Integrals Class 12 Notes Maths Chapter 7 - CBSE

Chapter:7

What are Integrals ?

https://drive.google.com/file/d/1ewITgZITwid7cx4r-HmrxzLDtBDg8zRE/view

Integration

The process of integration is an opposite process to
that of differentiation.

∫ f (x) dx = F(x) + C
Where ∫ —› symbol of integration
f(x) —› integrand
dx —› element of integration
F —› primitive
C —› constant of integration In differential calculus, we are given a function and we are required to differentiate it, but in integral calculus we required to find a function whose differential coefficient (or derivative) is given.
Hence if,

d(F(x) + c) = f(x)/dx

We write that, ∫ f(x)dx = F(x) + c

Standard Integrals

$$\text{(i)\space }\int 0\space \text{dx}=\text{C(constant)}\\ \text{(ii)}\int 1\text{dx} = x+\text{C}\\ \text{(iii)}\int x^n\text{dx}= \frac{x^{n+1}}{n+1}+\text{C} \space\text{if} ≠ -1\\ \text{(iv)} \int\frac{1}{x}\text{dx}=\text{log}_e\text{x}+\text{C}\\ \text{(v)}\int e^x\text{dx}=e^x+\text{C}$$

$$\text{(vi)}\int a^x\text{dx}=\frac{a^x}{\text{log}_ea}+\text{C}\\ \text{(vii)} \int \text{sin}\space x \space \text{dx\space=\space -cos}\space x+\text{C}\\ \text{(viii)}\int \text{cos}\space x \space\text{dx}=\text{sin}\space x+\text{C}\\ \text{(ix)}\int \text{sec}^2x\space\text{dx}= \text{tan}\space x+\text{C}\\ \text{(x)}\int\text{cosec}^2x \space \text{dx} =-\text{cot}\space x+\text{C}$$

$$\text{(xi)}\int \text{sec}\space x .\text{tan}\space x\space \text{dx}=\text{sec}\space x+\text{C}\\ \text{(xii)} \int \text{cosec}\space x\space.\space \text{cot}\space x\space\text{dx} =-\text{cosec x}+\text{C}\\\text{(xiii)}\int \text{cot}\space x\space\text{dx} = \text{log}|\text{sin} \space x|+\text{C}\\ \text{(xiv)}\int\text{tan}\space x \space\text{dx}=-\text{log}|\text{cos}\space x|+\text{C}\\\text{(xv)}\int\text{sec}\space x\space\text{dx}= \text{log}|\text{sec}\space x + \text{tan \space x}|+\text{C}=\text{log}|\text{tan}(\frac{\pi}{2}+\frac{x}{2})|+ \text{C}\\ \text{(xvi)} \int \text{cosec}\space x\space\text{dx}=\text{log}|\text{cosec}\space x-\text{cot}\space x|=\text{log}|\text{tan} \frac{x}{2}|+\text{C}$$

$$\text{(xvii)}\int\frac{1}{\sqrt{1-x^2}}\text{dx}=\text{sin}^{-1}x+\text{C}=-\text{cos}^{-1}x+\text{C}\\\text{(xviii)}-\int\frac{1}{\sqrt{1-x^2}}\text{dx}=\text{cos}^{-1}x+\text{C}=-\text{sin}^{-1}x+\text{C}\\\text{(xix)}\int\frac{\text{dx}}{x\sqrt{x^2-1}}=\text{sec}^{-1}x+\text{C}=-\text{cosec}^{-1}x+\text{C}\\\text{(xx)}-\int\frac{1}{x\sqrt{x^2-1}}\text{dx}=\text{cosec}^{-1}x+\text{C}=-\text{sec}^{-1}+\text{C}\\\text{(xxi)}\int\frac{1}{1+x^2}\text{dx}=\text{tan}^{-1}x+\text{C}=-\text{cot}^{-1}x+\text{C}\\\text{(xxii)}\int\frac{1}{1+x^2}\text{dx}=\text{cot}^{-1}x+\text{C}=-\text{tan}^{-1}x+\text{C}$$

Methods Of Integration

We need to develop additional techniques or methods for finding the integrals by reducing them into standard forms. Here are some important methods:
A. Integration by Substitution
B. Integration by Partial Fraction
C. Integration by Parts

A. Integration by Substitution

The given integral ∫f(x)dx can be transformed into another form by changing independent variables 'x' to 't' by substituting x=g(t). Consider I = ∫ f(x)dx Put x=g(t) so that dx/dt= g'(t) ==> dx=g'(t). dt → I= ∫ f(x)dx = ∫ f{g(t)}. g'(t)dt There are some useful substitution: (a) (i) If the integrand contains a t-ratio of f(x)or logarithm of f(x) or an exponential function in which the index is f(x), put f(x)=t. (ii) If the integrand is a rational function of ex, put ex = t. (b) For evaluating integrals ∫ sinnx.dx or ∫ cosnx.dx, where n = positive integer, (i) Put sin x = t, if the index of cos x is an odd positive integer. (ii) Put cos x = t, if the index of sin x is an odd positive integer. (iii)Express sinn x or cosn x in terms of cosines of multiples of angles by using 2sin² x = 1-cos 2x or 2cos² x = 1 + cos 2x if n is an even positive integer. (c) For evaluating integrals ∫ secn x dx or ∫ cosecn x dx where n = positive integer, (i) Put tan x = t, if the index of sec x is a positive even integer. (ii) Put cot x = t, if the index of cosec x is a positive even integer.

$$\text{(d)}\space \text{(i)}\space\text{If}\space \sqrt{a^2-x^2}\space\text{occurs in the integrand, put x = a sin}\space \alpha.\\ \qquad \text{(ii)}\space\text{If}\sqrt{a^2+x^2}\space\text{occurs in the integrand, put x = a tan} \space \alpha.\\ \qquad\text{(iii) If}\sqrt{x^2-a^2}\space\text{occurs in the integrand, put x = a sec}\space \alpha.$$

B. Partial fractions

Let f(x) =g(x)/h(x) be a proper rational function. First of all, we split up the denominator h(x) as the product of non-repeated linear or repeated linear factors or non-repeated or repeated quadratics (which cannot be split into real linear factors). Then f(x) can be written as the sum of the fractions in which the numerator is either a constant or real linear polynomials. For this the following points should be kept in mind.

(a) Corresponding to each non-repeated linear factor a x+ b, there is a partial fraction of the form A/(ax+b)

(b) Corresponding to each repeated linear factor (ax+b)2 there is sum of two terms of the form A/(ax+b)+B/(ax+b)²

(c) Corresponding to each non-repeated quadratic factor ax2+bx+c (which cannot be put as the product of linear factors) there is a partial factor of the form (ax+b)/ax²+bx+c

(d) Corresponding to each repeated quadratic factor (ax2+bx+c)2 (which cannot be put as the product of linear factors), there is a sum of two terms of the form (Ax+B)/(ax²+bx+c) +(Cx + D)/(ax² + bx + c)²

(e) When only even powers of x occur both in the numerator and denominator put x² = y the resulting algebraic function in y into partial fraction and then put y = x² in the resulting partial fraction.

C. Integration By Parts

Consider u and v as two functions of x, then integration
by parts says that; ∫ u.v dx = u { ∫ v dx} - ∫{du/dx. ∫ v dx } dx Where, u is known as first function and
v is known as second function.
v is known as second function.
i.e., Integral of product of two = (first function *
(integral of second function) - integral of
{(differentiation of first function) * (integral of
second function)}
We decide the function as first which comes first in
order from top to bottom.
INVERSE —› I
LOGARITHMIC —› L
ALGEBRIC —› A
TRIGONOMETRIC —› T
EXPONENTIAL —› E
Or which comes first in word 'ILATE'.

Fundamental Theorem Of Calculus

The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function. The two operations are inverses of each. other apart from a constant value which depends where one starts to compute area. The first part of the theorem, called first fundamental
theorem of calculus. Let f be a continuous real-valued function defined on a closed interval [a,b] and the function F(x) is defined by

$$\text{F(x)} = \int^x_a \text{f(t)}\space \text{dt}$$

then F’(x) = f(x)
The second part of the theorem called second fundamental theorem of calculus. Let f be a continuous real-valued function on a closed interval [a,b], then

$$ \int^b_a \text{f(x)}\text{dx}= \text{F(b)}-\text{F(a)}$$

Definite Integral

The definite integral has a unique value. A definite integral is denoted by

$$ \int^b_a \text{f(x)}\text{dx},$$

where a is called the lower limit of the integral and b is called the upper limit of the integral.

Standard Integrals

$$\text{(i)} \int^b_a \text{f(x)}\text{dx}=\text{F(b)\space-\space F(a)}\space \text{where}\int \text{f(x)dx=\space \text{F(x)\space +\space\text{C}}}\\\text{(ii)} \int^b_a\text{f(x)dx}=\int^b_a\text{f(t)}\text{dt}\\\text{(iii)}\int^b_a \text{f(x)}\text{dx}=-\int^a_b \text{f(x)dx}\\\text{(iv)}\int^b_a\text{f(x)dx} = \int^c_a\text{f(x)dx}\space+\space \int^b_c\text{f(x)dx},\\ \text{where the point c lies between a and b}\\\text{(v)}\int^b_a\text{f(x)dx}=\int^b_a\text{f(a\space+\space b-x)dx}\\\text{(vi)}\int^a_0 \text{f(x)dx}=\int^a_0 \text{f(a-x)dx}\\\text{(vii)}\int^\text{2a}_0\text{f(x)dx}=\int^a_0\text{{f(x)dx + f(2a-x)}}\text{dx}\\\text{(viii)} \int^{2a}_0\text{f(x)dx}= 2\int^a_0\text{f(x)dx,}\space \text{if}\space \text{f(2a-x)}= \text{f(x)}\space \text{and} \space 0, \text{if} \space \text{f(2a-x)}=-\text{f(x)}\\ \text{(ix)}\int^a_{-a} \text{f(x)dx}=2\int^a_0 \text{f(x)dx,}\space \text{if} \space \text{f(-x)}=\text{f(x)} \space \text{and} \space 0, \text{if f(-x) = -f(x)}$$

Evaluation Of Definite Integrals By Substitution

$$\text{To evaluate}\int^b_a \text{f(x)dx,} \space \text{by substitution, the following steps are useful}$$

Consider the integral without limits and substitute, y=f(x) or x=g(y) to reduce the given integral to a known form.
lntegrate the new integrand with respect to the new variable without mentioning the constant of integration.
Resubstitute for the new variables and write the answer in terms of the original variable.
Find the values of answer obtained in III step at he given limits of integral and find the difference of the values at the upper and lower limits.

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