Class 12 Maths Formula Ch7 Indefinite Integral
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INDEFINITE INTEGRALS
Class 12 Maths Formula Ch7 Indefinite Integral
IMPORTANT FORMULAS USED IN INTEGRALS
Integration: In chapter 5 we have learnt the method of finding the
derivatives of a function. Further there are formulas which help us to reverse
the process of derivatives. This process is called antiderivative or
Integration.
Types of Integrals: Integrals are of two types Indefinite
Integral and Definite Integral.
Let us see the problems of
the following types:
a) the problem of
finding the function whenever the derivative of the function is given.
b) the problem of finding the area of the graph of a function under the given
conditions.
Problem 'a' leads to the form of integral called
Indefinite Integral. Problem 'b' leads to the form of integral called Definite
Integral.
Integral Calculus
Indefinite Integral and Definite Integral together constitute the Integral
Calculus.
👉In this blog we will study all that
formulas which are helpful in finding the indefinite integrals.
Indefinite Integrals:
The formulas which gives
antiderivative are called indefinite integrals.
Integration is the
inverse process of differentiation.
Before start the formulas first study the following table.
Symbols/Terms/Phrases 
Meaning 
\[\int
f(x)dx\] 
Integral
of f with respect to x 
\[f(x)\:
in\: \int f(x)dx\] 
Integrand 
\[x\:
\: in\: \int f(x)dx\] 
Variable
of integration 
Integrate 
Find
the integral 
Constant
of integration 
Any
real number C, considered as constant function, 
Integration as the reverse process of differentiation
Geometrical Interpretation:
Indefinite integral represents the integral as family of curves
Consider the curve f(x) = 2x, antiderivative of 2x w. r. t. x is x^{2} + C = y say
f’(x) = x^{2} + C = y (say)
For C = 0, y = x^{2}
For C= 1, y = x^{2} + 1
For C = 1, y = x^{2} – 1
For C = 3, y = x^{2}  3
Thus, we get the family of parabolas whose vertex moves on y axis for different values of C. which can be seen in figure below. This gives the geometrical interpretation of indefinite integral.
Thus, we may conclude : indefinite integral gives the family of curves members of which can be
obtained by shifting any one of them parallel to itselfImportant Formulas used for finding Integration
\[\int dx=x+C\]\[\int kdx=kx+C\]\[\int 0dx=C\]\[\int x^{n}dx=\frac{x^{n+1}}{n+1} + C\]\[\int \frac{1}{x}dx=log x+C\]\[\int e^{x}dx=e^{x}+C\]\[\int e^{ax}dx=\frac{e^{ax}}{a}+C\]\[\int a^{x}dx=\frac{a^{x}}{loga}+C\]\[\int [f(x)]^{n}.f^{'}(x)dx=\frac{[f(x)]^{n+1}}{n+1} + C\]\[\int \frac{f^{'}(x)}{f(x)}dx=log\left  f(x) \right +C\]\[\int \frac{f^{'}\left ( x \right )}{\left [ f(x) \right ]^{n}}dx=\frac{\left [ f(x) \right ]^{n+1}}{n+1}\]
STANDARD INTEGRALS

INTEGRAL WITH TRIGONOMETRY
\[\int sinxdx=cosx+C\]
\[\int sin(ax)dx=\frac{cos(ax)}{a}+C\]
\[\int cosxdx=sinx+C\]
\[\int tanxdx=log\left  secx \right  + C\] or
\[\int tanxdx=log\left  cosx \right +C\]
Explanation:
Explanation:
*****************************************
\[Explanation: \:\: \int
cosecx dx=logcosecxcotx\]\[=log\left \frac{cosecscots}{1}\times
\frac{cosecx+cotx}{cosecx+cotx} \right \] \[=log\left 
\frac{cosec^{2}xcot^{2}x}{cosecx+cotx} \right \] \[ =log\left 
\frac{1}{cosecx+cotx} \right \] \[=log\left ( cosecx+cotx \right
)^{1}=log\left  cosecx+cotx \right \]\[\int cosecx
dx=logcosecxcotx =log\left  \frac{1}{sinx}\frac{cosx}{sinx} \right \]
\[=log\left  \frac{1cosx}{sinx} \right =log\left 
\frac{2sin^{2}\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}} \right \]
\[=log\left  \frac{sin\frac{x}{2}}{cos\frac{x}{2}} \right  =log\left 
tan\frac{x}{2} \right \]*********************************\[\int sec^{2}xdx=tanx+C\]\[\int
cosec^{2}xdx=cotx+C\]\[\int secx tanx dx=secx+C\]\[\int cosecx cotx
dx=cosecx+C\]
INTEGRATION WITH INVERSE TRIGONOMETRIC FUNCTIONS 

\[\int \frac{1}{\sqrt{1x^{2}}}dx=sin^{1}x+C\] 
\[\int
\frac{1}{\sqrt{1x^{2}}}dx=cos^{1}x+C\] 
\[\int \frac{1}{1+x^{2}}dx=tan^{1}x+C\] 
\[\int \frac{1}{1+x^{2}}dx=cot^{1}x+C\] 
\[\int
\frac{1}{x\sqrt{x^{2}1}}dx=sec^{1}x+C\] 
\[\int
\frac{1}{x\sqrt{x^{2}1}}dx=cosec^{1}x+C\] 
Methods of Integrations:
There are mainly three
methods of integration
i) Integration by
substitution.
ii) Integration by
partial fraction.
iii) Integration by parts.
Integration by Substitution(Example)
Integrate tanx by substitution method\[\int tanx dx=\int
\frac{sinx}{cosx}dx\]\[Putting\: \: cosx\: =t,\: \Rightarrow\: sinx
dx=dt\]\[\int tanx dx=\int \frac{dt}{t}=log\: t+C\]\[Now\: replacing\: \:
t\:\: by\: cosx\: we\: get\]\[\int tanx dx=log cosx+C\]
Now we shall explain the method of finding the integral of the type\[I=\int \frac{px+q}{ax^{2}+bx+c}dx\] For finding this integral we put \[px+q=A\frac{d}{dx}\left ( ax^{2}+bx+c \right )+B\] Now find the value of A and B by comparing the coefficient of x and the constant term. A and B are thus formed and hence and hence the integral is reduced to the one of the known form
Chapter 7 Exercise 7.4 Q
No. 18
\[I=\int \frac{5x2}{3x^{2}+2x+1}dx\]\[Let\:
5x2=A\frac{d}{dx}(3x^{2}+2x+1)+B\]\[5x2=A(6x+2)+B
…………….(1)\]\[5x2=6Ax+2A+B\] Comparing the coefficient of x we
get\[5=6A\: \:
\Rightarrow \; \; A=\frac{5}{6}\] Comparing the constant terms we
get\[2A+B=2\]Putting the value of A in this equation we get \[2\times
\frac{5}{6}+B=2\: \Rightarrow B=2\frac{5}{3}= \frac{11}{3}\]Putting values
of A and B in equation (1) we get\[5x2=\frac{5}{6}(6x+2)\frac{11}{3}\] \[I=\int
\frac{\frac{5}{6}(6x+2)\frac{11}{3}}{3x^{2}+2x+1}dx\] Now separating the
denominator and solving the integral.\[I=\int
\frac{\frac{5}{6}(6x+2)}{3x^{2}+2x+1}dx\frac{11}{3}\int\frac{dx}{3x^{2}+2x+1}\]\[I=\frac{5}{6}Log(3x^{2}+2x+1)\frac{11}{9}\int\frac{dx}{\left
( x+\frac{1}{3} \right )^{2}+\left (\frac{\sqrt{2}}{3} \right
)^{2}}\]\[=\frac{5}{6}Log(3x^{2}+2x+1)\frac{11}{3\sqrt{2}}\: tan^{1}\left
(\frac{3x+1}{\sqrt{2}} \right )\]
INTEGRATION BY PARTIAL FRACTION
\[\frac{px+q}{(xa)(xb)}=\frac{A}{xa}+\frac{B}{xb}\] 
\[\frac{px+q}{(xa)^{2}}=\frac{A}{xa}+\frac{B}{\left
( xa \right )^{2}}\] 
\[\frac{px^{2}+qx+r}{(xa)^{2}\left
( xb \right )}=\frac{A}{xa}+\frac{B}{\left ( xa \right
)^{2}}+\frac{C}{xb}\] 
\[\frac{px^{2}+qx+r}{(xa)\left
( xb \right )\left ( xc \right )}=\frac{A}{xa}+\frac{B}{\left ( xb\right
)}+\frac{C}{xc}\] 
\[\frac{px^{2}+qx+r}{\left
( xa \right )\left ( x^{2}+bx+c \right
)}=\frac{A}{xa}+\frac{Bx+C}{x^{2}+bx+c}\] 
INTEGRATION BY PARTS
\[\int uv dx=u\int
vdx\int \left ( \frac{d}{dx}u\int vdx \right )dx\]
Method of selecting the first function and second function
I...... Inverse Trigonometric Functions
L..... Logarithmic Function
A..... Algebraic Function(Polynomial Function
T......Trigonometric Function
E......
Exponential Function
Integration With Exponential Function
\[\int e^{x}\left [ f(x)+f^{'}(x) \right ]dx=e^{x}f(x)+C\]
************************************
\[\int \frac{1}{\sqrt{a^{2}x^{2}}}dx=sin^{1}\left ( \frac{x}{a} \right )+C = cos^{1}\left ( \frac{x}{a} \right )+C\]\[\int \frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\tan^{1}\left ( \frac{x}{a} \right )+C = \frac{1}{a}\cot^{1}\left ( \frac{x}{a} \right )+C\]\[\int \frac{1}{x\sqrt{x^{2}a^{2}}}dx=\frac{1}{a}sec^{1}\left ( \frac{x}{a} \right )+C = \frac{1}{a}cosec^{1}\left ( \frac{x}{a} \right )+ C\]
Comparison between Differentiation and Integration
1.Both are the operations on the functions.
2.As all functions are not differentiable , similarly all functions are also not integrable.
3.Derivative of a function when it exist is unique. The integral of a function is not so. Any two integrals of a function may be differ by a constant.
4.When a polynomial function is differentiated, the result is a polynomial whose degree is 1 less than the degree of the polynomial. When a polynomial function is integrated, the result is a polynomial whose degree is 1 more than that of the polynomial.
5.We can speak of the derivative at a point. We never speak of the integral at a point. Here in integral we can speak that the integral of a function over an interval on which the integral is definite.
6.Differentiation is a
process involving limits. So is integration, as we will be seen in Exercise
7.8.
7.The process of differentiation and integration are reverse of each other.
Presentation on Integral
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