# Indefinite-Integration-Formulas-Handbook-MathonGo

```ULTIMATE FORMULA HANDBOOK
MATHONGO
Basic Formulas
 cosec 2 xdx   cot x
 dx  x
x
n
 sec x tan xdx  sec x
n 1
 x dx  n  1 ( n   1)

x
 cosec x cot xdx   cosec x
dx  sec  1 x
2
x 1
a x dx 
ax
,a  0
log e a
 e x dx  e x
1
dx  log | x |
x

 cos dx  sin x
1
1 x
2

 sec 2 xdx  tan x
1
1 x
2
 tan x d x  log
sec x  c
 cot x dx  lo g
sin x  c
dx  sin  1 x
 sin xdx   cos x
 sec x dx  log

1
dx  tan  1 x
x

sec x  tan x  c  log tan     c
2
4
 cos ec x dx  log
cos ec x  cot x  c  tan x 2  c
Method of Substitution
Type 1

Type 3
Type 4
f &acute;( x)
dx  log f ( x)  c
f ( x)
Type 2
n
  f ( x )
f &acute;( x ) dx 
f ( x ) n 1
 c
n 1
 f  g ( x )  g &acute;( x ) dx . In this type, we substituteg (x) = t, then, Hence integral reduces to  f (t) dt
P( x).(ax  b) n
or
Working Rule :
P ( x)
(ax  b)n
where P(x) is a polynomial in x and n is a positive rational number.
Put z = ax + b
Type 5
sin f(x) or cos f(x) then put z = f(x)
Type 6 :

s in m x c o s n x d x
Working Rule :
(i)
If power of sin x is odd positive integer, put z = cos x
(ii)
If power of cos x is odd positive integer, put z = sin x
(iii) If powers of both sin x and cos x are odd positive integers, put z = sin x or z = cos x.
(iv)
If powers of neither cos x nor sin x is odd positive integer, see the sum of powers of sin x and cos
x.
(a)
If the sum of powers is even negative integer, put z = tan x.
(b)
If the sum of powers (m + n) is even positive integer and m, n are integers, express the
integrand as the algebraic sum of sines and cosines of multiple angles.
INDEFINITE INTEGRATION
ULTIMATE FORMULA HANDBOOK
m
x sec n xdx or
m
x sec n xdx : similar can be derived for the other pair
Type 7 :
 tan
.
 tan
For
MATHONGO
 cot
m
xco sec n xdx
(i)
If power of secx is even positive integer, put z = tanx.
(ii)
If power of secx is not even positive integer, then see the power of tanx.
(a)
If power of tanx is odd positive integer, put z = secx.
(b)
If power of tanx is even positive integer, then put sec2x – 1 in place of tan2x and then
substitute z = tanx.
(iii) If power of tanx is zero and power of secx is odd positive integer greater than 1, then method of
integration by parts is used.
SOME
STANDARD SUBSTITUTIONS
1
1.
2.
a
2
 x
2
or
1

a2  x2
2
a  x
1
or

2
a2  x2
x = a sin  or a cos 
x = a tan  or a cot 
6.
x
a  x or
7.
 x  a

 or
b  x 
4.
2
or
x2  a 2
a  x
a  x or
x  a
2
x = a sec  or a cosec 
x  a
x  b
8.
a x
a  x x = a cos 2
x
 a  b  x 
x = a cos2  + b sin2 
1
3.
1 x
x = a tan2  or x = a cot2 
x
or
x
 a x  b
x = a sec2  – b tan2 
5.
x
a  x or
a  x
x = a sin2  or x = a cos2 
x
1
9.
 a x  b
x
x – a = t2 or x – b = t2
SOME STANDARD FORMULAS DERIVED FROM SUBSTITUTION
Set-I
Set-II
1.
 x2
3.
 x2
1.

3.

dx
 a
dx
 a

2
dx
2
a x
2
dx
2
1
x  a
log
 c
2a
x  a

2
x  a
2
2.
 a2
2.

4.

dx
 x

2
1
a  x
log
 c
2a
a  x
1
x
 c
tan  1
a
a
 sin  1
x
c
a
 log x 
x2  a 2  c
dx
2
x  a
2
dx
x x2  a2
INDEFINITE INTEGRATION
 log x 

x2  a 2  c
1
x
sec  1
c
a
a
ULTIMATE FORMULA HANDBOOK
Set-III 1.
2.
3.
MATHONGO
a2
x
sin  1  c
2
2

a 2  x 2 dx 
x
2
a2  x2 

x 2  a 2 dx 
x
2
x2  a2 

x 2  a 2 dx 
x
2
x 2  a 2 dx 
a2
lo g x 
2
x
2
x2  a2  c
x2  a2 
a2
lo g x 
2
x2  a2  c
INTEGRATION BY PARTS
Integral of product of two functions
= (1st function) &times; (Integral of 2nd function) – Integral of {(differential of 1st function) &times; Integral of 2nd function}
d

f ( x).  g( x) dx  dx
In symbols :  f ( x). g ( x) dx  f ( x).  g ( x) dx   
 dx

or
 u.vdx  u  vdx –  u&acute;( vdx)dx
where
I stands for Inverse circular function
L stands for Logarthmic function
A stands for Algebraic function
T stands for Trigonometrical functions
and E stands for Exponential function
(ii)
If both the functions are trigonometrical, take that function as v whose integral is simpler.
(iii) If both the functions are algerbraic take that functions as u whose d c is simpler.
Standard Forms derived using By Parts
x
 f ( x )  f &acute;  x   dx  e x f ( x )  c
(i)
e
(iii)
 [ xf &acute;( x)  f ( x)]dx  x f ( x )  c
(iv)
(v)
e
e
ax
ax
sin  bx  c  dx = eax
cos  bx  c  dx = eax
(ii)
mx
e
mf (x)  f &acute; x  dx  emx f (x)  c
b

sin  bx  c  tan1 
a

c
a2  b2
b

cos  bx  c  tan1 
a

c
a2  b2
INDEFINITE INTEGRATION
ULTIMATE FORMULA HANDBOOK
MATHONGO
METHOD : INTEGRATION BY PARTIAL FRACTION
We have divided this method into 2 types, depending upon the denominator.
1.
If denominator has non repeated factors
2.
If denominator has repeating factors
Type 1 : For non-Repeating roots
When denominator can be expressed as non repeating factors
i.e. D(x) = (x – 1) (x – 2)...
(for linear factors)
2
= (ax + bx + c) (px + qx + r)... (for quadratic factors)
Type 2
When repeating factors are present i.e. when denominator is of the form
k1
D(x) = (x – )
(x – )k2 ... {for linear factor}
= (ax2 + bx + c)k1 (px2 + qx + c)k {for quadratic
(1)
If function is linear.
N( x)
i.e.
(2)
2
( x  a) ( x  b) ( x  c)
B1
C3
A
B2
C1
C2



= ( x  a)  ( x  b) 
2
2
( x  c) ( x  c)
( x  b)
( x  c)3
3
N( x)
i.e of the form
2
(ax  bx  c) ( px2  qx  c)2
=
Ax  B
2
ax  bx  c

P1 x  Q1
 px
2
 qx  r

P2 x  Q2
  px
2
 qx  r 2

INTEGRATION OF RATIONAL &amp; IRRATIONAL FUNCTIONS
Integral of the form
 ax
dx
2
 bx  c
,

dx
2
and
ax  bx  c

ax 2  bx  c dx
For evaluating such integral we make the coefficients of x2 in ax2 + bx + c as one. Complete the square
by adding and subtracting the square of half of the coefficient of x to get the form
2

c
b2  
b 



a  x 


 a 4 a2  
2a 



Integrals of the form  ax
px  q
2
 bx  c
dx,

px  q
ax 2  bx  c
dx
and ( px  q) ax 2  bx  c dx
For evaluating such integrals we choose suitable constants A and B such that
d

px  q  A 
( ax 2  bx  c)   B
 dx

INDEFINITE INTEGRATION
ULTIMATE FORMULA HANDBOOK
px 2  qx  r

Integrals of the Form :
MATHONGO
ax 2  bx  c
dx
For evaluating such integrals we choose suitable constants, A, B and C such that
 d

(ax2  bx  c   C
px2 + qx + r = A (ax2 + bx + c) + B 
dx


Integrals of the form :
x
x2  1
4
 kx 2  1
dx
For evaluating such integrals, divide the numerator and denominator by x2. Complete the square of
2
2
1
1


denominator to get the form  x    a or  x    a
x
x


Then the integral can be evaluated by using the method of substitution.
Special Integration
Type I
Type II
Type III
x
x
x
x2  q
4
Divide numerator &amp; denominator by x2
 px2  q
dx
4
 px2  q
x2  r
4
write this in form
 px2  q
dx
1
2
q 
x
2
 
q  x2 

4
dx
 a  b cos x
or
dx
x  px  q


express x2 + r as
2
x2 + r = l x  q + m
where l + m = 1
&amp;
 a  b sin x
or
q l  m  r
dx
 a  b cos x  c sin x
2 tan
Working Rule :

2
Integration of Trigonometric Functions
Type I
q
Put sin x 
x
2
x
2
cos x 
2 x
1  tan
2
1  tan 2
1  tan 2
x
2
whichever is needed and then put z = tan
x
2
and
INDEFINITE INTEGRATION
x
2
 q

ULTIMATE FORMULA HANDBOOK

Type II
MATHONGO
sin x
dx,
a sin x  b co s x

cos x
d x , or
a sin x  b co s x

p sin x  q co s x
dx
c s in x  b co s x
Step – 1 : Put Numerator = A (dinominator) + B (derivative of denominator.) where a  0, b  0
Step – 2 : Then equate the coefficients of sinx and cosx to find A and B.
a sin x  b cos x  c
 p sin x  q cos x  r
Type 3.
(i)
Write Numerator =  (Diff. of denominator) + &micro; (Denominator) + v
i.e. a sin x + b cos x + c =  (p cosx – q sinx) + &micro; (psin x + q cosx + r) + v
Type 4.
 a sin
1
2
2
x  b cos x
dx .
1
 a  b sin
2
x
dx ,
1
 a  b cos
2
x
dx,
1
  a sin x  b cos x
(i)
Divide numerator and denominator both by cos2x
(ii)
Replace sec2 x, if any, in denominator by 1 + tan2 x
2
dx,
 a  b sin
1
2
x  cos2 x
dx
(iii) Put tan x = t so that sec2 x dx = dt
This substitution reduces the integral in the form
Integrals of the form :
 at
1
2
 bt  c
dt
dx
P
Q
where P and Q are linear or quadratic expression in x
1.
Q is linear and P is linear or quadratic., For evaluating such integrals, put Q = t2
2.
Q is quadratic and P is linear., For evaluating such integrals, put P =
3.
Both P and Q are pure quadratic., For evaluating such integrals, put x 
1
t
1
.
t
Integration of Irrational Functions
Types of functions (intergrand)
Approach
1.
  ax  b  a / n 
f  x, 
  a, b, c, d, , n  R 

  cx  d 

Substitute :
2.
f x,  ax  b

a/n
,  ax  b
/m

ax  b
 tn
cx  d
ax + b = tp, where p is L.C.M. of m and n.
INDEFINITE INTEGRATION
ULTIMATE FORMULA HANDBOOK
3.

f  x 


a 2  x2
MATHONGO
n
 
Workrule : x 
m + p  N, m + p &gt; 1
Workrule : a + bx = tx
a2  x2  t
1
4.
x m  a  bx 
p
1
5.
6.
m
L1  x 
n
L 2  x 
(i)
If n &gt; m;
L1 ( x)
t
L 2 ( x)
(ii)
If n &lt; m;
L 2 ( x)
t
L1 ( x)
(i)
If p  I, substitute x = ts
where s is L.C.M. of denominator of m &amp; n.
(ii)
If
xm(a + bxn)pdx
m 1
is an Integer, substitute a + bxn = ts
n
is the denominator of fraction p.
(iii)
m 1
 p substitute ax–n + b = ts
n
s is denominator of rational number p.
If
INDEFINITE INTEGRATION
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