Algebraic Expressions

 (a + b)2 = a2 + 2ab + b2
 (a – b)2 = a2 – 2ab + b2
 a2 – b2 = (a – b) (a + b)
 a2 + b2 = (a + b)2 -2ab
 (a + b)3 = a3 + b3 + 3ab (a + b) = a3+3a2b + 3ab2 + b3
 (a – b)3 = a3 – b3 – 3ab (a – b) = a3-3a2b + 3ab2 - b3
 a3 – b3 = (a – b) (a2 + ab + b2)
 a3 + b3 = (a + b) (a2 – ab + b2)

Trignometric formulae
sin θ x cosec θ = 1 cos θ x sec θ = 1 tan θ x cot θ = 1
1 1 1
sin θ = cos θ = tan θ =
cosec θ sec θ cot θ
1 𝟏 𝟏
cosec θ = sec θ = cot θ =
sin θ cos θ tan θ

sin (− θ) = − sin θ
cos (−θ) = cos θ
tan (−θ) = − tan θ
cot (−θ) = − cot θ
sec (−θ) = sec θ
cosec (−θ) = − cosec θ

Vaibhav Sir :- 8425924179

Angle 0° 30° 45° 60° 90° 180° 270° 360°

T.F
Sin 0 1 1 √3 1 0 −1 0
2 √2 2
Cos 1 √3 1 1 0 −1 0 1
2 √2 2
tan 0 1 1 √3 ∞ 0 ∞ 0
√3
Cot ∞ √3 1 1 0 ∞ 0 ∞
√3
Cosec ∞ 2 √2 2 1 ∞ −1 ∞
√3
Sec 1 2 √2 2 ∞ −1 ∞ 1
√3

IMP Values
Sin 0 = 0 cos 0 = 1
Sin 𝛑 = 0 cos 𝛑 = − 1
Sin 2𝛑 = 0 cos 2𝛑 = 1
Sin 3𝛑 = 0 cos 3𝛑 = − 1
Sin 4𝛑 = 0 cos 4𝛑 = 1
𝜋 𝜋
Sin = 1 cos = 0
2 2
3𝜋 3𝜋
Sin =−1 cos =0
2 2
5𝜋 5𝜋
Sin =1 cos =0
2 2
7𝜋 7𝜋
Sin =−1 cos =0
2 2

Vaibhav Sir :- 8425924179

 sin cos Tan
𝜋
–θ +Cos θ +Sin θ +Cot θ
2
𝜋
+θ +Cos θ −Sin θ −Cot θ
2
𝛑–θ +Sin θ −Cos θ −Tan θ
𝛑+θ −Sin θ −Cos θ +Tan θ
3𝜋 −Cos θ −Sin θ +Cot θ
−θ
2
3𝜋 −Cos θ +Sin θ −Cot θ
+θ
2
2𝛑 – θ −Sin θ +Cos θ −Tan θ
2𝛑 + θ +Sin θ +Cos θ +Tan θ

Compound Angle Formulae

 cos ( A + B ) = cos A.cos B – sin A.sin B
 cos ( A – B ) = cos A.cos B + sin A.sin B
 sin ( A + B ) = sin A.cos B + cos A.sin B
 sin ( A – B ) = sin A.cos B − cos A.sin B

Defactorization Formulae
 cos ( A + B ) + cos ( A – B ) = 2 cos A. cos B
 cos ( A + B ) − cos ( A – B ) = − 2 sin A. sin B
 sin ( A + B ) + sin ( A – B ) = 2 sin A. cos B
 sin ( A + B ) − sin ( A – B ) = 2 cos A. sin B

Factorization Formulae
𝐴+𝐵 𝐴−𝐵
 cos A + cos B = 2 cos ( ) cos ( )
2 2
𝐴+𝐵 𝐴−𝐵
 cos A – cos B = − 2 sin ( ) sin ( )
2 2
𝐴+𝐵 𝐴−𝐵
 sin A + sin B = 2 sin ( ) cos ( )
2 2
𝐴+𝐵 𝐴−𝐵
 sin A – sinB = 2 cos ( ) sin ( )
2 2

Vaibhav Sir :- 8425924179

 Trigonometric identity
sin2 θ + cos2 θ = 1 sin2 θ =1- cos2 θ cos2 θ =1- sin2 θ
1+tan2 θ = sec2 θ 1 = sec2 θ − tan2 θ tan2 θ = sec2 θ - 1
1 + cot2 θ = cosec2 θ 1 = cosec2 θ − cot2 θ cot2 θ = cosec2 θ - 1
tan A+tan B tan A−tan B
tan (A+B) = tan (A-B) =
1 − 𝑡𝑎𝑛 𝐴.𝑡𝑎𝑛 𝐵 1 + 𝑡𝑎𝑛 𝐴.𝑡𝑎𝑛𝐵
cot A.cot B−1 cot A.cot B+1
cot (A+B) = cot (A-B) =
cot A+cot B cot A−cotB

Double angle formula

2 𝑡𝑎𝑛 𝜃
Sin 2 θ = 2 sin θ.cos θ = 2 𝑠𝑖𝑛2 θ = 1 - cos 2 θ
1 + 𝑡𝑎𝑛2 𝜃
1 − Cos 2 θ
Cos 2 θ = Cos2 θ − 𝑠𝑖𝑛2 θ = 2Cos2 θ – 1 𝑠𝑖𝑛2 θ =
2
1 − 𝑡𝑎𝑛2𝜃
Cos 2 θ = 1 − 2 𝑠𝑠𝑠2 θ = 2 𝑐𝑜𝑠2 θ = 1 + cos 2
1 + 𝑡𝑎𝑛2𝜃
θ
2 𝑡𝑎𝑛 𝜃 1 + Cos 2 θ
tan 2 θ = 𝑐𝑜𝑠2 θ =
1 − 𝑡𝑎𝑛2𝜃 2
Half Angle Formulae
𝜃
𝜃 𝜃 2 𝑡𝑎𝑛 2 𝜃
sin θ = 2sin cos = 𝜃 2𝑠𝑖𝑛2 = 1 − cos θ
2 2 1 + 𝑡𝑎𝑛2 2 2

𝜃 𝜃 𝜃 𝜃 1 − cos θ
cos θ = cos2 – sin2 = 2cos2 − 1 𝑠𝑖𝑛2 =
2 2 2 2 2
𝜃
𝜃 1 − 𝑡𝑎𝑛2 2 𝜃
cos θ = 1 - 2 𝑠𝑖𝑛2 = 𝜃 2𝑐𝑜𝑠2 = 1+ cos θ
2 1 + 𝑡𝑎𝑛2 2 2

𝜃
2 𝑡𝑎𝑛 2 𝜃 1 + 𝑐𝑜𝑠 𝜃
tan θ = 𝜃 𝑐𝑜𝑠2 =
1 − 𝑡𝑎𝑛2 2 2 2

Triple Angle Formulae

3 𝑠𝑖𝑛𝜃 – 𝑠𝑖𝑛 3𝜃
Sin 3θ = 3 sinθ – 4sin3θ sin3θ =
4
3 𝑐𝑜𝑠 𝜃 + 𝑐𝑜𝑠 3𝜃
Cos 3θ = 4 cos3θ – 3 cosθ cos3θ =
4
3 𝑡𝑎𝑛 𝜃 − 𝑡𝑎𝑛3 𝜃
Tan 3θ =
1 − 3 𝑡𝑎𝑛2 𝜃

Vaibhav Sir :- 8425924179

 Logarithms
y = 𝑎𝑥 ……………. 𝑥 = 𝑙𝑜𝑔𝑎𝑦
𝑙𝑜𝑔𝑎𝑎 = 1
log 1 = 0
log e = 1
𝑚
loga ( ) = loga m – loga n
𝑛

loga mn= 𝑛. loga m

𝑙𝑜𝑔 𝑎
logba =
𝑙𝑜𝑔 𝑏
logba × logab = 1
Inverse Trignometric Formulae
1) sin−1 (sin θ) = θ sin (sin−1x) = x
2) cos−1 (cos θ) = θ cos (cos−1x) = x
3) tan−1 (tan θ) = θ tan (tan−1x) = x
4) cot−1 (cot θ) = θ cot (cot−1x) = x
5) sec−1 (sec θ) = θ sec (sec−1x) = x
6) cosec−1 (cosec θ) = θ cosec (cosec−1x) = x
𝜋 𝜋
7) sin−1 (cos θ) = sin−1 [𝑠𝑖𝑛 ( − 𝜃)] = – θ
2 2
𝜋 𝜋
8) cos−1 (sin θ) = cos−1 [𝑐𝑜𝑠 ( − 𝜃)] = – θ
2 2
𝜋 𝜋
9) tan−1 (cot θ) = tan−1 [𝑡𝑎𝑛 ( − 𝜃)] = – θ
2 2
𝜋 𝜋
10) cot−1 (tan θ) = cot−1 [𝑐𝑜𝑡 ( − 𝜃)] = – θ
2 2
𝜋 𝜋
11) sec−1 (cosec θ) = sec−1 [𝑠𝑒𝑐 ( − 𝜃)] = – θ
2 2
𝜋 𝜋
12) cosec−1 (sec θ) = cosec−1 [𝑐𝑜𝑠𝑒𝑐 ( − 𝜃)] = – θ
2 2
1
13) sin−1 (x) =cosec−1 ( )
𝑥
1
14) cos−1 (x) =sec−1 ( )
𝑥

Vaibhav Sir :- 8425924179

 1
15) tan−1 (x) =cot−1 ( )
𝑥
1
16) cot−1 (x) =tan−1 ( )
𝑥
1
17) sec−1 (x) =cos−1 ( )
𝑥
1
18) cosec−1 (x) =sin−1 ( )
𝑥
𝜋
19) sin−1 (x) + cos−1 (x) =
2
𝜋
20) tan−1 (x) + cot−1 (x) =
2
𝜋
21) sec−1 (x) + cosec−1 (x) =
2

Trigonometric functions of angles of a triangle

A+B+C=𝛑
A+B=π–C
A+C=π–B
B+C=π–A
 sin ( A + B ) = sin ( π – C )
sin ( A + B ) = sin C A
 cos ( A + B ) = cos ( π – C )
cos ( A + B ) = − cos C
 sin ( B + C ) = sin ( π – A ) B C
sin ( B + C ) = sin A
 cos ( B + C ) = cos ( π – A )
cos ( B + C ) = − cos A
 sin ( A + C ) = sin ( π – B )
sin ( A + C ) = sin B
 cos ( A + C ) = cos ( π – B )
cos ( A + C ) = − cos B

Vaibhav Sir :- 8425924179

 𝐴+𝐵 𝜋 𝐶
 sin ( ) = sin ( − )
2 2 2
𝐴+𝐵 𝐶
sin ( ) = cos ( )
2 2
𝐴+𝐶 𝜋 𝐵
 sin ( ) = sin ( − )
2 2 2
𝐴+𝐶 𝐵
sin ( ) = cos ( )
2 2
𝐵+𝐶 𝜋 𝐴
 sin ( ) = sin ( − )
2 2 2
𝐴+𝐶 𝐴
sin ( ) = cos ( )
2 2
𝐴+𝐵 𝜋 𝐶
 cos ( ) = cos ( − )
2 2 2
𝐴+𝐵 𝐶
cos ( ) = sin ( )
2 2
𝐴+𝐶 𝜋 𝐵
 cos ( ) = cos ( − )
2 2 2
𝐴+𝐶 𝐵
cos ( ) = sin ( )
2 2
𝐵+𝐶 𝜋 𝐴
 cos ( ) = cos ( − )
2 2 2
𝐴+𝐶 𝐴
cos ( ) = sin ( )
2 2

Vaibhav Sir :- 8425924179

 1. FORMULAE OF DERIVATIVE
𝑑 𝑑
𝑘=0 Sin x = Cos x
𝑑𝑥 𝑑𝑥
𝑑 𝑑
𝑥=1 Cos x = − Sin x
𝑑𝑥 𝑑𝑥
𝑑 𝑑
xn = n.xn−1 Tan x = Sec2 x
𝑑𝑥 𝑑𝑥
𝑑 𝑑
ex = ex Cot x = − Cosec2 x
𝑑𝑥 𝑑𝑥
𝑑 𝑑
ax = ax . log a Sec x = Sec x. Tan x
𝑑𝑥 𝑑𝑥
𝑑 1 𝑑
log x = Cosec x = − Cosec x. Cot x
𝑑𝑥 𝑥 𝑑𝑥
𝑑 1 −1 𝑑 1
= Sin−1 x =
𝑑𝑥 𝑥 𝑥2 𝑑𝑥 √1 − 𝑥 2
𝑑 1 −𝑛 𝑑 −1
𝑛 =
Cos−1 x =
𝑑𝑥 𝑥 𝑥𝑛 + 1 𝑑𝑥 √1 − 𝑥 2
𝑑 1 𝑑 1
√𝑥 = 2√𝑥 Tan−1 x =
𝑑𝑥 𝑑𝑥 1 + 𝑥2
𝑑 −1
Cot−1 x =
𝑑𝑥 1 + 𝑥2
𝑑 1
Sec−1 x =
𝑑𝑥 𝑥√𝑥 2 − 1
𝑑 −1
Cosec−1 x =
𝑑𝑥 𝑥√𝑥 2 − 1

Composite function
𝑑 𝑑
[f(x)]n = n.f (x)n−1. f (x)
𝑑𝑥 𝑑𝑥
𝑑 𝑑
𝑒 𝑓(𝑥) = 𝑒 𝑓(𝑥) f (x)
𝑑𝑥 𝑑𝑥
𝑑 𝑑
sin[f (x)] = cos[f(x)] f (x)
𝑑𝑥 𝑑𝑥

Vaibhav Sir :- 8425924179

 Rule :
𝑑 𝑑 𝑑
(u+v)= u+ v
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑 𝑑 𝑑
(u−v)= u− v
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑 𝑑 𝑑
u.v = u. v + v. 𝑢
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑 𝑑
𝑑 𝑢 𝑣 𝑢−𝑢 𝑣
𝑑𝑥 𝑑𝑥
=
𝑑𝑥 𝑣 𝑣2

Derivaties of Parametric Function

𝑑𝑦
𝑑𝑦 𝑑𝑡
= 𝑑𝑥
𝑑𝑥
𝑑𝑡

Higher order Derivative

𝑑𝑦 𝑑2𝑦 𝑑3𝑦 𝑑 4𝑦 𝑑5𝑦 𝑑 𝑛𝑦
= 2 = 3 = 4 = 5 =
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑛

1 2 3 4 5 n
2. Application of Derivative
 To find equation of tangent and normal
𝑑𝑦
Steps : 1) Find from above equation
𝑑𝑥
𝑑𝑦
2) Find value of given point P (x1,x2)
𝑑𝑥
𝑑𝑦
=m
𝑑𝑥

3) Equation of Tangent y – y1 = m x (x – x1)

1
Equation of Normal y – y1 = − (x – x1)
𝑚
𝑑𝑦
Where = m (slope) at P (x1, y1)
𝑑𝑥

 Rate Measure
𝑑𝑠 𝑑𝑣
V= a=
𝑑𝑡 𝑑𝑡

Vaibhav Sir :- 8425924179

 Aproximation
𝑓(𝑎+ℎ) −𝑓(𝑎)
f’(a) = lim
ℎ→0 ℎ

hf’(a) = 𝑓(𝑎 + ℎ) − 𝑓(𝑎)

f(a) + hf’(a) = f(a+h)
0° → Degree
‘ → minutes
“ → seconds
1° = 60’
1’ = 60”
To convert minutes to Degree ÷60
To convert second to minutes ÷60
To convert second to Degree ÷360
To convert Degree to minutes ×60
To convert minutes to second ×60
To convert Degree to second ×360
 Increasing and Decreasing function
Conditions f(x) → f’(x)
1) f’(x) > 0 +ve
f(x) is Increasing
2) f’(x) < 0 −ve
f(x) is Decreasing

Vaibhav Sir :- 8425924179

 To find maxima & minima for f(x)
Steps : 1) Find f’(x)
2) f’(x) = 0, Find points (x1,x2)
3) Find f’’(x) at point
4) If f’’(x) < 0 function is maximum at x1
If f’’(x) > 0 function is minimum at x2
5) To find minimum and maximum value
Find f (x1)
 Rolle’s Theorem :
If f(x) is a real valued function defined on a variable such that,
Conditions :-
i) f(x) is continuous in closed interval [a,b]
ii) f(x) is differentiable in open interval (a,b)
iii) f(a) = f(b) then there exist a real number c ϵ (a,b) such that
f ’(x) =0
[a,b] = a ≤ x ≤ b
(a,b) = a < x < b

Vaibhav Sir :- 8425924179

 Lagrange’s mean value Theorem
If f(x) is a real valued function defined on a real variable such that
Conditions :-
i) f(x) is continuous in closed interval [a,b]
ii) f(x) is differentiable in open interval (a,b) then there exist a real
number c ϵ (a,b) such that
𝑓(𝑏) − 𝑓(𝑎)
f ’(c) =
𝑏−𝑎
𝑦2 −𝑦1
M slope =
𝑥2 −𝑥1

 Increasing
a.b = 0 (a.b → + ve)
Case (i) a → + ve b → + ve
a>0 b>0
Case (ii) a → − ve b → − ve
a<0 b<0
 Decreasing
a.b < 0 (a.b → − ve)
Case (i) a → + ve b → − ve
a>0 b<0
Case (ii) a → − ve b → + ve
a<0 b>0

Vaibhav Sir :- 8425924179

 3. Indefinite Integration
∫ 1 𝑑𝑥 = x + c
𝑥𝑛 + 1
∫ 𝑥 𝑛 dx = 𝑛+1
1
∫ 𝑥 dx = log x + c

∫ 𝑒 𝑥 dx = 𝑒 𝑥 + c
𝑎𝑥
∫ 𝑎 𝑥 dx = 𝑙𝑜𝑔 𝑎
+c

∫ 𝑐𝑜𝑠 𝑥 dx = sin x + c
∫ 𝑠𝑖𝑛 𝑥 dx = − cos x + c
∫ 𝑠𝑒𝑐 2 x dx = tan x + c
∫ 𝑐𝑜𝑠𝑒𝑐 2 x dx = − cot x + c
∫ 𝑠𝑒𝑐 𝑥. 𝑡𝑎𝑛 𝑥 dx = sec x + c
∫ 𝑐𝑜𝑠𝑒𝑐 𝑥. 𝑐𝑜𝑡 𝑥 dx = − cosec x + c
∫ 𝑡𝑎𝑛 𝑥 dx = log | 𝑠𝑒𝑐 𝑥 | + c
∫ 𝑠𝑒𝑐 𝑥 dx = log |𝑠𝑒𝑐 𝑥 + 𝑡𝑎𝑛 𝑥| + c
∫ 𝑐𝑜𝑡 𝑥 dx = log |𝑠𝑖𝑛 𝑥 | + c
∫ 𝑐𝑜𝑠𝑒𝑐 𝑥 dx = log |𝑐𝑜𝑠𝑒𝑐 𝑥 − 𝑐𝑜𝑡 𝑥| + c
1
∫ √1 − 𝑥2 dx = sin−1x + c = − cos−1x + c
1
∫ 1 + 𝑥2 dx = tan−1x + c = − cot−1 + c
1
∫ 𝑥√𝑥2 − 1 dx = sec−1x + c = − cosec−1x + c

Vaibhav Sir :- 8425924179

 Theorem 1
∫[𝑓(𝑥) + 𝑔(𝑥)] dx = ∫ 𝑓(𝑥) dx + ∫ 𝑔(𝑥) dx
Theorem 2
∫[𝑓(𝑥) − 𝑔(𝑥)] dx = ∫ 𝑓(𝑥) dx − ∫ 𝑔(𝑥) dx
Theorem 3
∫ 𝑘 𝑓(𝑥) dx = k ∫ 𝑓(𝑥) dx
Chain rule
𝑒 𝑎𝑥 + 𝑏
∫ 𝑒 𝑎𝑥 + 𝑏 dx = 𝑑 +c
𝑎𝑥 +𝑏
𝑑𝑥

[𝑓(𝑥)]𝑛 + 1
∫ 𝑓 ′(𝑥)[f (x)]n dx = 𝑛+1
+c
𝑓 ′(𝑥)
∫ dx = log [f(x)] + c
𝑓(𝑥)
𝑓 ′𝑥
∫ √𝑓(𝑥) dx = 2√𝑓(𝑥) + c

Some Special Integrals

1 𝑥
∫ √𝑎2 −𝑥2 dx = Sin−1 𝑎 + c
1 1 𝑥
∫ 𝑎2 +𝑥2 dx = 𝑎 tan−1 𝑎 + c
1 1 𝑥
∫ 𝑥√𝑥2 − 𝑎2 dx = 𝑎 Sec−1 𝑎 + c
1 1 𝑥−𝑎
∫ 𝑥2 −𝑎2 dx = 2𝑎 log [𝑥+𝑎] + c
1 1 𝑎+𝑥
∫ 𝑎2 −𝑥2 dx = 2𝑎 log [𝑎−𝑥] + c
1
∫ √𝑥2 + 𝑎2 dx = log [𝑥 + √𝑥 2 + 𝑎2 ] + c
1
∫ √𝑥2 − 𝑎2 dx = log [𝑥 + √𝑥 2 − 𝑎2 ] + c

Vaibhav Sir :- 8425924179

 Expressions Substitutions
√1 − 𝑥 2 x = sin θ or x = cos θ
√1 + 𝑥 2 x = tan θ or x = cot θ
√𝑥 2 − 1 x = sec θ or x = cosec θ
𝑎+𝑥 𝑎−𝑥 x = a cos 2θ or x = a cos θ
√𝑎−𝑥 or √𝑎+𝑥
1+𝑥 1−𝑥 x = cos 2θ or x = cos θ
√ or √
1−𝑥 1+𝑥
𝑎+𝑥 2 𝑎−𝑥 2 x2 = a cos 2θ or x2 = a cos θ
√ or √
𝑎−𝑥 2 𝑎+𝑥 2
2𝑥 1 − 𝑥2 x = tan θ
or
1 + 𝑥2 1 + 𝑥2
3x – 4x3 or 1−2x2 x = sin θ
4x3− 3x or 2x2−1 x = cos θ
3𝑥 − 𝑥 3 x = tan θ
1 − 3𝑥 2
2𝑓(𝑥) 1 − [𝑓(𝑥)]2 f(x) = tan θ
or
1 + [𝑓(𝑥)]2 2
1 + [𝑓(𝑥)]

Some special integtrals

Type 1
1 1
∫ 𝑎𝑥2 + 𝑏𝑥 + 𝑐 & ∫ √𝑎𝑥2 + 𝑏𝑥 + 𝑐

 Make coefficient of x2 is 1 by taking ‘a’ common

1 2
 Add and Subtract ( . 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥) in equation.
2
 Make standard integration form
Type 2
1
𝑎𝑠𝑖𝑛2 𝑥+ 𝑏𝑐𝑜𝑠 2 𝑥+𝑐

 ÷ by sin2x or cos2x
 Replace sec2x into 1 + tan2x

Vaibhav Sir :- 8425924179

 Cosec2x into 1 + cot2x
 Put tan x = t or cot x = t
1
You will get form ∫
𝑎𝑡 2 +𝑏𝑡+𝑐

Type 3
CASE 1
1
∫ 𝑎 𝑠𝑖𝑛 𝑥+𝑏 𝑐𝑜𝑠 𝑥+𝑐 dx
𝑥 2𝑑𝑡
Put tan = t dx =
2 1+𝑡 2
2𝑡 1−𝑡 2
Sin x = cos x =
1+𝑡 2 1+ 𝑡 2

CASE 2
1
∫ 𝑎 𝑠𝑖𝑛 2𝑥+𝑏 𝑐𝑜𝑠 2𝑥+𝑐 dx
𝑑𝑡
Put tan x = t dx =
1+𝑡 2
2𝑡 1−𝑡 2
Sin x = cos x =
1+𝑡 2 1+ 𝑡 2

INTEGRATION BY PARTS
LIATE → Series
L – Logarithmic function
I - Inverse function
A – Algebric function
T – Trigonometric function
E – Exponential function
FIS – IDFIL (trick)
𝑑𝑢
∫ 𝑢. 𝑣 𝑑𝑥 = u∫ 𝑣 𝑑𝑥 −∫ [𝑑𝑥 ∫ 𝑣 𝑑𝑥] dx
𝑥 𝑎2 𝑥
∫ √𝑎2 − 𝑥 2 dx = 2 √𝑎2 − 𝑥 2 + 2
sin−1 + c
𝑎

Vaibhav Sir :- 8425924179

 𝑥 𝑎2
∫ √𝑥 2 + 𝑎2 dx = 2 √𝑥 2 + 𝑎2 + 2
log|𝑥 + √𝑥 2 + 𝑎2 |
𝑥 𝑎2
∫ √𝑥 2 − 𝑎2 dx =
2
√𝑥 2 − 𝑎2 −
2
log|𝑥 + √𝑥 2 − 𝑎2 |

∫ 𝑒 𝑥 [𝑓(𝑥) + 𝑓 ′(𝑥)] dx = ex. f(x) + c

∫ 𝑒 𝑓(𝑥) . 𝑓 ′(𝑥) dx = ef(x) + c
𝑒 𝑎𝑥
∫ 𝑒 𝑎𝑥 .cos (bx + c) dx = 𝑎2 +𝑏2 [a. cos (bx+c) + b. sin (bx+c)]
𝑎𝑥 𝑒 𝑎𝑥
∫ 𝑒 .sin (bx + c) dx = 𝑎2 +𝑏2 [a. sin (bx+c) − b. cos (bx+c)]

Form Of Rational Fraction Form of Partial Function

𝑝𝑥 + 𝑞 𝐴
+𝐵
𝑥−𝑎
(𝑥 − 𝑎)(𝑥 − 𝑏)𝑎 ≠ 𝑏
𝑝𝑥 + 𝑞 𝐴
+
𝐵
𝑥−𝑎 (𝑥−𝑎)2
(𝑥 − 𝑎)2
𝑝𝑥 2 + 𝑞𝑥 + 𝑟 𝐴
+
𝐵
+
𝐶
𝑥−𝑎 𝑥−𝑏 𝑥−𝑐
(𝑥 − 𝑎)(𝑥 − 𝑏)(𝑥 − 𝑐)
𝑝𝑥 2 + 𝑞𝑥 + 𝑟 𝐴
+
𝐵
+
𝐶
𝑥−𝑎 (𝑥−𝑎)2 𝑥−𝑏
(𝑥 − 𝑎)2 (𝑥 − 𝑏)
𝑝𝑥 2 + 𝑞𝑥 + 𝑟 𝐴
+
𝐵𝑥+𝐶
𝑥−𝑎 𝑥 2 +𝑏𝑥+𝑐
(𝑥 − 𝑎)(𝑥 2 + 𝑏𝑥 + 𝑐)

𝑝𝑥+𝑞
∫ 𝑎𝑥2 +𝑏𝑥+𝑐 dx
𝑑
Step 1 : (px+q) = A. (𝑎𝑥 2 + 𝑏𝑥 + 𝑐) + B
𝑑𝑥
𝑑
A.𝑑𝑥(𝑎𝑥 2 +𝑏𝑥+𝑐) + B
Step 2 : ∫ dx
(𝑎𝑥 2 +𝑏𝑥+𝑐)
𝑑
𝑑𝑥
(𝑎𝑥 2 +𝑏𝑥+𝑐) 1
Step 3 : 𝐴 ∫ dx + B ∫ dx
(𝑎𝑥 2 +𝑏𝑥+𝑐) 𝑎𝑥 2 +𝑏𝑥+𝑐
𝑝𝑥+𝑞
∫ √𝑎𝑥2 +𝑏𝑥+𝑐 dx

Vaibhav Sir :- 8425924179

 𝑑
Step 1 : (px+q) = A. (𝑎𝑥 2 + 𝑏𝑥 + 𝑐) + B
𝑑𝑥
𝑑
A. (𝑎𝑥 2 +𝑏𝑥+𝑐) + B
𝑑𝑥
Step 2 : ∫ dx
√(𝑎𝑥 2 +𝑏𝑥+𝑐)
𝑑
𝑑𝑥
(𝑎𝑥 2 +𝑏𝑥+𝑐) 1
Step 3 : 𝐴 ∫ dx + B ∫ dx
√(𝑎𝑥 2 +𝑏𝑥+𝑐) √(𝑎𝑥 2 +𝑏𝑥+𝑐)

4.Indefinite Integral
𝑎
∫0 𝑓(𝑥) 𝑑𝑥 = 0
𝑏 𝑎
∫𝑎 𝑓(𝑥) 𝑑𝑥 = − ∫𝑏 𝑓(𝑥) 𝑑𝑥
𝑐 𝑏 𝑐
∫𝑎 𝑓(𝑥) 𝑑𝑥 = ∫𝑎 𝑓(𝑥) 𝑑𝑥 + ∫𝑏 𝑓(𝑥) 𝑑𝑥
𝑏 𝑏
∫𝑎 𝑓(𝑥) 𝑑𝑥 = ∫𝑎 𝑓(𝑎 + 𝑏 − 𝑥) dx
𝑎 𝑎
∫0 𝑓(𝑥) dx = ∫0 𝑓(𝑎 − 𝑥) dx
2𝑎 𝑎 𝑎
∫0 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑥) 𝑑𝑥 + ∫0 𝑓(2𝑎 − 𝑥) dx
𝑎 𝑎
∫−𝑎 𝑓(𝑥) dx = 2∫0 𝑓(𝑥) dx ……if f(x) is even
=0 ……if f(x) is odd

5. Application of definite integral

Area under curve and axis is given by
𝑏
A = ∫𝑎 𝑓(𝑥) dx
Area between two curve is given by
𝑏 𝑐
A = ∫𝑎 𝑓(𝑥) + ∫𝑏 𝑓(𝑥) 𝑑𝑥

Vaibhav Sir :- 8425924179

 6. differential equations
Order : Highest order of the Derivative
Degree : Power of highest order Derivative.
Order and Degree is defined only when it is free from radicals or
fractions.
1 𝑑2𝑦 𝑑𝑦 𝑑𝑦
For Eg. √,∛,∜, , ,sin( ),cos( )
𝑑𝑦/𝑑𝑥 𝑑𝑥 2 𝑑𝑥 𝑑𝑥

HOMOGENOUS DIFFERENTIAL EQUATION :

A differential equation f(x,y)dx + g(x,y) dy = 0 is said to be
homogenous differential equation:
Steps : -
𝑑𝑦 −𝑓(𝑥,𝑦)
1) Express the homogenous D.E. in the form = …..eqn (i)
𝑑𝑥 𝑔(𝑥,𝑦)
𝑑𝑦 𝑑𝑦
2) Put y = vx and =v+x in eqn (i) and cancel x from R.H.S.
𝑑𝑥 𝑑𝑥
𝑑𝑦
The equation reduces to the form v + x + = − f(v)
𝑑𝑥

3) Take v on R.H.S. and separate the variables v and x.

4) Integrate both sides in obtain the solution in v and x.
5) To obtain the required solution in terms of x and y we resubstitute
𝑦
the value of v i.e v = .
𝑥

Linear differential equation

Steps : -
𝑑𝑦
1) Write the given D.E. in the form + Py = Q, where P,Q are
𝑑𝑥
constants or functions of x only.
2) Find integrating factor I.F = 𝑒 ∫ 𝑝 𝑑𝑥

Vaibhav Sir :- 8425924179

3) Write the solution of the given differential equations as
y (I.F) = ∫(𝑄 . 𝐼. 𝐹. ) dx + c
This is the required general solution of the D.E.
𝒅𝒚
Note : -If the D.E. is in the form + px = Q ,Where P and Q are
𝒅𝒙
constants or functions of y only, the I.F = 𝐞∫ 𝐩 𝐝𝐱 and the general
solution of the D.E. is given by x (I.F) = ∫(𝑸 . 𝑰. 𝑭. ) dx +

7. PROBABILITY DISTRIBUTION
Random Experiment :
An experiment whose outcome can’t be predicted with certainly is
called as Random Experiment.
Sample Space :
The set of all possible outcome of a random experiment is called as
Sample space for that experiment.
Event : Any subject of a sample space is known as Event.
1) Equally likely Events.
2) Exhaustive events.
3) Mutually exclusive events.
A coin is tossed – Head / Tail
A bomb is dropped – Hit / Not Hit
An arrow is fired – Hit / Not Hit
A exam attempted – Pass / Fail

Vaibhav Sir :- 8425924179

IMP SAMPLE SET
A coin tossed
S = {H , t} n(s) = 2
A 2 coin tossed
S = { HH,HT,TH,TT } n(s) = 4
A 3 coin tossed
S = { HHH,HHT,HTH,HTT,THH,THT,TTH,TTT } n(s) = 8
A 4 coin tossed
S = { HHHH, HHHT, HHTH, HTHH,THHH, HHTT, TTHH, HTTH,
THTH,HTHT,THHT,TTTH,TTHT,THTT,HTTT,TTTT }
n(s) = 16
Tosses Of Dice :
S = { 1,2,3,4,5,6 } n(s) = 6
Tosses of 2 Dice :
S = { (1,1) , (1,2) , (1,3) , (1,4) ,(1,5) , (1,6) ,
(2,1) , (2,2) , (2,3) , (2,4) , (2,5) ,(2,6) ,
(3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) ,
(4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) ,
(5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) ,
(6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) }
n(s) = 36
Probability of event X
𝒏(𝒙)
P(x) =
𝒏(𝒔)

Vaibhav Sir :- 8425924179

  Random Variables
It represents a possible numerical value from a random experiment
Random Variables

Discrete Continous
Random Variables
 Probability = ∑ 𝑃(𝑥𝑖) = 1
 Mean E(x) = ∑ 𝑥𝑖 . 𝑝𝑖
 Variance v(x) = ∑(𝑥 2 ) – [∑(𝑥)]2
 Standard Deviation σx = √𝑣(𝑥)
Always p + q = 1
Where p = Probability of success , q = Probability of failure
8.BIONOMIAL DISTRIBUTION
n 𝑛! n
Cx = C0 = nCn = 1
(𝑛−𝑥)! 𝑥!

P (X = x) = nCx (p)x (q)n−x (binomial distribution)

P+q=1
Where p = Probability of Success , q = Probability of Failure
Mean / expected value
Mean / μ = E(x) = n.p
Variance
V (x) = n.p.q
Standard Deviation

S.D. = √𝑣(𝑥)
S.D. = √𝑛. 𝑝. 𝑞

Vaibhav Sir :- 8425924179