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Integration by Parts – In this section we will be looking at Integration by Parts. Of all the techniques we’ll be looking at in this class this is the technique that students are most likely to run into down the road in other classes...
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INTEGRAL CALCULUS MATHEMATICIA
(Combo Edition).
INDEFINITE INTEGRALS & ITS FORMS
OVERVIEW
In this chapter, we shall learn
Meaning of Integral of a function Schools/Coaching
Various forms and methods of Integrals Institutions &
(a) Integration as inverse Process of differentiation Tutors can have this
(b) Integration by Substitution method Book in the
(c) Integration of simple Algebraic functions Discounted Price.
(d) Integration of Trigonometric functions For inquiry, please
(e) Integration by Partial Fraction method WhatsApp @
(f) Integration By Parts +91 9650350480.
(g) Miscellaneous types of Integrals
(h) Real life Applications based problems of Integration
INTRODUCTION
Integral Calculus is motivated by the problem of defining and calculating the area of the region bounded
by the graph of the functions.
If a function f is differentiable in an interval I, i.e., its derivative f ′ exists at each point of I, then a
natural question arises that given f ′ at each point of I, can we determine the function? The functions that
could possibly have given function as a derivative are called ‘Anti-derivatives’ or ‘Primitive’ of the
function. Further, the formula that gives all these anti derivatives is called the indefinite integral of the
function and such process of finding anti derivatives is called integration.
The development of integral calculus arises out of the efforts of solving the problem of finding a
function whenever its derivative is given and, the problem of finding the area bounded by the graph of a
function under certain conditions. These two problems lead to the two forms of the integrals, e.g.,
indefinite and definite integrals, which together constitute the Integral Calculus.
There is a connection, known as the Fundamental Theorem of Calculus, between indefinite integral and
definite integral which makes the definite integral as a practical tool for science and engineering. The
definite integral is used to solve many interesting problems from various disciplines like probability,
economics and, finance.
IMPORTANT TERMS, DEFINITIONS & RESULTS
01. Meaning of integral of a function :
d
If differentiation of a function F (x) is f (x) i.e., if F x f x , then we say that one integral or
dx
primitive or anti-derivative of f (x) is F (x) and in symbols, we write
f x dx F x .
Therefore, we can say that integration is the inverse process of differentiation.
10th Mathematics Study Notes Mahayodha Academy
,02. List of formulae for Integral Calculus :
FORMULAE FOR INDEFINITE INTEGRALS
n x n 1 1
A. x dx n 1
k, n 1 B. x dx log x k
x 1 x ax 1
C. a dx a k D. e dx e ax k
log a a
1 1
E. sin ax dx a cos ax k F. cos ax dx a sin ax k
G. tan xdx log sec x k H. cot xdx log sin x k
OR log cos x k OR log cosec x k
I. sec xdx log sec x tan x k J. cosec xdx log cosec x cot x k
x x
OR log tan k OR log tan k
4 2 2
2 2
K. sec xdx tan x k L. cosec xdx cot x k
M. sec x tan xdx sec x c N. cosec x cot xdx cosec x k
1 1 1 x
O. x dx sec 1 x k P. a 2 2
dx tan 1 k
x 12 x a a
1 1
S. dx log x x 2 a 2 k T. dx log x x 2 a 2 k
2 2 2 2
x a x a
1 x
U. dx sin 1 k
a2 x2 a
To keep on tips :
x a2 1 1
x 2 a 2 dx x 2 a 2 log x x 2 a 2 k dx k
V. 2 2 x 2
x
1
x a2 dx 2 x k
x 2 a 2 dx x 2 a 2 log x x 2 a 2 k x
W. 2 2 2
xdx x 3/ 2 k
x 2 a2 3
x
X. a 2 x 2 dx a x 2 sin 1 k
2 2 a
1 1
Y. ax b dx a log ax b k , where ‘a’ is any non-zero constant (and k is integral constant)
Z. dx x k , where λ is a constant (and, k is the integral constant).
10th Mathematics Study Notes Mahayodha Academy
,Integral Calculus
03. Introduction to the terms and symbols used in Integration :
Terms Symbols Meaning of Terms Symbols
f (x)dx Integral of f with respect to x
f (x) in f (x)dx Integrand
x in f (x)dx Variable of integration
Integrate / Evaluate Find the integral
Integral value of f A function F such that F(x) f (x)
Constant of Integration Any real number denoted by C or k
Integration As An Inverse Process Of Differentiation
If derivative of a function is given and we have to find its primitive (original function) then, we have to
follow inverse process of finding the differentiation. This process of finding the primitive is called the
antidifferentiation or integration.
Let’s illustrate with a few examples.
d
(i) We know that (3x 2 ) 6x . So, the antiderivative of 6x is 3x 2 .
dx
d
Also note that (3x 2 1) 6x . So, the antiderivative of 6x is 3x 2 1 .
dx
d
Even, (3x 2 2) 6x . So, the antiderivative of 6x is 3x 2 2 .
dx
d 2 4 2 4
And, 3x 6x . So, the antiderivative of 6x is 3x .
dx 5 5
We can think of many more antiderivatives of this function.
d
(ii) We know that (cos x) sin x . So, the antiderivative of – sin x is cos x .
dx
d
Also note that (cos x 2) sin x . So, the antiderivative of – sin x is cos x 2 .
dx
d
Even, (cos x 3) sin x . So, the antiderivative of – sin x is cos x 3 .
dx
Here too we can think of many more antiderivatives of this function.
From above examples, we find that the antiderivatives of 6x are 3x 2 , 3x 2 1 , 3x 2 2 etc. That is,
integration of 6x isn’t unique. Actually, there may be infinite antiderivatives for 6x. So, for obtaining
the antiderivative of 6x, we add the constant C (which is called the constant of integration). Here C is a
parameter and we get different integrals of the given function for different values of C.
d
In fact, (3x 2 C) 6x . So, the antiderivative of 6x is 3x 2 C . Here C may be any real constant.
dx
d
Similarly, (cos x C) sin x . So, the antiderivative of – sin x is cos x C .
dx
d
Hence, F x C f x implies that, f x dx F x C .
dx
Hence the primitive of f (x) is F (x) C.
10th Mathematics Study Notes Mahayodha Academy
, WORKED OUT ILLUSTRATIVE EXAMPLES
2
1
Ex01. What is the antiderivative of x ?
x
2
1
Sol. Let I x dx
x
1
I x 2 dx
x
1
I x 2 2x log x C , where C is the integral constant.
2
(x 3 8)(x 1)
Ex02. Evaluate : 2 dx .
x 2x 4
(x 3 8)(x 1)
Sol. Let I 2 dx
x 2x 4
(x 3 23 )(x 1)
I dx
x 2 2x 4
(x 2)(x 2 2x 4)(x 1)
I dx
x 2 2x 4
I (x 2)(x 1)dx
I x 2 x 2 dx
1 1
I x 3 x 2 2x C .
3 2
Ex03. If f (x) 6x 2 2 , find f (x) . Given that f (x) 6, when x 1 .
Sol. We have f (x) 6x 2 2
d
f (x) f (x) 6x 2 2
dx
d
Integrating both sides, we get : f (x) dx (6x 2 2)dx
dx
3
f (x) 2x 2x C
As f (x) 6, when x 1 so, we have
f (1) 2(1)3 2(1) C
6 2 2 C C 10
Hence the function is, f (x) 2x 3 2x 10 .
elog e x
e 4loge x e 3loge x 2loga sec x
Ex04. Find the value of (a) dx (b) 3loge x 2loge x dx (c) a dx, a 0 .
x e e
loge x
e
Sol. (a) Let I dx
x
x
I dx
x
1
I dx
x
I 2 x C .
10th Mathematics Study Notes Mahayodha Academy
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