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Integrals Class 12 Notes CBSE Maths Chapter 7 (Free PDF Download)

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Revision Notes for CBSE Class 12 Maths Chapter 7 (Integrals) - Free PDF Download

If you find solving maths problems difficult and beyond your capabilities, you would change your mind once you access Class 12 Maths Integrals Revision Notes prepared by the Vedantu team's scholars. Our subject matter experts make even a complicated problem look simple by breaking it down into small steps, which will give you a good foundation. Calculus is an important and tricky topic; hence you need professional help that will prepare you for your CBSE exams and other entrance exams like IIT, NEET, etc. Based entirely on the CBSE curriculum, our Revision Notes Class 12 Chapter 7 will give you precisely what you need to ace your examination.

CBSE Class 12 Maths Revision Notes 2024-25 - Chapter Wise PDF Notes for Free

In the table below we have provided the PDF links of all the chapters of CBSE Class 12 Maths whereby the students are required to revise the chapters by downloading the PDF. 


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Integrals Class 12 Notes Maths - Basic Subjective Questions

1. Find $\int(1-x) \sqrt{x} d x$.

Ans. $\int(1-x) \sqrt{x} d x=\int(\sqrt{x}-x \sqrt{x}) d x$

$=\int\left(x^{1 / 2}-x^{3 / 2}\right)d x=\frac{2}{3} x^{3 / 2}-\frac{2}{5} x^{5 / 2}+C \\$ 

$\left [ \because \int x^n d x=\frac{x^{n+1}}{n+1},n\neq -1 \right ]$


2. Evaluate $\int \frac{2}{1+\cos 2 x} d x$.

Ans. $\int \frac{2}{1+\cos 2 x} d x=\int \frac{2}{2 \cos ^2 x} d x$

$$=\int \sec ^2 x d x=\tan x+C .$$


3. Find $\int_4^5 e^x d x$.

Ans. $\int_4^5 e^x d x=\left[e^x\right]_4^5=e^5-e^4=e^4(e-1)$.


4. Find $\int_0^{\pi / 4} \tan x d x$.

Ans. $\int_0^{\pi / 4} \tan x d x=[\log |\sec x|]_0^{\pi / 4}$

$$ \begin{aligned} & =\log \left|\sec \frac{\pi}{4}\right|-\log |\sec 0| \\ & =\log \sqrt{2}-\log 1 \\ & =\log \sqrt{2} . \end{aligned} $$


5. Evaluate $\int_0^1 \frac{d x}{\sqrt{1-x^2}}$

Ans. $\int_0^1 \frac{d x}{\sqrt{1-x^2}}=\left[\sin ^{-1} x\right]_0^1$

$$\begin{aligned}& =\sin ^{-1}(1)-\sin ^{-1}(0) \\& =\frac{\pi}{2}-0=\frac{\pi}{2}\end{aligned}$$


Section-B (2 Marks Questions)

6. Evaluate $\int\left(\operatorname{cosec}^2 x-\cot x\right) e^x d x$.

Ans.  $\int\left(\operatorname{cosec}^2 x-\cot x\right) e^x d x$

$=-\int\left(\cot x-\operatorname{cosec}^2 x\right) e^x d x \\$

$=-\int\left[\cot x+\left(-\operatorname{cosec}^2 x\right)\right] e^x d x=-e^x \cot x+C \\$

${\left[\therefore  \int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x \cdot f(x)\right]}$



7. Evaluate $\int \sin ^3 x d x$.

Ans.\begin{aligned} & \int \sin ^3 x d x=\int \frac{3 \sin x-\sin 3 x}{4} d x \\ &= \frac{1}{4} \int 3 \sin x d x-\frac{1}{4} \int \sin 3 x d x \\ &= \frac{1}{4}\left(-3 \cos x+\frac{\cos 3 x}{3}\right)+C . \end{aligned}


8. Evaluate $\int \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}} d x$.

Ans. $\int \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}} d x=\int \sqrt{\frac{2 \sin ^2 x}{2 \cos ^2 x}} d x$

$$ \begin{aligned} & =\log \left|\sec \frac{\pi}{4}\right|-\log |\sec 0| \\ & =\log \sqrt{2}-\log 1 \\ & =\log \sqrt{2} . \end{aligned} $$


9. Write the value of $\int \frac{2-3 \sin x}{\cos ^2 x} d x$.

Ans.

$\int \frac{2-3 \sin x}{\cos ^2 x} d x$

$=\int\left(\frac{2}{\cos ^2 x}-\frac{3 \sin x}{\cos ^2 x}\right) d x $

$=\int \left ( 2\;sec^{2}x-3\;sec\;x\;tan\;x \right )dx$

$=2 \int \sec ^2 x d x-3 \int \sec x \tan x d x \\$ 

$=2 \tan x-3 \sec x+C$


10. Evaluate $\int \frac{2 \cos x}{\sin ^2 x} d x$

Ans. $\int \frac{2 \cos x}{\sin ^2 x} d x=\int 2 \operatorname{cosec} x \cot x d x$

$=-2 \operatorname{cosec} x+C \\

$\left [ \because \int \operatorname{cosec} x \cot x d x=-\operatorname{cosec} x \right ]$



11. Write the value of $\int_{\frac{-\pi}{4}}^0 \frac{1+\tan x}{1-\tan x} d x$

$$I=\int_{\frac{-\pi}{4}}^0 \frac{1+\tan x}{1-\tan x} d x=\int_{\frac{-\pi}{4}}^0 \tan\left(\frac{\pi}{4}+x\right) d x$$

Ans. $\begin{aligned}& =\left[\log \left|\sec\left(\frac{\pi}{4}+x\right)\right|\right]_{\dfrac{-\pi}{4}}^0 \\& =\log \sqrt{2}=\frac{1}{2} \log 2\end{aligned}$


12. Find $\int \log x d x$.

Ans. $\int(\log x .1) d x=\log x \int 1 d x-\int\left[\frac{d}{d x}(\log x) \int 1 d x\right] d x=(\log x) x-\int \frac{1}{x} x d x=x \log x-x+C$.


13. If $\int_0^a \frac{1}{4+x^2} d x=\frac{\pi}{8}$, then find the value of $a$.

Ans.

We have, $\int_0^a \frac{1}{4+x^2} d x=\frac{\pi}{8}$

$$\begin{aligned} & \Rightarrow \frac{1}{2}\left[\tan ^{-1} \frac{x}{2}\right]_0^a=\frac{\pi}{8} \\ & \Rightarrow \tan ^{-1} \frac{a}{2}-0=\frac{\pi}{4} \\ & \Rightarrow \frac{a}{2}=1 \Rightarrow a=2 \\ & \Rightarrow \tan ^{-1} \frac{a}{2}=\frac{\pi}{4} \end{aligned}$$


PDF Summary - Class 12 Maths Integrals Notes (Chapter 7)


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

  1. $\quad \dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {\int {\text{f}} ({\text{x}}){\text{dx}}} \right] = f({\text{x}})$

  2. $\int {{f^\prime }} (x){\text{dx}} = \int {\dfrac{{\text{d}}}{{{\text{dx}}}}} [f({\text{x}})]{\text{dx}} = f({\text{x}}) + {\text{c}}$

  3. $\quad \int {\text{k}} f({\text{x}}){\text{dx}} = {\text{k}}\int f ({\text{x}}){\text{dx}}$, where ${\text{k}}$ is any constant

  4. 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{ } $

  1. 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.

  1. $\quad \int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne  - 1$.

  2. $\quad \int {\dfrac{1}{x}} dx = \log |x| + C$

  3. $\int {{{\text{e}}^{\text{x}}}} {\text{dx}} = {{\text{e}}^{\text{x}}} + {\text{C}}$

  4. $\int {{{\text{a}}^{\text{x}}}} {\text{dx}} = \dfrac{{{{\text{a}}^{\text{x}}}}}{{{{\log }_{\text{e}}}{\text{a}}}} + {\text{C}}$.

  5. $\quad \int {\sin } xdx =  - \cos x + C$

  6. $\quad \int {\cos } xdx = \sin x + C$

  7. $\int {{{\sec }^2}} xdx = \tan x + C$

  8. $\quad \int {{{\operatorname{cosec} }^2}} xdx =  - \cot x + C$

  9. $\quad \int {\sec } x\tan xdx = \sec x + C$

  10. $\int {\operatorname{cosec} } x\cot xdx =  - \operatorname{cosec} x + C$.

  11. $\int {\tan } xdx =  - \log |\cos x| + C = \log |\sec x| + C$.

  12. $\int {\cot } xdx = \log |\sin x| + C$

  13. $\int {\sec } xdx = \log |\sec x + \tan x| + C$

  14. $\int {\operatorname{cosec} } xdx = \log |\operatorname{cosec} x - \cot x| + C$

  15. $\int {\dfrac{{{\text{dx}}}}{{\sqrt {1 - {{\text{x}}^2}} }}}  = {\sin ^{ - 1}}{\text{x}} + {\text{C}};|{\text{x}}| < 1$

  16. $\int {\dfrac{{dx}}{{1 + {x^2}}}}  = {\tan ^{ - 1}}x + C$

  17. $\int {\dfrac{{{\text{dx}}}}{{{\text{x}}\sqrt {{{\text{x}}^2} - 1} }}}  = {\sec ^{ - 1}}|{\text{x}}| + {\text{C}};|{\text{x}}| > 1$


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.

  1. 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$

  1. 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, $\int {{x^{n - 1}}} {f^\prime }\left( {{x^n}} \right)dx = \int {{f^\prime }} (t)\dfrac{{dt}}{n} = \dfrac{1}{n}$

$\int {{f^\prime }} ({\text{t}}){\text{dt}} = \dfrac{1}{{\text{n}}}f({\text{t}}) = \dfrac{1}{{\text{n}}}f\left( {{{\text{x}}^{\text{n}}}} \right) + {\text{c}}$

  1. 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, ${\text{(Tex translation failed)}}$

  1. 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

  1. $\int {\dfrac{{dx}}{{{x^2} + {a^2}}}}  = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C$

  2. $\quad \int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^2} - {{\text{a}}^2}}}}  = \dfrac{1}{{2{\text{a}}}}\log \left| {\dfrac{{{\text{x}} - {\text{a}}}}{{{\text{x}} + {\text{a}}}}} \right| + {\text{C}}$

  3. $\quad \int {\dfrac{{dx}}{{{a^2} - {x^2}}}}  = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + C$

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

  5. $\quad \int {\dfrac{{dx}}{{\sqrt {{x^2} + {a^2}} }}}  = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C$

  6. $\int {\dfrac{{dx}}{{\sqrt {{x^2} - {a^2}} }}}  = \log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$

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

  8. $\quad \int {\sqrt {{x^2} + {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} + {a^2}}  + \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C$

  9. $\quad \int {\sqrt {{x^2} - {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} - {a^2}}  - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$


2.1.2 Integrals of the Form

  1. $\int f \left( {{{\text{a}}^2} - {{\text{x}}^2}} \right){\text{dx}}$,

  2. $\int f \left( {{a^2} + {x^2}} \right)dx$,

  3. $\int f \left( {{x^2} - {a^2}} \right)dx$,

  4. $\int f \left( {\dfrac{{a - x}}{{a + x}}} \right)dx$,

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

  1. $\int {\dfrac{{dx}}{{a{x^2} + bx + c}}} $

  2. $\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}} }}} $

  3. $\int {\sqrt {a{x^2} + bx + c} } dxa$,

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

  1. $\int {\dfrac{{px + q}}{{a{x^2} + bx + c}}} dx$

  2. $\int {\dfrac{{{\text{px}} + {\text{q}}}}{{\sqrt {a{x^2} + bx + c} }}} {\text{dx}}$

  3. $\int {(px + q)} \sqrt {a{x^2} + bx + c} dx$

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

  1. Choose the first and second functions in such a way that the derivative and integral of the first function may be determined quickly.

  2. 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.

  3. 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.

  4. The rule of parts integration can be applied several times if necessary.

  5. 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.

  1. 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.

  2. 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 

  1. $\int {\dfrac{{dx}}{{a + b\cos x}}} $

  2. $\int {\dfrac{{dx}}{{a + b\sin x}}} $

  3. $\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

  1. $\int {\dfrac{{dx}}{{a + b{{\cos }^2}x}}} $

  2. $\int {\dfrac{{dx}}{{a + b{{\sin }^2}x}}} $

  3. $\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 $\tan x = z$.


Solved Examples:

Example 1 

 Evaluate: ${\text{ }}\int {\left( {{x^3} + 5{x^2} - 4 + \dfrac{7}{x} + \dfrac{2}{{\sqrt x }}} \right)} dx$

Sol: ${\text{ }}\int {\left( {{x^3} + 5{x^2} - 4 + \dfrac{7}{x} + \dfrac{2}{{\sqrt x }}} \right)} dx$

$ = \int {{x^3}} dx + \int 5 {x^2}dx - \int 4 dx + \int {\dfrac{7}{x}} dx + \int {\dfrac{2}{{\sqrt x }}} dx$

$ = \int {{x^3}} dx + 5 \cdot \int {{x^2}} dx - 4 \cdot \int 1  \cdot dx + 7 \cdot \int {\dfrac{1}{x}} dx + 2 \cdot \int {{x^{ - 1/2}}} dx$

$ = \dfrac{{{x^4}}}{4} + 5 \cdot \dfrac{{{x^3}}}{3} - 4x + 7\log |x| + 2\left( {\dfrac{{{x^{1/2}}}}{{1/2}}} \right) + C$

$ = \dfrac{{{x^4}}}{4} + \dfrac{5}{3}{x^3} - 4x + 7\log |x| + 4\sqrt x  + C$


Example 2

Evaluate: $\int {{{\text{e}}^{{\text{x}}\log a}}}  + {{\text{e}}^{{\text{a}}\log x}} + {{\text{e}}^{{\text{a}}\log a}}{\text{dx}}$

Sol. We have,

$\int {{{\text{e}}^{{\text{x}}\log a}}}  + {{\text{e}}^{{\text{a}}\log x}} + {{\text{e}}^{{\text{a}}\log a}}{\text{dx}}$

$ = \int {{{\text{e}}^{\log {a^{\text{x}}}}}}  + {{\text{e}}^{\log {x^{\text{a}}}}} + {{\text{e}}^{\log {{\text{a}}^{\text{a}}}}}{\text{dx}}$

$ = \int {\left( {{a^x} + {x^a} + {a^a}} \right)} dx$

$ = \int {{a^x}} dx + \int {{x^a}} dx + \int {{a^a}} dx$

$ = \dfrac{{{a^x}}}{{\log a}} + \dfrac{{{x^{a + 1}}}}{{a + 1}} + {a^a} \cdot x + C$


Example 3

Evaluate: $\int {\dfrac{{{x^4}}}{{{x^2} + 1}}} dx$

Sol. $\int {\dfrac{{{x^4}}}{{{x^2} + 1}}} dx$

$ = \int {\dfrac{{{x^4} - 1 + 1}}{{{x^2} + 1}}} dx = \int {\dfrac{{{x^4} - 1}}{{{x^2} + 1}}}  + \dfrac{1}{{{x^2} + 1}}dx$

$ = \int {\left( {{x^2} - 1} \right)} dx + \int {\dfrac{1}{{{x^2} + 1}}} dx = \dfrac{{{x^3}}}{3} - x + {\tan ^{ - 1}}x + C$


Example 4

Evaluate: $\int {\dfrac{{{2^x} + {3^x}}}{{{5^x}}}} dx$

Sol. $\int {\dfrac{{{2^x} + {3^x}}}{{{5^x}}}} dx$

$ = \int {\left( {\dfrac{{{2^x}}}{{{5^x}}} + \dfrac{{{3^x}}}{{{5^x}}}} \right)} dx$

$ = \int {\left[ {{{\left( {\dfrac{2}{5}} \right)}^x} + {{\left( {\dfrac{3}{5}} \right)}^x}} \right]} dx = \dfrac{{{{(2/5)}^x}}}{{{{\log }_c}2/5}} + \dfrac{{{{(3/5)}^x}}}{{{{\log }_c}3/5}} + C$


Example 5

Evaluate: $\int {{{\text{x}}^3}} {\sin ^4}{\text{dx}}$

Sol. We have

${\text{I}} = \int {{{\text{x}}^3}} {\sin ^4}{\text{dx}}$

Let ${x^4} = t\quad  \Rightarrow \quad d\left( {{x^4}} \right) = dt$

$ \Rightarrow \quad 4{x^3}dx = dt \Rightarrow \quad dx = \dfrac{1}{{4{x^3}}}dt$


NCERT Class 12 Revision Notes Maths Chapter 7 Solution PDF

You can study together in a group or access the solutions offline for a quick revision of formulas and solutions with the  Revision Notes Class 12 Maths Chapter 7 available in PDF form on the official website of Vedantu. These NCERT Maths Class 12 Chapter 7 Integrals notes can also be printed out so that you can refer to them as per your convenience.


NCERT Solutions Chapter 7 Class 12 Maths Revision Notes 

Definition of Integration - Integration is adding parts to find the whole and is used in mathematics to find areas, volumes, etc. The integration process is mathematically an inverse of differentiation. So if the differential coefficient of a function F(x) is f(x), which means:

d[F(x)]/dx = f(x) then we can call F(x) as the antiderivative or integral of f(x). It is denoted as ∫f(x)dx = F(x)

The notation ∫ is for integration and f(x) is referred to as the integrand. In this equation, x is the variable of integration and dx represents the integration with respect to x.


Indefinite Integral - In Class 12 Maths Revision Notes solution Chapter 7, you would learn that an indefinite integral is a function that takes another function's antiderivative function. Let us clarify this with the example discussed above where:

If d[F(x)]/dx = f(x) then ∫f(x)dx = F(x). We also know that for any arbitrary constant K:

d[F(x) + K]/dx = d[F(x)]/dx + 0 = f(x) . Or, we can say:

 ∫f(x)dx = F(x) + K

This goes to show that both F(x) and F(x) + K are integrals of the same function f(x). Hence, by changing K's value, we can get different integrals of f(x). By this virtue, the integral of f(x) is not a definite value; hence F(x) is called the indefinite integral of f(x).


Properties of Indefinite Integral

  • The derivative and the integrand are equal to each other in an indefinite integral.

(∫f(x)dx)’ = f(x)

  • If we calculate the indefinite integral of the sum of two functions, it is equal to the sum of the integrals of two functions:

∫[f(x) + g(x)]dx = ∫f(x)dx + ∫g(x)dx

  • If we calculate the indefinite integral of the difference of two functions, it is equal to the difference of the integrals of two functions:

∫[f(x) - g(x)]dx = ∫f(x)dx - ∫g(x)dx

  • If there is a constant in the integral, it can be moved outside of the integration.

∫Cf(x)dx = C∫f(x)dx

  • If ∫f(x)dx = F(x) + K, then ∫f(ax + y)dx = 1/a F(ax + y) + K


Some Special Integrals

  • ∫dx/(x2 + y2) = 1/y tan-1 x/y + K

  • ∫dx/(x2 - y2) = 1/2y * log |x - y/x + y| + K

  • ∫dx/(y2 - x2) = 1/2y * log |y + x/y - x| + K

  • ∫dx/√(y2 - x2) = Sin-1 x/y + K

  • ∫dx/√(x2 + y2) = log |x + √(x2 + y2)| + K


Method of Integration by Parts 

Integration by parts is the method of integration of the product of two functions. Let us say g and z are two functions of x, then:

∫(gz)dx = g ∫ zdx - ∫(dg/dx ∫zdx)dx

Which Can Be Explained In Words As Follows:

Integral of the product of 2 functions = first function * integral of the 2nd function - integral of the differential of the first functions * integral of the second function.


NCERT Class 12 Revision Notes Maths Chapter 7 Solution PDF

Enhance your study experience by collaborating with peers or utilizing offline resources for swift formula and solution reviews. Explore the PDF format of Revision Notes for Class 12 Maths Chapter 7, Integrals, available on Vedantu's official website. These accessible notes can be conveniently printed for personalized reference, ensuring flexibility in your study routine.


What are the Benefits of Referring to Vedantu’s Revision Notes for Class 12 Chapter 7 - Integrals?

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  • Provides quick, clear summaries of key concepts.

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  • Efficient tool for last-minute exam prep.

  • Enhances retention of crucial information.

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  • Prioritizes important topics and questions.

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  • Boosts student confidence for exams.


Conclusion

CBSE Class 12 Maths Chapter 7, 'Integrals,' revision notes offer a comprehensive understanding of crucial concepts. Covering essential properties and characteristics, these well-organized materials facilitate efficient revision for enhanced retention. With practical examples and real-life applications, students gain a deeper insight into the subject. These revision notes play a vital role in establishing a good foundation in Integrals, contributing to academic excellence. Proven to be invaluable, they empower students to succeed in their studies by providing the necessary clarity and support.


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FAQs on Integrals Class 12 Notes CBSE Maths Chapter 7 (Free PDF Download)

1. How would you solve integrals of trigonometric formulas like; ∫dx/y + z Cos(x)

The working rule for solving such integrals is:

  1. In place of Cos(x) put (1 - tan2 x/2)/(1 + tan2x/2) and for Sin x replace with (2tan x/2)/(1 + tan2 x/2). This is done so that the integrand becomes a function of tan x/2.

  2. Now replace tan x/2 with ½ Sec2 x/2 dx = dz

  3. Now integrate the rational algebraic function of z

  4. In the result replace z with tan x/2

2. What is the difference between a definite integral and an indefinite integral?

In an indefinite integral there are no limits of integration whereas a definite integral is bound by lower and upper limits which are constant. 

A Definite Integral is denoted as - ∫ab f(x), Here a and b are constants.

An Indefinite Integral is denoted as - ∫f(x)

3. What are the important questions for Chapter 7 of Class 12 Maths?

Chapter 7 of Class 12 Maths is an important Chapter in terms of your final board examination as well as for other competitive exams. This Chapter explains the concept of Integration, which includes integral calculus and its properties. There are many important questions in this Chapter that come from very important concepts and topics. These include evaluation of integrals, determination of integrals, functions of integrals, anti-derivatives, and various methods of integrations.

4. Where can I find revision notes for Chapter 7 of Class 12 Maths?

The board question papers generally rely on NCERT textbooks and the concepts it covers. Therefore, studying from NCERT textbooks can ensure that you will score good marks. For Chapter 7 of Class 12 Maths, you can revise notes based on NCERT Solutions on Vedantu. It provides all important questions and solutions. These solutions and questions are extremely important for all CBSE students from the viewpoint of examinations. Also, the solutions PDF is available for free download on the Vedantu mobile app.

5. How can I score well in Chapter 7 of Class 12 Maths?

To score well in Chapter 7 of Class 12 Maths, you need to keep practising. You need to have a personal notebook where you can write all the notes. This will help you to remember the concepts and topics that you learned. This will help you prepare thoroughly, and it will ensure that you don’t forget all the important concepts and topics you learned. It will cater you with knowledge and a better understanding of the subject. Making notes will also be helpful during your revision time.

6. What are the basics of integration, according to Chapter 7 of Class 12 Maths?

Chapter 7 “Integrals” of Class 12 Maths is an easy chapter and will help you score good marks. Integration refers to the addition of parts to find the whole. This technique is used to find volumes, areas, and so on. This technique is a mathematical inverse of differentiation. There are many important formulas in this chapter that you must learn and understand for you to know the basics of this chapter. Apart from this, you must understand the different types of methods in Integration. The students can also refer to the Revision Notes for Chapter 7 of Class 12 Maths free of cost from the Vedantu website or from the Vedantu app.

7. What are the properties of Indefinite Integral, according to Chapter 7 of Class 12 Maths?

Indefinite Integral is the function that uses another function's antiderivative function. Some of the properties include:

  • In an Indefinite Integral, the integrand and the derivative are equal to each other.

  • The Indefinite Integral of the sums of two functions equals the integral of the sums of two functions. 

  • The difference between the two functions of indefinite integrals is equal to that of the two functions. 

  • An integral always has a constant, which can be moved outside of the integration.