Class 12 Maths Formula Ch-7 Indefinite Integral

INDEFINITE INTEGRALS

Class 12 Maths Formula Ch-7 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 anti-derivative 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 anti-derivative 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,

 

Integral of a function is defined as follows:\[If\: \frac{d}{dx}F(x)=f(x),\ then\: \: \int f(x)dx=F(x)+C\] where C ∊ R is called the constant of integration.

Integration as the reverse process of differentiation

Geometrical Interpretation:

Indefinite integral represents the integral as family of curves

Consider the curve f(x) = 2x,  anti-derivative of 2x w. r. t.  x is x² + C = y say

f’(x) = x² + C = y (say)

For C = 0, y = x²                                            

For C= 1, y = x² + 1

For C = -1, y = x² – 1                                    

For C = -3, y = x² - 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 itself

Important 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: 

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\[Explanation: \:\: \int cosecx dx=log|cosecx-cotx|\]\[=log\left |\frac{cosecs-cots}{1}\times \frac{cosecx+cotx}{cosecx+cotx} \right |\] \[=log\left | \frac{cosec^{2}x-cot^{2}x}{cosecx+cotx} \right |\] \[ =log\left | \frac{1}{cosecx+cotx} \right |\] \[=log\left ( cosecx+cotx \right )^{-1}=-log\left | cosecx+cotx \right |\]\[\int cosecx dx=log|cosecx-cotx| =log\left | \frac{1}{sinx}-\frac{cosx}{sinx} \right |\] \[=log\left | \frac{1-cosx}{sinx} \right |=log\left | \frac{2sin^{2}\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}} \right |\] \[=log\left | \frac{sin\frac{x}{2}}{cos\frac{x}{2}} \right | =log\left | tan\frac{x}{2} \right |\]*********************************\[\int sec^{2}xdx=tanx+C\]\[\int cosec^{2}xdx=-cotx+C\]\[\int secx tanx dx=secx+C\]\[\int cosecx cotx dx=-cosecx+C\]

INTEGRATION WITH INVERSE TRIGONOMETRIC FUNCTIONS
\[\int \frac{1}{\sqrt{1-x^{2}}}dx=sin^{-1}x+C\]	\[\int \frac{-1}{\sqrt{1-x^{2}}}dx=cos^{-1}x+C\]
\[\int \frac{1}{1+x^{2}}dx=tan^{-1}x+C\]	\[\int \frac{-1}{1+x^{2}}dx=cot^{-1}x+C\]
\[\int \frac{1}{|x|\sqrt{x^{2}-1}}dx=sec^{-1}x+C\]	\[\int \frac{-1}{|x|\sqrt{x^{2}-1}}dx=cosec^{-1}x+C\]

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

\[I=\int \frac{5x-2}{3x^{2}+2x+1}dx\]\[Let\: 5x-2=A\frac{d}{dx}(3x^{2}+2x+1)+B\]\[5x-2=A(6x+2)+B …………….(1)\]\[5x-2=6Ax+2A+B\] 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\[5x-2=\frac{5}{6}(6x+2)-\frac{11}{3}\] \[I=\int \frac{\frac{5}{6}(6x+2)-\frac{11}{3}}{3x^{2}+2x+1}dx\] 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

INTEGRATION BY PARTS

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

\[\int e^{x}\left [ f(x)+f^{'}(x) \right ]dx=e^{x}f(x)+C\]

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\[\int \frac{1}{\sqrt{a^{2}-x^{2}}}dx=sin^{-1}\left ( \frac{x}{a} \right )+C = -cos^{-1}\left ( \frac{x}{a} \right )+C\]\[\int \frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\tan^{-1}\left ( \frac{x}{a} \right )+C = -\frac{1}{a}\cot^{-1}\left ( \frac{x}{a} \right )+C\]\[\int \frac{1}{x\sqrt{x^{2}-a^{2}}}dx=\frac{1}{a}sec^{-1}\left ( \frac{x}{a} \right )+C = -\frac{1}{a}cosec^{-1}\left ( \frac{x}{a} \right )+ C\] 

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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