
$ =\int{{{f}_{1}}}(x)dx\pm \int{{{f}_{2}}}(x)dx\pm \int{{{f}_{3}}}(x)dx\pm \text{ }\!\!~\!\!\text{ }\ldots \text{ }\!\!~\!\!\text{ } $
If $\int f (x)dx = F(x) + c$ then $\int f (ax \pm b)dx = \dfrac{1}{a}F(ax \pm b) + c$
1.2 Standard Formula of Integration
The definition of an integral has the following conclusions as a direct result.
$\quad \int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne - 1$.
$\quad \int {\dfrac{1}{x}} dx = \log |x| + C$
$\int {{{\text{e}}^{\text{x}}}} {\text{dx}} = {{\text{e}}^{\text{x}}} + {\text{C}}$
$\int {{{\text{a}}^{\text{x}}}} {\text{dx}} = \dfrac{{{{\text{a}}^{\text{x}}}}}{{{{\log }_{\text{e}}}{\text{a}}}} + {\text{C}}$.
$\quad \int {\sin } xdx = - \cos x + C$
$\quad \int {\cos } xdx = \sin x + C$
$\int {{{\sec }^2}} xdx = \tan x + C$
$\quad \int {{{\operatorname{cosec} }^2}} xdx = - \cot x + C$
$\quad \int {\sec } x\tan xdx = \sec x + C$
$\int {\operatorname{cosec} } x\cot xdx = - \operatorname{cosec} x + C$.
$\int {\tan } xdx = - \log |\cos x| + C = \log |\sec x| + C$.
$\int {\cot } xdx = \log |\sin x| + C$
$\int {\sec } xdx = \log |\sec x + \tan x| + C$
$\int {\operatorname{cosec} } xdx = \log |\operatorname{cosec} x - \cot x| + C$
$\int {\dfrac{{{\text{dx}}}}{{\sqrt {1 - {{\text{x}}^2}} }}} = {\sin ^{ - 1}}{\text{x}} + {\text{C}};|{\text{x}}| < 1$
$\int {\dfrac{{dx}}{{1 + {x^2}}}} = {\tan ^{ - 1}}x + C$
$\int {\dfrac{{{\text{dx}}}}{{{\text{x}}\sqrt {{{\text{x}}^2} - 1} }}} = {\sec ^{ - 1}}|{\text{x}}| + {\text{C}};|{\text{x}}| > 1$
2. Methods of Integration
2.1 Method of Substitution
By using the suitable substitution, the variable $x$ in $\int f (x)dx$ is changed into another variable $t$ so that the integrand $f(x)$ is changed into ${\text{F}}({\text{t}})$ is an algebraic sum of standard integrals or a standard integral. There is no universal formula for determining a suitable substitute, and experience is the greatest guidance in this regard.
The Following Ideas, on the Other Hand, will be Beneficial.
If the integrand is of the form ${f^\prime }(ax + b)$, then we
put $ax + b = t$ and $dx = \dfrac{1}{a}dt$
Thus, $\int {{f^\prime }} (ax + b)dx = \int {{f^\prime }} (t)\dfrac{{dt}}{a}$
$ = \dfrac{1}{a}\int {{f^\prime }} (t)dt = \dfrac{{f(t)}}{a} = \dfrac{{f(ax + b)}}{a} + c$
When the integrand is of the form ${x^{n - 1}}{f^\prime }\left( {{x^n}} \right)$, we put ${x^n} = t$ and $n{x^{n - 1}}dx = dt$
Thus, $\int {{x^{n - 1}}} {f^\prime }\left( {{x^n}} \right)dx = \int {{f^\prime }} (t)\dfrac{{dt}}{n} = \dfrac{1}{n}$
$\int {{f^\prime }} ({\text{t}}){\text{dt}} = \dfrac{1}{{\text{n}}}f({\text{t}}) = \dfrac{1}{{\text{n}}}f\left( {{{\text{x}}^{\text{n}}}} \right) + {\text{c}}$
When the integrand is of the form ${[f(x)]^n} \cdot {f^\prime }(x)$, we put $f(x) = t$ and ${f^\prime }(x)dx = dt$
Thus, ${\text{(Tex translation failed)}}$
When the integrand is of the form $\dfrac{{{f^\prime }(x)}}{{f(x)}}$, we put
$f({\text{x}}) = {\text{t}}$ and ${f^\prime }({\text{x}}){\text{dx}} = {\text{dt}}$
Thus, $\int {\dfrac{{{f^\prime }({\text{x}})}}{{f({\text{x}})}}} {\text{dx}} = \int {\dfrac{{{\text{dt}}}}{{\text{t}}}} = \log |{\text{t}}| = \log |f({\text{x}})| + {\text{c}}$
2.1.1 Some Special Integrals
$\int {\dfrac{{dx}}{{{x^2} + {a^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C$
$\quad \int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^2} - {{\text{a}}^2}}}} = \dfrac{1}{{2{\text{a}}}}\log \left| {\dfrac{{{\text{x}} - {\text{a}}}}{{{\text{x}} + {\text{a}}}}} \right| + {\text{C}}$
$\quad \int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + C$
$\int {\dfrac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = {\sin ^{ - 1}}\dfrac{x}{a} + C$
$\quad \int {\dfrac{{dx}}{{\sqrt {{x^2} + {a^2}} }}} = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C$
$\int {\dfrac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} = \log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$
$\int {\sqrt {{a^2} - {x^2}} } dx = \dfrac{x}{2}\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a} + C$.
$\quad \int {\sqrt {{x^2} + {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} + {a^2}} + \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C$
$\quad \int {\sqrt {{x^2} - {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} - {a^2}} - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$
2.1.2 Integrals of the Form
$\int f \left( {{{\text{a}}^2} - {{\text{x}}^2}} \right){\text{dx}}$,
$\int f \left( {{a^2} + {x^2}} \right)dx$,
$\int f \left( {{x^2} - {a^2}} \right)dx$,
$\int f \left( {\dfrac{{a - x}}{{a + x}}} \right)dx$,
Working Rule
Integral Substitution
$\int f \left( {{{\text{a}}^2} - {{\text{x}}^2}} \right){\text{dx}},\quad {\text{x}} = {\text{a}}\sin \theta $ or ${\text{x}} = {\text{a}}\cos \theta $
$\int f \left( {{a^2} + {x^2}} \right)dx,\quad x = a\tan \theta $ or $x = a\cot \theta $
$\int f \left( {{x^2} - {a^2}} \right)dx,\quad x = a\sec \theta $ or $x = a\operatorname{cosec} \theta $
$\int f \left( {\dfrac{{a - x}}{{a + x}}} \right)dx$ or $\int f \left( {\dfrac{{a + x}}{{a - x}}} \right)dx\quad x = a\cos 2\theta $
2.1.3 Integrals of the Form
$\int {\dfrac{{dx}}{{a{x^2} + bx + c}}} $
$\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} }}} $
$\int {\sqrt {a{x^2} + bx + c} } dxa$,
Working Rule
(i)Make the coefficient of ${x^2}$ unity by taking the coefficient of ${{\text{x}}^2}$ outside the quadratic.
(ii) Complete the square in the terms involving $x$, i.e. write $a{x^2} + bx + c$ in the form
$a\left[ {{{\left( {x + \dfrac{b}{{2a}}} \right)}^2}} \right] - \dfrac{{\left( {{b^2} - 4ac} \right)}}{{4a}}$
(iii) One of the nine special integrals is used to transform the integrand.
(iv) Then just integrate the function.
2.1.4 Integrals of the Form
$\int {\dfrac{{px + q}}{{a{x^2} + bx + c}}} dx$
$\int {\dfrac{{{\text{px}} + {\text{q}}}}{{\sqrt {a{x^2} + bx + c} }}} {\text{dx}}$
$\int {(px + q)} \sqrt {a{x^2} + bx + c} dx$
$\int {\dfrac{{px + q}}{{a{x^2} + bx + c}}} dx$
$\int {\dfrac{{{\text{px}} + {\text{q}}}}{{\sqrt {a{x^2} + bx + c} }}} {\text{dx}}$
$\int {(px + q)} \sqrt {a{x^2} + bx + c} dx$
Integral Working Rule
$\int {\dfrac{{{\text{px}} + {\text{q}}}}{{{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}}}} {\text{dx}}$ Put ${\text{px}} + {\text{q}} = \lambda (2{\text{ax}} + {\text{b}}) + \mu $ or ${\text{px}} + {\text{q}} = \lambda $ (derivative of quadratic) $ + \mu $.
When the coefficient of x and the constant term on both sides are compared, we obtain
${\text{p}} = 2{\text{a}}\lambda $ and ${\text{q}} = {\text{b}}\lambda + \mu \Rightarrow \lambda = \dfrac{{}}{{2{\text{a}}}}$ and $\mu = \left( {{\text{q}} - \dfrac{{{\text{bp}}}}{{2{\text{a}}}}} \right)$.
Then integral becomes
$\int {\dfrac{{px + q}}{{a{x^2} + bx + c}}} dx$
$ = \dfrac{{2{\text{ax}} + {\text{b}}}}{{2{\text{a}}}}\int {\dfrac{{{\text{dx}}}}{{{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}}}} {\text{dx}} + \left( {{\text{q}} - \dfrac{{{\text{bp}}}}{{2{\text{a}}}}} \right)\int {\dfrac{{\text{d}}}{{{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}}}} $
$ = \dfrac{{dx}}{{2a}}\log \left| {a{x^2} + bx + c} \right| + \left( {q - \dfrac{{bp}}{{2a}}} \right)\int {\dfrac{{dx}}{{a{x^2} + bx + c}}} $
$\int {\dfrac{{{\text{px}} + {\text{q}}}}{{\sqrt {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} }}} {\text{dx}}$
In this case the integral becomes
$\int {\dfrac{{{\text{px}} + {\text{q}}}}{{\sqrt {a{x^2} + bx + c} }}} dx = $
$\dfrac{p}{{2a}}\int {\dfrac{{2ax + b}}{{\sqrt {a{x^2} + bx + c} }}} dx + \left( {q - \dfrac{{bp}}{{2a}}} \right)\int {\dfrac{{dx}}{{\sqrt {a{x^2} + bx + c} }}} $
$ = \dfrac{p}{a}\sqrt {a{x^2} + bx + c} + \left( {q - \dfrac{{bp}}{{2a}}} \right)\int {\dfrac{{dx}}{{\sqrt {a{x^2} + bx + c} }}} $
$\int {(px + q)} \sqrt {a{x^2} + bx + c} dx$
In this example, the integral is transformed to
$\int {(px + q)} \sqrt {a{x^2} + bx + c} dx = \dfrac{p}{{2a}}\int {(2ax + b)} \sqrt {a{x^2} + bx + c} dx$
$ + \left( {{\text{q}} - \dfrac{{{\text{bp}}}}{{2{\text{a}}}}} \right)\int {\sqrt {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} } {\text{dx}}$
$ = \dfrac{{\text{p}}}{{3{\text{a}}}}{\left( {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} \right)^{3/2}} + \left( {{\text{q}} - \dfrac{{{\text{bp}}}}{{2{\text{a}}}}} \right)\int {\sqrt {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} } {\text{dx}}$
2.1.5 Integrals of the Form
$\int {\dfrac{{{\text{P}}({\text{x}})}}{{\sqrt {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} }}} {\text{dx}}$, where ${\text{P}}({\text{x}})$ is a polynomial in ${\text{x}}$ of degree ${\text{n}} \geqslant 2$.
Working Rule: Write
$ {\int {\dfrac{{{\text{P}}({\text{x}})}}{{\sqrt {a{x^2} + bx + c} }}} dx = } $
$ { = \left( {{a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_{n - 1}}{x^{n - 1}}} \right)\sqrt {a{x^2} + bx + c} + k\int {\dfrac{{dx}}{{\sqrt {a{x^2} + bx + c} }}} } $
where ${\text{k}},{{\text{a}}_0},{{\text{a}}_1}, \ldots {{\text{a}}_{{\text{n}} - 1}}$ are constants that may be found by separating the above relation and equating the coefficients of various powers of$x$ on both sides.
2.1.6 Integrals of the Form
$\int {\dfrac{{{x^2} + 1}}{{{x^4} + k{x^2} + 1}}} dx\quad $ or $\quad \int {\dfrac{{{x^2} - 1}}{{{x^4} + k{x^2} + 1}}} dx$
where ${\text{k}}$ is a constant positive, negative or zero.
Working Rule
(i) ${{\text{x}}^2}$ is divided into the numerator and denominator.
(ii) Put $x - \dfrac{1}{x} = z$ or $x + \dfrac{1}{x} = z$, Whatever substitution yields the numerator of the resultant integrand on differentiation
(iii) In ${\text{Z}}$, evaluate the resultant integral.
(iv) In terms of X, express the outcome.
2.1.7 Integrals of the Form
$\int {\dfrac{{{\text{dx}}}}{{{\text{P}}\sqrt {\text{Q}} }}} $, where ${\text{P}},{\text{Q}}$ are linear or quadratic functions of ${\text{x}}$.
Integral Substitution
$\int {\dfrac{1}{{(ax + b)\sqrt {cx + d} }}} dx$
$cx + d = {z^2}$
$\int {\dfrac{{dx}}{{\left( {a{x^2} + bx + c} \right)\sqrt {px + q} }}} \quad px + q = {z^2}$
$\int {\dfrac{{dx}}{{(px + q)\sqrt {a{x^2} + bx + c} }}} \quad px + q = \dfrac{1}{z}$
$\int {\dfrac{{dx}}{{\left( {a{x^2} + b} \right)\sqrt {c{x^2} + d} }}} \quad x = \dfrac{1}{z}$
3. Method of Partial Fractions For Rational Functions
By resolving the integrand into partial fractions, Integrals of the type $\int {\dfrac{{p(x)}}{{g(x)}}} $ may be integrated.
The Following is How we Proceed:
First of all check the degree of $p({\text{x}})$ and $g({\text{x}})$.
If degree of $p({\text{x}}) \geqslant $ degree of $g({\text{x}})$, then divide $p({\text{x}})$ by $g$(x) till its degree is less, i.e. put in the form $\dfrac{{p({\text{x}})}}{{g({\text{x}})}} = {\text{r}}({\text{x}}) + \dfrac{{f({\text{x}})}}{{g({\text{x}})}}$ where degree of $f({\text{x}}) < $ degree of$g({\text{x}})$
CASE 1: When there are non-repeated linear components in the denominator. i.e.
$g({\text{x}}) = \left( {{\text{x}} - {\alpha _1}} \right)\left( {{\text{x}} - {\alpha _2}} \right) \ldots \left( {{\text{x}} - {\alpha _{\text{n}}}} \right)$
In such a case write $f({\text{x}})$ and $g$ as:
$\dfrac{{f({\text{x}})}}{{g({\text{x}})}} = \dfrac{{{{\text{A}}_1}}}{{\left( {{\text{x}} - {\alpha _1}} \right)}} + \dfrac{{{{\text{A}}_2}}}{{\left( {{\text{x}} - {\alpha _2}} \right)}} + \ldots + \dfrac{{{{\text{A}}_{\text{n}}}}}{{\left( {{\text{x}} - {\alpha _{\text{n}}}} \right)}}$
After extracting L.C.M., compare the coefficients of various powers on both sides to get the constants ${A_1},{A_2}, \ldots {A_n}$.
CASE 2: When there are both repeated and non-repeated linear factors in the denominator.
i.e.$g({\text{x}}) = {\left( {{\text{x}} - {\alpha _1}} \right)^2}\left( {{\text{x}} - {\alpha _3}} \right) \ldots \left( {{\text{x}} - {\alpha _{\text{n}}}} \right)$
In such a case write $f(x)$ and $g(x)$ as:
$\dfrac{{f({\text{x}})}}{{g({\text{x}})}} = \dfrac{{{{\text{A}}_1}}}{{{\text{x}} - {\alpha _1}}} + \dfrac{{{{\text{A}}_2}}}{{{{\left( {{\text{x}} - {\alpha _1}} \right)}^2}}} + \dfrac{{{{\text{A}}_3}}}{{{\text{x}} - {\alpha _3}}} + \ldots + \dfrac{{{{\text{A}}_{\text{n}}}}}{{\left( {{\text{x}} - {\alpha _{\text{n}}}} \right)}}$
After extracting L.C.M., compare the coefficients of various powers on both sides to get the constants ${A_1},{A_2}, \ldots {A_n}$.
Note: corresponding to a linear factor that has been repeated${({\text{x}} - {\text{a}})^{\text{r}}}$ in the denominator, a sum of $r$ partial fractions of the type $\dfrac{{{{\text{A}}_1}}}{{{\text{x}} - {\text{a}}}} + \dfrac{{{{\text{A}}_2}}}{{{{({\text{x}} - {\text{a}})}^2}}} + \ldots + \dfrac{{{{\text{A}}_{\text{r}}}}}{{{{({\text{x}} - {\text{a}})}^r}}}$ is taken.
CASE 3: When there is a non-repeated quadratic component in the denominator that cannot be factored further:
$g({\text{x}}) = \left( {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} \right)\left( {{\text{x}} - {\alpha _3}} \right)\left( {{\text{x}} - {\alpha _4}} \right) \ldots \left( {{\text{x}} - {\alpha _{\text{n}}}} \right)$
In such a case express $f(x)$ and $g(x)$ as:
$\dfrac{{f(x)}}{{g(x)}} = \dfrac{{{A_1}x + {A_2}}}{{a{x^2} + bx + c}} + \dfrac{{{A_3}}}{{x - {\alpha _3}}} + \ldots + \dfrac{{{A_n}}}{{x - {\alpha _n}}}$
After extracting L.C.M., compare the coefficients of various powers on both sides to get the constants ${A_1},{A_2}, \ldots {A_n}$.
CASE 4: When there is a repeating quadratic component in the denominator that cannot be factored further:
i.e. $g(x) = {\left( {a{x^2} + bx + c} \right)^2}\left( {x - {\alpha _5}} \right)\left( {x - {\alpha _6}} \right) \ldots \left( {x - {\alpha _n}} \right)$
In such a case write $f({\text{x}})$ and $g({\text{x}})$ as
$ {\dfrac{{f(x)}}{{g(x)}} = \dfrac{{{A_1}x + {A_2}}}{{a{x^2} + bx + c}} + \dfrac{{{A_3}x + {A_4}}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}} + } $
$ {\dfrac{{{A_5}}}{{x - {\alpha _5}}} + \ldots + \dfrac{{{A_n}}}{{\left( {x - {\alpha _n}} \right)}}} $
After extracting L.C.M., compare the coefficients of various powers on both sides to get the constants ${A_1},{A_2}, \ldots {A_n}$.
CASE 5: If $x$ will be the even power then
(i) Put ${{\text{x}}^2} = {\text{z}}$ in the integrand.
(ii) Resolve the resulting rational expression in ${\text{z}}$ into partial fractions
(iii) Put ${\text{z}} = {{\text{x}}^2}$ again in the partial fractions and then integrate both sides.
4. Method of Integration by Parts
Integration by parts refers to the process of combining the output of two functions.
e.g. if $u$ and $v$ are two functions of $x$, then $\int {(uv)} {\text{dx}} = u \cdot \int v \;{\text{dx}} - \int {\left( {\dfrac{{{\text{d}}u}}{{{\text{dx}}}} \cdot \int v {\text{dx}}} \right)} {\text{dx}}$
To put it another way, the integral of the product of two functions = first function $ \times $ integral of the second - integral of (differential of first $ \times $ integral of the second function).
Working Hints
Choose the first and second functions in such a way that the derivative and integral of the first function may be determined quickly.
In case of integrals of the form $\int f ({\text{x}}) \cdot {{\text{x}}^{\text{n}}}{\text{dx}}$, take ${{\text{x}}^{\text{n}}}$ as the first function and $f({\text{x}})$ as the second function.
In case of integrals of the form $\int {{{(\log x)}^{\text{n}}}} \cdot 1{\text{dx}}$, take 1 as the second function and ${(\log x)^n}$ as the first function.
The rule of parts integration can be applied several times if necessary.
If the two functions are of different types, the first function can be chosen as the one with the first initial in the word "ILATE," where
I - Inverse Trigonometric function
L - Logarithmic function
A - Algebraic function ${\text{T}} - $ Trigonometric function
${\text{E}} - $ Exponential function.
If both functions are trigonometric, use the second function as the second function with a simple integral. If both functions are algebraic, choose the one with the simpler derivative as the initial function.
If the integral is an inverse trigonometric function of an algebraic expression in x, simplify the integrand first using an appropriate trigonometric substitution before integrating the new integrand.
4.1 Integrals of the Form $\int {{{\text{e}}^x}} \left[ {f({\text{x}}) + {f^\prime }({\text{x}})} \right]{\text{dx}}$
Working Rule
(i) Divide the integral into two equal halves.
(ii) Only use parts to integrate the first integral, i.e.
$ = \left[ {f({\text{x}}) \cdot {{\text{e}}^{\text{x}}} - \int {{f^\prime }} ({\text{x}}) \cdot {{\text{e}}^{\text{x}}}{\text{x}}} \right] + \int {{{\text{e}}^{\text{x}}}} {f^\prime }({\text{x}}){\text{dx}}$
$ = {{\text{e}}^{\text{x}}}f({\text{x}}) + {\text{C}}$
4.2 Integrals of the Form:
After integrating by parts, the initial integrand reappears.
Working Rule
(i) Use the part-by-part integration technique twice.
(ii) If we integrate by parts a second time, we'll get the same integrand as before, therefore we'll set it equal to $I$.
(iii) On one side, transpose and gather words involving $I$, and assess $I$.
5. Integral of the Form (Trigonometric Formats)
5.1
$\int {\dfrac{{dx}}{{a + b\cos x}}} $
$\int {\dfrac{{dx}}{{a + b\sin x}}} $
$\int {\dfrac{{dx}}{{a + b\cos x + c\sin x}}} $
Working Rule
(i) Put $\cos x = \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}$ and $\sin x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}$ so that the given integrand becomes a function of $\tan \dfrac{x}{2}$.
(ii) Put $\tan \dfrac{{\text{x}}}{2} = {\text{z}} \Rightarrow \dfrac{1}{2}{\sec ^2}\dfrac{{\text{x}}}{2}{\text{dx}} = {\text{dz}}$
(iii) Integrate the resulting rational algebraic function of $z$
(iv) In the answer, put $z = \tan \dfrac{x}{2}$.
5.2 Integrals of the Form
$\int {\dfrac{{dx}}{{a + b{{\cos }^2}x}}} $
$\int {\dfrac{{dx}}{{a + b{{\sin }^2}x}}} $
$\int {\dfrac{{dx}}{{a{{\cos }^2}x + b\sin x\cos x + {{\operatorname{csin} }^2}x}}} $
Working Rule
(i) Divide the numerator and denominator by ${\cos ^2}x$.
(ii) In the denominator, replace sec'x, if any, by $1 + {\tan ^2}x$.
(iii) Put $\tan x = z \Rightarrow {\sec ^2}xdx = dz$
(iv) Integrate the resulting rational algebraic function of $z$.
(v) In the answer, put $z = \tan x$.
5.3 Integrals of the Form
$\int {\dfrac{{a\cos x + b\sin x}}{{c\cos x + d\sin x}}} dx$
Working Rule
(i) Put Numerator $ = \lambda $ (denominator) $ + \mu $ (derivative of denominator) $a\cos x + b\sin x = \lambda (c\cos x + d\sin x) + \mu ( - c\sin x + d\cos x)$
(ii) Equate coefficients of $\sin x$ and $\cos x$ on both sides and find the values of $\lambda $, and $\mu $.
(iii) Divide the provided integral into two parts and evaluate each independently, as follows:
$\int {\dfrac{{a\cos x + b\sin x}}{{c\cos x + d\sin x}}} dx = $
$\lambda \iint {{\text{ld}}}{\text{x}} + \mu \int {\dfrac{{ - {\text{csinx}} + {\text{d}}\cos {\text{x}}}}{{{\text{a}}\cos {\text{x}} + {\text{b}}\sin {\text{x}}}}} {\text{dx}} = \lambda {\text{x}} + \mu \log |{\text{a}}\cos {\text{x}} + {\text{b}}\sin {\text{x}}|$
(iv) Substitute the values of $\lambda $ and $\mu $ found in step 2 .
5.4 Integrals of the Form $\int {\dfrac{{a + b\cos x + \operatorname{csin} x}}{{e + f\cos x + g\sin x}}} dx$
Working Rule
(i) Put Numerator $ = I$ (denominator) $ + {\text{m}}$ (derivative of denominator) $ + {\text{n}}$
$a + b\cos x + c\sin x = l(e + f\cos x + g\sin x) + m$
$( - f\sin x + g\cos x) + n$
(ii) Equate coefficients of $\sin x,\cos x$ and constant term n on both sides and find the values of $l,\;{\text{m}},{\text{n}}$.
(iii) The next step is to divide the provided integral into$3$ distinct integrals and evaluate each one separately, i.e.
$\int {\dfrac{{a + b\cos x + \operatorname{csin} x}}{{e + f\cos x + g\sin x}}} dx$
$l\int 1 {\text{dx}} + {\text{m}}\int {\dfrac{{ - {\text{f}}\sin {\text{x}} + {\text{g}}\cos {\text{x}}}}{{{\text{e}} + {\text{f}}\cos x + {\text{g}}\sin x}}} {\text{dx}} + $
${\text{n}}\int {\dfrac{{{\text{dx}}}}{{{\text{e}} + {\text{f}}\cos x + {\text{g}}\sin x}}} $
$ = {\text{Lx}} + {\text{m}}\log |{\text{e}} + {\text{f}}\cos {\text{x}} + {\text{g}}\sin {\text{x}}| + {\text{n}}\int {\dfrac{{d{\text{x}}}}{{{\text{e}} + {\text{f}}\cos {\text{x}} + {\text{g}}\sin x}}} {\text{dx}}$
(iv) Substitute the values of $l,\;{\text{m}},{\text{n}}$ found in Step (ii).
5.5 Integrals of the Form $\int {{{\sin }^{\text{m}}}} x{\cos ^{\text{n}}}{\text{xdx}}$
Working Rule
(i) If the power of $\sin x$ is an odd positive integer, put $\cos x = t$.
(ii) If the power of $\cos {\text{x}}$ is an odd positive integer, put $\sin {\text{x}} = {\text{t}}$.
(iii) If the power of $\sin {\text{x}}$ and $\cos {\text{x}}$ are both odd positive integers, put $\sin x = t$ or $\cos x = t$
(iv) If the power of $\sin x$ and $\cos x$ are both even positive integers, use De' Moivre's theorem as follows:
Let, $\quad \cos x + i\sin x = z$. Then $\cos x - i\sin x = {z^{ - 1}}$
Adding these, we get $z + \dfrac{1}{z} = 2\cos x$ and $z - \dfrac{1}{z} = 2i\sin x$
By De Moivre's theorem, we have
${z^n} + \dfrac{1}{{{z^n}}} = 2\cos nx$ and ${z^n} - \dfrac{1}{{{z^n}}} = 2i{\sin ^n}x$
$\therefore $
${\sin ^m}x{\cos ^n}x = \dfrac{1}{{{{(2i)}^m}}} \cdot \dfrac{1}{{{2^n}}}{\left( {z + \dfrac{1}{z}} \right)^n}{\left( {z - \dfrac{1}{z}} \right)^m}$
$ = \dfrac{1}{{{2^{m + n}}}} \cdot \dfrac{1}{{{i^m}}}{\left( {z + \dfrac{1}{z}} \right)^n}{\left( {z - \dfrac{1}{z}} \right)^m}$
Using the Binomial theorem, extend each of the components on the R.H.S. Then, equidistant from the start and end, group the words. As a result, all such pairings may be expressed as the sines or cosines of various angles. Term by term, continue to integrate.
(v) If the sum of powers of $\sin x$ and $\cos x$ is an even negative integer, put $\tan x = z$.
Indefinite and Definite Integrals:
Indefinite integrals are those integrals that do not have any limit of integration. It has an arbitrary constant. Definite integrals are those integrals that have an upper and lower limit. Definite integral has two different values for the upper limit and lowers limit when they are evaluated. The final value of a definite integral is the value of integral to the upper limit minus the value of the definite integral for the lower limit.
\[\int_a^b {f\left( x \right).dx} = F\left( b \right) - F\left( a \right)\]
In the above equation,
F (b) and F (a) are the integral functions for the upper and lower limit respectively.
f (x) is the integrand
dx is the integrating agent
‘b’ is the upper limit of the definite integral
‘a’ is the lower limit of the definite integral
Properties of Definite Integrals:
Property 1: Definite integrals between the same limits of the same function with different variables are equal.
\[\int_a^b {f\left( p \right),dp} = \int_b^a {f\left( q \right)dp} \]
Property 2: The value of a definite integral is equal to its negative when the upper and lower limits of the integral function are interchanged.
\[\int_a^b {f\left( p \right),dp} = - \int_b^a {f\left( p \right)dp} \]
Property 3: The definite integral of a function is zero when the upper and lower limits are the same.
\[\int_a^a {f\left( p \right).dp = 0} \]
Property 4: A definite integral can be written as the sum of two definite integrals. However, the following conditions must be considered.
The lower limit of the first addend should be equal to the lower limit of the original definite integral.
The upper limit of the second addend should be equal to the upper limit of the original definite integral.
The upper limit of the first addend should be equal to the lower limit of the second addend integral.
\[\int_a^b {f\left( b \right).dp = \int_a^c {f\left( p \right).dp + \int_c^b {f\left( p \right).dp} } } \]
Other Properties of Definite Integrals:
Property 5:
\[\int_a^b {f\left( p \right).dp = \int_a^b {f\left( {a + b - p} \right).dp} } \]
Property 6:
\[\int_0^a {f\left( p \right).dp = \int_a^b {f\left( {a - p} \right).dp} } \]
Property 7:
\[\int_0^{2a} {f\left( p \right).dp} = \int_0^a {f\left( p \right).dp + \int_0^a {f\left( {2a - p} \right).dp{\text{ }}if{\text{ }}f\left( {2a - p} \right) = f\left( p \right)} } \]
Fundamental Theorem of Calculus: Integrals & Anti Derivatives
The Fundamental Theorem of Calculus theorem that shows the relationship between the concept of derivation and integration, also between the definite integral and the indefinite integral— consists of 2 parts, the first of which, the Fundamental Theorem of Calculus, Part 1, and second is the Fundamental Theorem of Calculus, Part 2.
A. Fundamental Theorem of Calculus Part 1
Part 1 of Fundamental theorem creates a link between differentiation and integration.
By that, the first fundamental theorem of calculus depicts that, if f is continuous on the closed interval [a,b] and F is the unknown integral of f on [a,b], then
\[\int_{a}^{b} f(x) dx = F(x)|_{a}^{b} = F(b) - F(a)\].
This typically states the definite integral over an interval [a,b] is equivalent to the antiderivative calculated at ‘b’ minus the antiderivative assessed at ‘a’. This provides the link between the definite integral and the indefinite integral (antiderivative).
This outcome, while taught initially in primary calculus courses, is literally an intense outcome linking the purely algebraic indefinite integral and the purely evaluative geometric definite integral.
Given f is
continuous on interval [a, b]
F is any function that satisfies F’(x) = f(x)
Then
\[\int_{a}^{b} f(X)dx = F(b) - F(a)\]
B. Fundamental Theorem of Calculus Part 2
Within the theorem the second fundamental theorem of calculus, depicts the connection between the derivative and the integral— the two main concepts in calculus. The part 2 theorem is quite helpful in identifying the derivative of a curve and even assesses it at definite values of the variable when developing an anti-derivative explicitly which might not be easy otherwise.
The theorem bears ‘f’ as a continuous function on an open interval I and ‘a’ any point in I, and states that if F is demonstrated by
F(x) = \[\int_a^{x} f(t)dt\],
then
F'(x)=f(x) at each point in I.
The above expression represents that The fundamental theorem of calculus by the sides of curves shows that if f(z) has a continuous indefinite integral F(z) in an area R comprising of parameterized curve $\gamma:z=z(t) for \alpha < = t < = \beta$, then
\[\int_\gamma f(z)dz = F(z(\beta))-F(z(\alpha))\].
\[\frac{d}{dx} \int_{a}^{x} f(t)dt = f(x)\]
Derivative of an integral.
Derivative matches the upper limit of integration.
Lower limit of integration is a constant.
Important Formulas of Class 12 Chapter 7 You Shouldn’t Miss!
1. Basic Integration Formulas:
\[\int k \, dx = kx + C\]
\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{for } n \neq -1\]
\[\int \frac{1}{x} \, dx = \ln|x| + C\]
2. Integration of Trigonometric Functions:
\[\int \sin x \, dx = -\cos x + C\]
\[\int \cos x \, dx = \sin x + C\]
\[\int \sec^2 x \, dx = \tan x + C\]
\[\int \csc^2 x \, dx = -\cot x + C\]
\[\int \sec x \tan x \, dx = \sec x + C\]
\[\int \csc x \cot x \, dx = -\csc x + C\]
3. Integration by Substitution:
\[\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du\]
$\text{(where }u=g(x)\text{)}$
4. Integration by Parts:
\[\int u \, dv = uv - \int v \, du\]
5. Definite Integrals:
\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]
$\text{(where }F\text{ is the antiderivative of }f)$
Importance of Integrals Class 12 Notes
Fundamental Concepts: Integrals are the inverse process of differentiation, essential for understanding the broader concepts of calculus. They help in solving problems related to area, volume, and other quantities that require summing infinitesimal parts.
Application in Real Life: Integrals are widely used in fields such as physics, engineering, and economics. For example, they are crucial in calculating areas under curves, solving problems related to motion, and finding accumulated quantities.
Advanced Topics: A strong grasp of integrals is necessary for tackling more advanced topics in calculus, including differential equations and multi-variable calculus, which are important for higher studies.
Problem Solving: Integration techniques like substitution, integration by parts, and partial fractions are critical tools for solving complex problems and simplifying functions, which is a significant aspect of the Class 12 curriculum.
Exam Preparation: Integrals are a major component of the Class 12 syllabus and often feature prominently in exams. Understanding their properties, methods, and applications helps in achieving better results and developing a deeper appreciation of mathematics.
Tips for Learning the Class 12 Maths Chapter 7 Integrals
Here are some useful tips for learning Class 12 Maths Chapter 7, Integrals:
Understand the Basics:
Ensure you have a strong grasp of fundamental concepts such as differentiation and basic integration formulas. This will make learning integrals easier.
Master Key Formulas:
Memorise and understand the key integration formulas, including basic integration, trigonometric integrals, and techniques such as substitution and integration by parts.
Practice Regularly:
Solve a variety of problems to reinforce your understanding. Practice integrating different types of functions to become familiar with various techniques and approaches.
Use Visual Aids:
Graph functions and their integrals to visually understand the concept of areas under curves and the geometric interpretation of integration.
Learn Integration Techniques:
Focus on different methods such as integration by substitution, integration by parts, and partial fractions. Practice applying these techniques to different types of integrals.
Review Examples:
Study solved examples from your textbook and notes to see step-by-step solutions. Pay attention to the methods used and try to solve similar problems on your own.
Work on Definite Integrals:
Understand how to compute definite integrals and apply the Fundamental Theorem of Calculus. Practice finding the area under curves using definite integrals.
Conclusion
Class 12 Maths Chapter 7: Integrals is crucial for your success in calculus and future mathematical studies. Our FREE PDF notes offer a comprehensive guide to understanding integration concepts, techniques, and applications. By utilising these notes, you can strengthen the concepts of integrals, improve problem-solving skills, and enhance your exam preparation. Download the PDF to take a significant step towards achieving excellence in your Class 12 Mathematics course.
Related Study Materials for Class 12 Maths Chapter 7 Integrals
Students can also download additional study materials provided by Vedantu for Class 12 Maths Chapter 7 Integrals:
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FAQs on Integrals Class 12 Notes: CBSE Maths Chapter 7
1. What are the key concepts summarised in the Class 12 Integrals revision notes?
The revision notes for Class 12 Integrals cover the definition of integration, standard and advanced integration techniques (such as substitution, integration by parts, and partial fractions), properties of definite and indefinite integrals, key formulas, and applications of integrals as per the CBSE syllabus. These notes also outline the relationship between integration and differentiation (Fundamental Theorem of Calculus).
2. How should students structure their revision for Class 12 Maths Chapter 7: Integrals?
A recommended revision sequence includes:
- Reviewing fundamental integration formulas and definitions first.
- Practicing basic and advanced techniques (substitution, integration by parts, partial fractions).
- Solving examples related to definite and indefinite integrals.
- Applying key properties and theorems, such as the Fundamental Theorem of Calculus.
- Completing practice questions on the application of integrals (area, other real-life uses).
3. What are the main differences between definite and indefinite integrals?
Indefinite integrals give a general antiderivative and always include a constant of integration, whereas definite integrals are evaluated over specific limits and yield a numerical value representing the area under a curve. Definite integrals do not include the constant 'C' but are calculated as F(b) - F(a), where F(x) is the antiderivative of f(x).
4. How do the revision notes for Integrals support quick and efficient exam preparation?
The notes provide concise explanations of all core concepts, summarise essential integration formulas, and offer problem-solving strategies that are in line with the CBSE Class 12 syllabus. They help students focus only on what is required for board exams and eliminate unnecessary information, making revision faster and more targeted.
5. Which core formulas should be prioritised during last-minute revision of Integrals?
Students should focus on memorising and understanding the following formulas:
- ∫xn dx = xn+1⁄(n+1) + C
- ∫1/x dx = ln|x| + C
- Standard trigonometric integrals (e.g., ∫sinx, ∫cosx, etc.)
- ∫ex dx = ex + C
- Basic properties of definite integrals (additivity, reversal of limits, etc.)
6. What misconceptions do students often have regarding integration techniques?
Common misconceptions include:
- Confusing integration by substitution with integration by parts.
- Omitting the constant of integration 'C' in indefinite integrals.
- Misapplying limits in definite integrals and not evaluating the antiderivative at both upper and lower limits.
- Assuming all integration rules or substitutions work universally, which is not always true—choice of technique often depends on the integrand's form.
7. How are integrals applied in real-world scenarios and other mathematical topics?
Integrals are used to calculate areas under curves, determine accumulated quantities (such as distance from velocity), and solve physics and engineering problems. In mathematics, integrals are foundational for differential equations, probability distributions, and multi-variable calculus.
8. What is the importance of understanding the Fundamental Theorem of Calculus for Class 12 students?
The Fundamental Theorem of Calculus links differentiation and integration, showing that integration can be reversed by differentiation and vice versa. This concept is crucial for solving both definite and indefinite integrals and is frequently tested in board exams, forming the basis for higher-level calculus.
9. Which methods should be compared when deciding how to integrate a function in exams?
Students should compare the suitability of substitution, integration by parts, and partial fraction decomposition based on the function's form:
- Use substitution for composite or chain rule patterns.
- Use integration by parts when the function is a product of two different types (e.g. algebraic and trigonometric).
- Use partial fractions for rational functions where denominator factorisation is possible.
10. How can revision notes be used most effectively during the exam season?
Integrate revision notes into your study routines by reviewing key formulas regularly, summarising core rules in short concept maps, and actively solving example problems from the notes. This revisiting strategy reinforces memory, clarifies difficult areas, and ensures readiness for board exam questions from the Integrals chapter.











