MATHSWORLD NOTES
INDEFINITE INTEGRALS
f(x) dx = Φ(x) + C
1.
d
(
dx
2.
d
dx
3.
d
n+1
4.
dx
5.
d
dx
6.
d
dx
7.
d
dx
8.
d
dx
9.
d
dx
10.
d
dx
) = x n, n ≠ - 1
( log x ) =
(
dx
d
xn + 1
(
n
x dx =
1
1
x
x
ex
) = ex
ax
x
) = a ;
dx =
xn + 1
n+1
a>0, a≠1
ex
+C
ax
x
a dx =
n≠-1
log │x│ + C
ex dx =
loge a
+ C,
loge a
+C
( - cos x ) = sin x
sin x dx =
- cos x
+C
( sin x ) = cos x
cos x dx =
sin x
+C
( tan x ) = sec2 x
sec2 x dx =
tan x
+C
( - cot x ) = cosec2 x
cosec2 x dx = - cot x + C
( sec x ) = sec x tan x
sec x tan x dx = sec x + C
( - cosec x ) = cosec x cot x
cosec x cot x dx = - cosec x + C
11.
d
dx
12.
d
dx
13.
d
dx
14.
d
dx
15.
d
dx
16.
d
dx
17.
d
( log sin x ) = cot x
cot x dx = log │sin x│ + C
( - log cos x ) = tan x
tan x dx = - log │cos x│ + C
( log (sec x + tan x ) ) = sec x
sec x dx = log │sec x + tan x│ + C
( sin-1 x
a
) =
( cos-1 x
a
) =-
(
dx
18.
d
d
(
d
dx
1
a
(
dx
20.
1
tan-1
a
dx
19.
cosec x dx = log │cosec x – cot x│ + C
( log (cosec x - cot x ) ) = cosec x
1
a
(
1
a
cot-1
sec-1
cosec-1
1
1
a2 – x2
a2 – x2
1
1
-
a2 – x2
x
) =
a
x
) =
a
x
a
a
1
a2 + x2
a2 + x2
-1
-1
x2 - a2
x
x2 - a2
[ k1.f1(x) ± k2.f2(x)……. ± kn.fn(x) ] dx =
x2
1
x
dx =
dx =
dx =
x2 - a2
1
x
x
x2 - a2
dx =
1
+C
a
-1
dx = cos
a2 + x2
-1
)=
–
1
1
x
x
a2
a2 + x2
) =
dx = sin-1
x
a
tan-1
a
1
cot-1
a
1
a
1
a
+C
x
+C
a
x
+C
a
sec-1
cosec-1
x
a
x
a
+C
+C
k1 f1(x) dx ± k2 f2(x) dx ……. ± kn fn(x) dx
2
METHODS OF INTEGRATION
1. Integration by substitution
2. Integration by parts
3. Integration of rational algebraic functions by using partial fractions
1. INTEGRATION BY SUBSTITUTION:
i)
If Φ(x) is a continuously differentiable function, then to evaluate integrals of
the form
f(Φ(x)) Φ’(x) dx,
Φ(x) = t
we substitute
The above integral reduces to
If
Φ’(x) = dt.
and
f(x) dx = Φ(x),
f(t) dt
then
f(ax + b) dx =
1
a
Φ(ax + b)
ii) In rational algebraic functions if the degree of numerator is greater
than or equal to the degree of denominator, then always divide the
numerator by denominator and use the result.
Nr.
Dr.
=Q +
R
Dr.
iii) To evaluate integrals of the form
sinm x dx
and
cosm x dx
where m ≤ 4,
we express sinm x and cosm x in terms of sines and cosines of multiples of x
by using trigonometrical identities given below.
sin2 x =
1 – cos 2x
cos2 x =
2
sin3 x =
1 + cos 2x
2
3 sin x – sin 3x
4
cos3 x =
cos 3x + 3 cos x
4
3
iv) To evaluate integrals of the form
sin mx sin nx dx,
sin mx cos nx dx,
cos mx cos nx dx,
cos mx sin nx dx
and
we use the trigonometrical identities given below.
2 sin A cos B = sin (A + B) + sin (A – B)
2 cos A sin B = sin (A + B) - sin (A – B)
2 cos A cos B = cos (A + B) + cos (A - B)
2 sin A sin B
= cos (A - B) - cos (A + B)
f’(x)
v) Integrals of the form
dx
f (x)
If the numerator in integrand is exact differential of the denominator then its
integral is logarithm of the denominator.
f’(x)
dx = log [ f (x) ] + C
f (x)
vi) Integrals of the form
[ f (x) ]n f’ (x) dx
[ f (x) ] f’ (x) dx =
[ f (x) ]n+1
n
,
n≠-1
n+1
vii) Integrals of the form
sinm x cosn x dx,
m, n Є N
Algorithm:
a) If the exponent of sin x is odd positive integer, put cos x = t.
b) If the exponent of cos x is odd positive integer, put sin x = t.
c) If the exponent of both sin x and cos x are odd positive integers,
put either sin x = t or cos x = t.
d) If the exponent of both sin x and cos x are even positive
integers, express them in terms of sines and cosines of multiples
of x by using trigonometric results or De’ Moivere’s theorem.
4
viii) Some Special Integrals:
Expression
Substitution
a2 + x2
x = a tan θ
or
a cot θ
a2 - x2
x = a sin θ
or
a cos θ
x2 - a2
x = a sec θ
or
a cosec θ
a–x
a+x
x–α
β- x
1
a2
dx =
x2
+
1
ii)
x2
-
a2
-
x2
1
a 2 – x2
v)
dx =
a2
1
iii)
iv)
(x – α) (x – β)
or
i)
x = a cos 2θ
a+x
a-x
or
1
dx =
1
a
1
2a
1
2a
dx = sin-1
x = α cos2 θ + β sin2 θ
tan-1
log |
log |
x
a
x
a
+C
x-a
x+a
a+x
a-x
|
+C
|
+C
+C
dx = log | x +
a2 + x 2
| +C
dx = log | x +
x2 - a2
| +C
a2 + x 2
vi)
1
x 2 - a2
5
1
ix) Integrals of the type
dx
ax2 + bx + c
Algorithm:
a) Make the coefficient of x2 unity, if it is not, by multiplying and dividing by it.
b) Add and subtract the square of the half of coefficient of x to express
ax2 + bx + c in the form
b
a[{x+
4ac – b2
2
} +
4a2
2a
c) Use the suitable formula from the following:
1
i)
a2
1
x2
-
dx =
a2
1
iii)
a
x2
+
ii)
1
dx =
dx =
a2 - x2
1
tan-1
log |
2a
1
2a
log |
x
]
+C
a
x-a
x+a
a+x
a-x
|
+C
|
+C
1
x) Integrals of the type
ax2 + bx + c
dx
Algorithm:
a) Make the coefficient of x2 unity, if it is not, by multiplying and dividing by it.
b) Find half of the coefficient of x.
c) Add and subtract (½ coeff. of x)2 inside the square root to express the
quantity inside the square root in the form.
{x+
b
2
} +
4ac – b2
4a
2a
4ac – b2
or
2
4a
2
- {x+
b
}2
2a
d) Use the suitable formula from the following:
1
i)
a2
–
dx = sin-1
x2
1
ii)
x
a
+C
dx = log | x +
a2 + x 2
| +C
dx = log | x +
x2 - a2
| +C
a2 + x 2
1
iii)
x2
-
a2
6
px + q
xi) Integrals of the type
ax2 + bx + c
dx
Algorithm:
a) Write the numerator in the following form:
d
px + q = λ {
dx
(ax2 + bx + c)} + μ
i.e. px + q = λ (2ax + b) + μ
b) Obtain the values of λ and μ by equating the coefficients of like powers
of x on both sides.
c) Replace px + q by λ(2ax + b) + μ in the given integral to get
px + q
ax2 + bx + c
dx = λ
2ax + b
ax2 + bx + c
dx + μ
1
ax2 + bx + c
dx
d) Integrate R.H.S. in step c and put the values of λ and μ obtained in step b.
P(x)
xii) Integrals of the type
ax2 + bx + c
dx
where degree of P(x) is ≥ 2
Divide the numerator by the denominator and express the integrand as
R(x)
Q(x) +
where R(x) is a linear function of x.
2
(ax + bx + c)
P(x)
ax2 + bx + c
dx =
R(x)
Q(x) dx +
ax2 + bx + c
dx
7
px + q
xiii) Integrals of the type
dx
ax2 + bx + c
Algorithm:
a) Write the numerator in the following form:
px + q = λ {
d
dx
(ax2 + bx + c)} + μ
i.e. px + q = λ (2ax + b) + μ
b)
c)
Obtain the values of λ and μ by equating the coefficients of like powers
of x on both sides.
Replace px + q by λ(2ax + b) + μ in the given integral to get
px + q
ax2 + bx + c
dx = λ
2ax + b
1
dx + μ
ax2 + bx + c
ax2 + bx + c
dx
d) Integrate R.H.S. in step c and put the values of λ and μ obtained in step b.
xiv) Integrals of the type
1
a
sin2
x+b
cos2
x+c
1
(a sin x + b cos x)2
1
1
dx,
a+b
dx,
sin2
x
1
dx,
a+b
sin2
x+c
dx,
a + b cos2 x
dx
cos2
x
Algorithm:
a) Divide the Numerator and the Denominator by cos2 x.
b) Replace sec2 x, if any, in Denominator by 1 + tan2 x.
c) Put tan x = t, so that sec2 x dx = dt. This reduces the integral in the
form
1
dt
at2 + bt + c
d) Evaluate this as explained earlier.
8
xv) Integrals of the type
1
1
1
dx,
dx,
a + b sin x
a sin x + b cos x
1
dx,
a + b cos x
dx
a sin x + b cos x + c
Algorithm:
1 - tan2 x/2
2 tan x/2
and
a) Put sin x =
1+
tan2
cos x =
1 + tan2 x/2
x/2
b) Replace
1 + tan2 x/2
in the numerator by
c) Put
tan x/2 = t
so that ½ sec2 x/2 dx = dt
This reduces the integral in the form
sec2 x/2
1
at2 + bt + c
dt
d) Evaluate this as explained earlier.
xvi) Integrals of the type
1
dx
a sin x + b cos x
Alternative Method:
Put a = r cos θ, b = r sin θ so that, r = √(a2 + b2)
xvii) Integrals of the type
a sin x + b cos x
and
θ = tan-1 (b/a)
dx
c sin x + d cos x
Algorithm:
a) a sin x + b cos x = λ (c cos x – d sin x) + μ (c sin x + d cos x)
b) Obtain the values of λ and μ by equating the coefficients of sin x and
cos x on both the sides.
c)
a sin x + b cos x
dx = λ log | c sin x + d cos x | + μ x + C
c sin x + d cos x
9
xviii) Integrals of the type
a sin x + b cos x + c
dx
p sin x + q cos x + r
Algorithm:
a) a sin x + b cos x + c = λ (p cos x – q sin x) + μ (p sin x + q cos x +r) + ν
b) Obtain the values of λ and μ by equating the coefficients of sin x and
cos x and the constant terms on both the sides.
c)
a sin x + b cos x + c
dx = λ log | p sin x + q cos x + r | + μ x +
p sin x + q cos x + r
1
dx
p sin x + q cos x + r
d) Evaluate the integral on RHS of Step c by using the method discussed
earlier.
2. INTEGRATION BY PARTS:
If u and v are two functions of x, then
u v dx = u v dx -
{
du
dx
v dx } dx
i.e. The integral of the product of two functions =
(First function) x (Integral of second function) –
Integral of {(Differentiation of first function x (Integral of second function)}
NOTE-1: Proper choice of first and second function
If one of the two functions is not directly integrable, we take it as the first
function and the remaining function is taken as the second function.
If there is no other function, then unity is taken as the second function.
If both the functions are easily integrable, then the first function is chosen in
such a way that the derivative of the function is a simple function and the
function thus obtained under the integral sign is easily integrable than the
original function.
NOTE-2: Choose the first function as in “ILATE”
10
ex { f (x) + f’ (x) } dx = ex f (x) + C
i)
ii)
eax
ax
e sin bx dx =
iii)
a2 + b2
(a sin bx – b cos bx) +
C
eax
ax
(a cos bx + b sin bx) +
a2 + b2 C
1
1 2
x
dx =
x a2 - x2 +
a sin-1
2
2
a
e cos bx dx =
iv)
a2 - x2
v)
2
a +x
vi)
2
x -a
2
2
dx =
dx =
1
2
1
2
x
x
vii) Integrals of the form
a2 + x2
2
x -a
2
+
1
2
1
-
2
ax2 + bx + c
+C
a2 log | x +
a2 + x2
| +C
a2 log | x +
x2 - a2
| +C
dx
Algorithm:
a) Make the coefficient of x2 as one by taking ‘a’ common to obtain
b
c
x2 +
x+
a
a
4ac – b2
b 2
b 2
} +
b) Add and subtract (
) to obtain { x +
2a
4a2
2a
c) Use any of the form given in (iv), (v) or (vi).
viii) Integrals of the form
Algorithm:
a) Express px + q as
px + q = λ {
d
dx
( px + q )
ax2 + bx + c
(ax2 + bx + c)} + μ
dx
i.e. px + q = λ (2ax + b) + μ
b) Obtain the values of λ and μ by equating the coefficients of like powers
of x on both sides.
c) Replace px + q by λ(2ax + b) + μ in the given integral to get
(px + q)
ax2 + bx + c dx = λ
+μ
(2ax + b) ax2 + bx + c dx
ax2 + bx + c dx
11
3.
INTEGRATION OF RATIONAL ALGEBRAIC
FUNCTIONS BY USING PARTICAL FRACTIONS:
CASE – I:
When denominator is expressible as the product of non-repeating linear
factors.
Let g(x) = (x – a1) (x – a2) ……(x – an).
Then,
f(x)
g(x)
A1
x – a1
=
A2
x – a2
+
+ …... +
An
x – an
where A1, A2, …… An are constants and can be determined by equating the
numerator on RHS to the numerator on LHS and then substituting
x = a1, a2, …, an.
CASE – II:
When denominator is expressible as the product of the linear factors such that
some of them are repeating.
Let g(x) = (x – a)k (x – a1) (x – a2) ……(x – an).
Then,
f(x)
g(x)
A1
x–a
=
+
B1
x – a1
A2
(x – a)2
+
+
B2
x – a2
+
A3
Ak
+ …... +
3
(x – a)
(x – a)k
+ …... +
Bn
x – an
CASE – III:
When some of the factors of denominator are quadratic but non – repeating.
Corresponding to each quadratic factor
ax2 + bx + c,
Ax + B
we assume partial fraction of the type
ax2 + bx + c
where A and B constants to be determined by comparing coefficients of
similar powers of x in the numerator of both the sides.
In practice it is advisable to assume partial fractions of the type
A (2ax + b)
ax2 + bx + c
B
+
ax2 + bx + c
12
CASE – IV:
When some of the factors of denominator are quadratic and repeating.
Corresponding to each quadratic factor
(ax2 + bx + c)k,
we assume 2 k partial fractions of the type
{
A0 (2ax + b)
A1
+
ax2 + bx + c
}
ax2 + bx + c
A1 (2ax + b)
+ {
(ax2 + bx + c)2
A2k-1 (2ax + b)
+…..+ {
A2k
+
(ax2 + bx + c)k
A2
+
}
(ax2 + bx + c)2
}
(ax2 + bx + c)k
4. INTEGRATION OF SOME SPECIAL IRRATIONAL ALGEBRAIC
FUNCTIONS:
Φ (x)
i) Integrals of the form
dx
P Q
Put Q = t2.
To evaluate
1
(ax + b)
dx
(cx + d)
Φ (x)
ii) Integrals of the form
dx
P Q
Put Q = t2.
1
To evaluate
iii) Integrals of the form
1
dx
(px2 + qx + r)
Φ (x)
iv) Integrals of the form
1
2
(ax + b)
(cx2 + d)
Put ax + b = 1 / t
where P and Q both are pure
quadratic expression in x.
dx
P Q
To evaluate
Put px + q = t2.
where P is a linear expression
and Q is a quadratic expression.
dx
P Q
(ax + b)
where P is a quadratic expression
and Q is a linear expression.
(px + q)
Φ (x)
To evaluate
Put cx + d = t2.
dx
(ax2 + bx + c)
Put P = 1 / t
where P and Q both are linear
functions of x.
dx
Put x = 1 / t and then c + d t2 = u2
13