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Coordinate Geometry
Course- Diploma
Semester-II
Subject- Advanced Mathematics
Unit- I
RAI UNIVERSITY, AHMEDABAD

Unit-I COORDINATE GEOMETRY
Introduction— The analytical geometry is also known as coordinate geometry, or Cartesian geometry. It is the
study of geometry using a coordinate system.
Application— Analytical geometry is widely used in physics and engineering, and is the foundation of most
modern fields of geometry, including algebraic, differential, discrete, and computational geometry. Usually the
Cartesian coordinate system is applied to manipulate equations for planes, straight lines, and squares, often in
two and sometimes in three dimensions. Geometrically, one studies the Euclidean plane (2 dimensions) and
Euclidean space (3 dimensions).
Cartesian coordinate system— Cartesian coordinate system is of two types—
1. Euclidean plane (2 dimensions) or 2-dimensional coordinate system or Plane geometry
2. Euclidean space (3 dimensions) or 3-dimensional coordinate system or Solid geometry
Plane geometry—A two dimensional coordinate system is as shows in the figure given below—There are two
perpendicular lines. The horizontal line is called as − and the vertical line is called as − . Each of
these axis is devide in unity of it. The point of intersection of coordinate axis is a point called as origin.
1.1.1 Point— A point in coordinate geometry is represented by an ordered pair ( , ) of numbers. It is an
exact location in the coordinate plane and denoted by a dot. Where and represents the perpendicular
distance of point from − axis and − axis respectively as shown—
1.1.2 Line segment, Line and Ray— If we join two points in such a way that distance between that points
become minimum, the that geometrical structure is known as Line segment. If we extend this line segment
indefinitely from both ends, it is called as Line. If we extend the line segment indefinitely from one end keeping
another end fixed, then such geometrical structure is called as Ray.
2-dimensional coordinate system
Representation of a point on coordinate system
Line Segment, Ray and Line

1.1.3 Distance between two points—
Distance between two points A( , ) and B( , ) is given by—
= ( − ) + ( − )
1.1.4 Dividing a line segment in the ratio m: n—
Let a Point ( , ) divides the line segment PQ in ratio m: n,
Where, Points P and Q are P ≡ ( , ) and Q ≡ ( , )
Internal Division—
=
+
+
=
+
+
External Division—
=
−
−
=
−
−
Mid Point—
=
+
=
+
Distance between two points
Internal Division
External Division
Mid Point
External Division

1.1.5 Area of the triangle—The area of the triangle having vertices A( , ), B( , ) and C( , ) is given
by—
1.1.6 Centroid— Let ( , ), ( , ) and ( , ) are vertices of a triangle, Then its Centroid is the
point of intersection of the Medians. It is given by—
=
+ +
=
+ +
1.1.7 Incenter— Let ( , ), ( , ) and ( , ) are vertices of a triangle, Then its Incenter is the point
of intersection of the angle bisectors. It is given by—
=
+ +
+ +
=
+ +
+ +
1.1.8 Circum-center— Let ( , ), ( , ) and ( , ) are vertices of a triangle, Then its Circum-
center is the point of intersection of the side perpendicular bisectors. It is given by—
= =
Centroid
Incenter
∆=
1
2 1 1 1
∆=
1
2
| ( − ) + ( − ) + ( − )|
Note—These three points will be collinear if ∆= .
Circum-center

1.1.9 Orthocenter— Let ( , ), ( , ) and ( , ) are vertices of a triangle, Then its Orthocenter is
the point of intersection of the altitudes of the triangle. It is given by—
= =
1.1.10 The Euler Line—The Euler line of a triangle is the line which passes through the orthocenter, circum-
center, and centroid of the triangle.
Orthocenter

EXERCISE-I
Question— A point divides internally the line- segment joining the points (8, 9) and (-7, 4) in the ratio
2 : 3. Find the co-ordinates of the point.
Question— A (4, 5) and B (7, - 1) are two given points and the point C divides the line segment AB externally
in the ratio 4 : 3. Find the co-ordinates of C.
Question— Find the ratio in which the line-segment joining the points (5, - 4) and (2, 3) is divided by the x-
axis. Also find the mid-point of these two points.
Question—Find the centroid, in centre, circumcentre and orthocenter of a triangle whose vertices are
(5, 3), (6, 1) and (7, 8). Also find the area of the triangle.

1.2.1 Straight line—The shortest path of line joining of any two point gives rise to a straight line. A general
equation of a straight line is given by—
+ + = .
1.2.2 Slope of a straight line—Slope of any straight line is the tangent of angle (− < ≤ ) make by the
positive −axis. Slope of a line is also called as gradient and it is denoted by m.
Slope of the line joining the points P( , ) and Q( , ) is given by—
1.2.3 Different form of equation of straight lines—
Slope-intercept form—The equation of a straight line having slope m and an intercept c on −axis is
= + .
Slope-Point form— The equation of a straight line passing through the fixed point( , ) and having slope m
is
− = ( − )
Two point form—The equation of the line passing through two fixed points A( , ) and B( , ) is
− =
−
−
( − )
Intercept form—The equation of the line cutting off intercepts a and b on −axis and −axis respectively is
+ =
Parametric form— The equation of a straight line passing through a fixed point P(x , y ) and making an angle
θ, 0 ≤ θ ≤ π with positive direction of x −axis is—
−
=
−
= =

Normal or perpendicular form—
1.2.4 Angle between two straight lines— The angle between two straight lines is given by—
Condition for coincident, parallel, perpendicular and intersecting of two straight lines—
1. coincident— Two lines + + = 0 and + + = 0 will be coincident if—
= =
2. Parallel—Two straight lines are said to be parallel if their slope will be equal.
(a) Lines + + = 0 and + + = 0 will be parallel if—
= ≠
(b) Lines = + and = + will be parallel if—
=
3. Perpendicular— Two straight lines are said to be perpendicular if the product of their slopes will be
equal to −1.
(a) The lines a x + b y + c = 0 and a x + b y + c = 0 will be perpendicular if—
+ = 0
(b) Lines = + and = + will be perpendicular if—
= −1
4. Intersecting— Two lines + + = 0 and + + = 0 intersects each other if—
≠
Length of perpendicular— Length of perpendicular drawn from a point P( , ) to a straight line + +
= 0 is—
=
| + + |
√ +
Distance between two parallel lines— Distance between two parallel lines + + = 0 and + +
= 0 is—
=
| − |
√ +
The equation of a straight line in terms of the length of
perpendicular p from the origin upon it and the angle
which this perpendicular makes with the positive
direction of −axis is—
+ =
=
−
+
=
−
+

EXERCISE-II
Question—Find the equation of a straight line which makes an angle of 135° with the x −axis and cuts y −axis
at a distance − 5 from the origin.
Question—Find the equation of the straight line which passes through the point (1, - 2) and cuts off equal
intercepts from axes.
Question—Find the equation of a straight line passing through (-3 , 2) and cutting an intercept equal in
magnitude but opposite in sign from the axes.
Question—If the coordinates of A and B be (1, 1) and (5, 7), then find the equation of the perpendicular
bisector of the line segment AB.
Question—Find the equation of the straight line passing through the point(1,2)
and having a slope of 3.
Question—Find the equation of the straight line passing through the point(3,2) and parallel to the line
2x + 3y = 1.
Question—Find the equation of the straight line passing through the point(4,1) and perpendicular to
3x + 4y + 2 = 0.
Question— Find the equation of a line through the points (1,2) and (3,1). What is its slope? What is its y
intercept?

1.3.1 Circle— A circle is the set of all points that are at the same distance r from a fixed point. A general
equation of a circle is—
+ + + + = .
This equation represents a circle having center at (− , − ) and radius = + −
There are several standard forms of equation of a circle as given below—
Center-radius form—The equation of a circle having center(ℎ, ) and radius r is—
( − ) + ( − ) =
Center at origin—The equation of a circle having center at origin and radius r is—
+ =
Diameter form—Let A( , ) and B( , ) are the extremities of a diameter of the circle. Then the equation
of the circle is— ( − )( − ) + ( − )( − ) =
Three point form—The equation of the circle passing through the points A( , ), B( , ) and C( , )is—
+ 1
+ 1
+ 1
+ 1
= 0.
Polar form— The polar equation of a circle centered at P(r , θ ) with radius R units −
r + r − 2rr cos( θ − θ ) = R
(− , − )
r

1.3.2 Tangent to a circle—There are two types of tangent to any circle—
1. Tangent through a point on the circle
2. Tangent from a point outside to the circle
Tangent through a point on the Circle—
Let P( , ) is a point on the circle. The equation of the tangent line is as given below—
Equation of circle Equation of tangent line
( − ℎ) + ( − ) = ( − ℎ)( − ℎ) + ( − )( − ) =
+ = + =
+ + 2 + 2 + = 0 + + ( + ) + ( + ) + = 0
Tangent from a point ( , ) outside to the circle—
If + + 2 + 2 + = 0 is the equation of the circle, then—
S ≡ + + 2 + 2 +
S ≡ + + 2 + 2 +
T ≡ + + ( + ) + ( + ) +
Normal line—A normal line always pass through the center of the circle. So, we can easily find out the
equation of normal by using 2-points form.
 A Pair of tangents lines from an external
point P( , ) is =
 The cord of contact with respect to external point
P( , ) is =

EXERCISE-III
Question—Find the center of the circle + + 20 − 10 − 19 = 0.
Question—Find the area of the circle 4 + 4 + 16 − 16 − 16 = 0.
Question—Find the equation of the circle having center at origin and radius 2.
Question—Find the equation of the circle having center(2,3) and radius 4.
Question—Find the equation of the circle having extremities of diameter (2,5) and (4,1).
Question—Find the equation of the circle Passing through the points (1,3), (3, −1) and (2,4).
Question—Find the equation of the tangent line to the circle x + y + 4x − 5y + 1 = 0 at the point(1,2).
Question—Find the equation of the tangent line to the circle + − 2 − 4 + 6 = 0 at the point(1,2).
Question—Find the equation of the normal line to the circle + + 2 − 4 = 0 at the point(1,1).
Question—Find the radius of the circle having tangents 2 + 3 + 6 = 0 and 2 + 3 − 3 = 0.

Reference—
1. en.wikipedia.org/wiki/Analytic_geometry
2. https://www.khanacademy.org
3. www.stewartcalculus.com
4. www.britannica.com
5. mathworld.wolfram.com
6. www.mathsisfun.com
7. www.purplemath.com
8. http://www.education.gov.za/
9. http://www.cut-the-knot.org/
10. math2.org

Complex number
Course- Diploma
Semester-II
Unit- II

Unit-II COMPLEX NUMBER
Introduction— Indian mathematician Mahavira (850 A.D.) was first to mention in his work
'Ganitasara Sangraha'; 'As in nature of things a negative (quantity) is not a square (quantity), it
has, therefore, no square root'. Hence there is no real number which satisfies the polynomial
equation + 1 = 0.
A symbol√−1, denoted by letter “i” was introduced by Swiss Mathematician, Leonhard Euler
(1707-1783) in 1748 to provide solutions of equation + 1 = 0. “i” was regarded as a
fictitious or imaginary number which could be manipulated algebraically like an ordinary real
number, except that its square was – 1. The letter “i” was used to denote√−1 , possibly
because “i” is the first letter of the Latin word 'imaginarius'.
2.1Definition— A number of the form = + , [ , ∈ & = √−1 ], is called a complex
number.
Figure 2.01 represents the complex plane. It consists of a Real axis and an Imaginary axis.
2. 2 Real & Imaginary part of a complex number—
Let = + is a complex number. Then it’s real part & imaginary part is given by—
( ) = ( ) =
2. 3 Representation of a complex number = + –
A complex number = + can be represented on the co-ordinate as given below—
Fig - 2.01
This system of representing complex number is
also called as Argand diagram.

2. 4 Modulus of a complex number— Let = + is a complex number,
then its modulus is given by—
| |= + .
Example— Find the modulus of the complex number = + .
Solution— Given complex number is z = 3 + 4i.
Since, modulus of z = x + iy is given by | |= +
Hence, required modulus of z is | |=√3 + 4 =√9 + 16 = √25
∴ | | = 5
2. 5 Argument of a complex number—
Let = + is a complex number,
then its argument is given by—
arg( ) = = tan .
Note— is measured in radians and positive in the counterclockwise sense.
For = , ( ) is not defined.
The value of that lies in the interval − ≤ ≤ is called as Principal value of the argument
of (≠ 0).
Hence, − ≤ ( ) ≤
Example— Find the argument of the complex number = + .
Solution— given complex number is z = 1 + i.
Since, argument of z = x + iy is given by arg( ) = tan and the point (1,1) lies in first
quadrant.
Hence, arg( ) = tan = tan (1) =
Example— Find the argument of the complex number = − + .
Solution— given complex number is z = −1 + i.
Since, argument of z = x + iy is given by arg( ) = tan and the point (−1,1) lies in second
quadrant.
Hence, arg( ) = tan = tan (−1) = − =
Example— Find the argument of the complex number = − − .
Solution— given complex number is z = −1 − i.
Since, argument of z = x + iy is given by arg( ) = tan and the point (−1, −1) lies in third
quadrant.
Hence, arg( ) = tan = tan (1) = − + =
Example— Find the argument of the complex number = − .
Solution— given complex number is z = 1 − i.
Fig - 2.02

Since, argument of z = x + iy is given by arg( ) = tan and the point (1, −1) lies in fourth
quadrant.
Hence, arg( ) = tan = tan (−1) = −
2.6 Cartesian representation of a complex number—
A complex number of the form = + is known as Cartesian representation of complex
number. This can also be written in paired form as ( , ). Cartesian representation of complex
number = + is shown in Fig- 1.02.
2.7 Polar representation of a complex number—
A complex number of the form = + can be represented in polar form as = ( +
). Where and is given by—
= + and
= tan
This is shown in Fig- 2.03.
2.8 Euler representation of a complex number—
A complex number of the form = + or = ( + ) can be represented in
“Euler form” form as---
=
Where and is given by—
= + , = tan
This is shown in Fig- 2.03.
2.9 Conversion of a complex number given in cartesian system to polar system—
If = + is a complex number given in Cartesian system then it can be written in polar
form by writing—
= ( + )
Where = + and = tan .
Example— Convert the following complex numbers in polar form—
( ) + ( ) + ( ) − + ( ) + √
Solution—
( ) Given complex number is = 3 + 4 ,
Hence, = √3 + 4 = √9 + 16 = √25 = 5 ∴ = 5 and
= tan = tan
Fig - 2.03

Thus, = 3 + 4 can be written in polar form as = 5( + ), where = tan
( ) Given complex number is = 1 + ,
Hence, = √1 + 1 = √1 + 1 = √2 ∴ = √2 and
= tan = tan =
Thus, = 1 + can be written in polar form as = √2( + )
( ) Given complex number is = −1 + ,
Hence, = (−1) + 1 = √1 + 1 = √2 ∴ = √2 and
= tan = tan =
Thus, = −1 + can be written in polar form as = √2( + )
( ) Given complex number is = 1 + √3 ,
Hence, = (1) + (√3) = √1 + 3 = √4 = 2 ∴ = 2 and
= tan = tan
√
=
Thus, = 1 + √3 can be written in polar form as = 2( + )
2.10 Conversion of a complex number given in polar system to cartesian system —
Example— Convert the followings complex numbers in Cartesian form—
( ) ( ) √2 ( )√3 ( ) √2
Solution—
( ) Given complex number is .
Here, = 1 = , let = + is Cartesian representation.
Then, = = 1. = 0 and
= = 1.
2
= 1
Fig - 2.04
If = (cos + ) is a complex
number given in Polar system then it
can be written in Cartesian system as—
= +
Where = and = .

Thus, (1, ) can be written in Cartesian form as = 0 +
( ) Given complex number is √2
Here, = √2 = , let = + is Cartesian representation.
Then, = = √2. = 1 and = = √2. = 1
Thus, √2 can be written in Cartesian form as = 1 +
Then, = = √3. =
√
and
= = √3. =
Thus, √3 can be written in Cartesian form as =
√
+
Then, = = √2. = −1 and
= = √2.
3
4
= 1
Thus, √2 can be written in Cartesian form as = −1 +
2.10 Conjugate of a complex number—
Example— Find the conjugate of the complex number = + .
Solution— Given complex number is z = 2 + 3i.
Since, complex conjugate of z = x + iy is given by = − .
Hence, required complex conjugate is = − .
2.11 Addition of two complex numbers—
Fig - 1.05
Let a complex number is given by = + , then the
conjugate of complex number is denoted by ̅ and it is given
by = − .
A complex number = + and its conjugate = − is
represented in Argand plane as shown in Fig- 2.05.
Hence,
⇒ ( ) = = ( + ) & ( ) = = ( − )

Let = + and = + are two complex numbers, then addition of these two
complex numbers and is given by—
= ( + ) + ( + )
Example— Find the addition of the complex numbers = + and = + .
Solution— given complex numbers are = 2 + 3 and = 4 + 2 .
Hence, + =(2 + 4) + (3 + 2)
∴ = 6 + 5
2.12 Subtraction of two complex numbers—
Let = + and = + are two complex numbers, then subtraction of complex
numbers from is given by—
= ( − ) + ( − )
Example— Subtract the complex numbers = + from = + .
Solution— given complex numbers are = 2 + 3 and = 4 + 2 .
Hence, − =(4 − 2) + (2 − 3)
∴ = 2 −
2.13 Multiplication of two complex numbers—
Let = + and = + are two complex numbers, then multiplication of complex
numbers and is given by—
= ( − ) + ( + )
Example—Multiply the complex numbers = + and = + .
Solution— Given complex numbers are = 3 + 4 and = 4 + 3 .
Hence, ( ) =(3 + 4 )(4 + 3 ) = (12 − 12) + (9 + 16)
∴ = 25
2.14 Division of two complex numbers—
Let = + and = + are two complex numbers, then division of complex
numbers by is given by—
= =
+
+
Now, multiplying and devide by = − .
Hence,
= = (
+
+
)(
−
−
)
Now, multiplying numerator and denominator,
= =
( + ) + ( − )
( + )
∴ = =
( + )
( + )
+
( − )
( + )

Example—Divide the complex numbers = + by = + .
Solution— given complex numbers are = 1 + and = 3 + 4 .
Hence,
= =
3 + 4
1 +
Now, multiplying and devide by = 1 −
Hence, = = =
( ) ( )
=
∴ = = +
2.15 De Moivre’s Theorem—
In mathematics, De Moivre's formula or De Moivre's identity states that for any complex
number = and integer " " it holds that—
( + ) = cos( ) + ( )
Example— Solve by using De Moivre’s Theorem—
(a) ( + ) ( ) ( + ) ( ) +
√
( )
Solution—
(a) Given that = ( + ) .
From De Moivre’s Theorem we know that—
( + ) = cos( ) + ( )
Hence, ( + ) = 2 + 2 .
(b) Let = (1 + )
Now, converting to polar form—
| | = √1 + 1 = √2 and = tan =
Hence, = √2( + )
Now, there is given (1 + )
So, (1 + ) = √2 +
( + ) = cos( ) + ( )
Hence, (1 + ) = 2 + = 2 cos + + +
= 2 −cos − = 2 −
√
−
√
= −4(1 + )
(c)Let = +
√
= +
( + ) = cos( ) + ( )

Hence, +
√
= + = +
= cos 3 + + 3 + = − −
= − −
√
( ) Let =
= ∗ =
( )
= ( + )
( + ) = cos( ) + ( )
Hence, = cos (24 ) + (24 )
2.16 Powers to " ”—
Let = , here, = 1 & = tan ( ) = tan ∞ =
Hence, = can be written in Cartesian form as—
=
2
+
2
Thus, = +
( + ) = cos( ) + ( )
Therefore, = +
Note—
= , = −1, = − , = 1
Hence, = , = −1, = − , = 1
Where, is an integer.
Example—Solve the following complex numbers—
(a) (b) (c) (d)
Solution—
(a) Given complex number is =
It can be written as = ( ∗ )
Hence, = −
(b) Given complex number is =
Hence, = −1

(c) Given complex number is =
Hence, = −1
(d) Given complex number is =
Hence, =
2.17 Laws of algebra of complex numbers—
The closure law— The sum of two complex numbers is a complex number, i.e., + is a
complex number for all complex numbers and .
The commutative law— For any two complex numbers and ,
+ = + .
The associative law— For any three complex numbers , , ,
( + ) + = + ( + ).
The distributive law— For any three complex numbers , , ,
(a) ( + ) = +
(b) ( + ) = +
The existence of additive identity— There exists the complex number 0 + 0 (denoted as 0),
called the additive identity or the zero complex number, such that, for every complex number
, + 0 = .
2.18 Properties of complex numbers—
1. =
2. If = + , then = +
3. = | |
4. ± = ±
5. =
6. = , ≠

EXERCISE-I
(a) Find the modulus of the following complex numbers—
1. 2 + 3
2. −1 + 2
3. 3 +
4. 3 − 4
5. 4 + 3
(b) Find the argument of the following complex numbers—
1. −1 +
2. 1 −
3. −1 −
4. √3 +
5. 1 + √3
6. 1 −
7. + (1 − )
8. (1 − ) +
9. −
10.1 + +
(c) Find the complex conjugate of the following complex numbers—
1. 12 + 3
2. √3 + 12
3. 4 + √3
4. 1 −
5. + (1 − )

6. (1 − ) +
7. −
8. 1 + +
EXERCISE-II
(a) If = 2 + 3 , = −3 + & = 1 + . Evaluate the followings—
1. + +
2. + 2 −
3.
4.
5.
(b) If = , = −1 + & = 1 + √3 . Evaluate the argument of the followings—
1. + +
2. + 2 −
3.
4.
5.
EXERCISE-III
(a) Convert the following complex numbers in polar form—
1. √3 +
2. 3 + 3
3.
4. −1 + √3
5. 2
(b) Convert the following complex numbers in Euler’s form—
1. √3 −
2. −3 + √3
3.
4. −1 + √3
5. 1 +
(c) Convert the following complex numbers in Euler’s form—
1.
2.
√
+
√
3. − −
√
4. −1 + √3
5. 1 +

EXERCISE-IV
(a) Solve the following complex numbers—
1.
√
+
√
2. − +
√
3.
4.
5. (( + )( − ))
(b) Solve the following Complex numbers—
1.
2.
3.
4.
5.
Crackers Attack
LEVEL-I
(a) Find the modulus of the following complex numbers—
1. 1 −
2. + (1 − )
3. (1 − ) +
4. −
5. 1 + +
(b) Find the argument of the following complex numbers—
1. 1 −
2. + (1 − )
3. (1 − ) +
4. −
5. 1 + +
LEVEL-II
(a) Find the square root of the following complex numbers—
1. 8 − 6
2. −15 + 8
3. (1 − ) +
4. −
5. 1 + +
Advanced Problems

LEVEL-I
1. Evaluate .
2. Express and in terms of Euler’s expressions.
3. Prove that sin + cos = 1, by using complex numbers.
4. If = + , evaluate ln z in terms of " " and " ".
LEVEL-II
1. Evaluate (i.e., find all possible values of ) 1 .
2. Evaluate (1 + )√ .
3. Evaluate √ .
Reference—
1. en.wikipedia.org/wiki/Complex_number
3. www.stewartcalculus.com
4. www.britannica.com
7. www.purplemath.com
8. www.mathwarehouse.com
9. www.clarku.edu
10. home.scarlet.be

Trigonometry and Mensuration
Course- Diploma
Semester-II
Unit- III

Unit-III Trigonometry and Mensuration
3.1 Trigonometry
3.1.1 Trigonometric Identities—
1. sin + cos = 1
2. sec − tan = 1
3. cosec − cot = 1
3.1.2 Compound angles formulas—
1. ( + ) = +
2. ( − ) = −
3. ( + ) = −
4. ( − ) = +
5. ( + ) =
6. ( − ) =
7. ( + ) =
8. ( − ) =
3.1.3 Double angle formulas—
1. 2 = 2 =
2. 2 = − = 2 − 1 = 1 − 2 =
3. 2 =
4. 2 =
3.1.4 Sub multiple angle formulas—
1. = 2
=
2. = −
= 2 − 1
= 1 − 2
=
3. =
4. =

Example— Find the values of
°
,
°
and
°
by using submultiple angle
formula.
Solution— We know that—
cos = 1 − 2
∴ =
∴ sin =
On putting = 45°
sin 22
°
=
°
= √
=
√
√
=
√
∴
°
=
√
Now, taking a right angle triangle as shown in fig—
By using Pythagoras theorem—
AB + BC = AC
BC = √AC − AB
BC = 2 + √2
Hence,
°
= =
+ √
tan 22
°
= =
√
√
=
√
√
=
√
√
×
√
√
=
√
√
°
= √ −
°
2 − √2
2
A
B
C

Example— Find the values of °, ° and ° by using sub multiple angle
formula.
Solution— We know that— sin = 2 sin cos
On putting = 72°
sin 72° = 2 sin 36° cos 36°
sin(90° − 18°) = 2(2 sin 18° cos 18°)(1 − 2 sin 18°)
cos 18° = 4 18° 18°(1 − 2 18°)
1 = 4 sin 18° (1 − 2 sin 18°)
8 sin 18° − 4 sin 18° + 1 = 0
(2 sin 18° − 1)(4 sin 18° + 2 sin 18° − 1) = 0
but sin 18° ≠ , therefore—
4 sin 18° + 2 sin 18° − 1 = 0
∴ sin 18° =
√
° =
√ −
Now, taking a right angle triangle as shown in fig—
By using Pythagoras theorem—
AB + BC = AC
BC = √AC − AB
BC = 10 + 2√5
Hence,
° = =
+ √
tan 18° = =
√
√
=
√
√
=
√
√
×
√
√
= 1 −
√
° = −
√
Selected + sign, since ° > 0
°
√5 − 1
4
B C
A

3.1.5 Properties of triangle—
1. sine rule—
2. Cosine rule—
3. Relation between area of triangle, angles and its sides—
Where, 2s = a + b + c
∆ = area of triangle
R = radius of circum circle
r = radius of incircle

4. Half angle formulae—
EXERCISE
Question— Find the values of sin 72 , cos 72 and tan 72 .
Question— Find the sine of the angles A,B and C of triangle ABC having sides = 3, = 4
and = 5.
Question— Find the in-circle radius of the triangle having its side 5cm, 12cm and 13cm.
Question— Find the circum radius of the circle inscribing the triangle having its sides equal to
8cm, 15cm and 17cm.
Question— Find the area of the triangle having its sides 4cm, 4cm and 6cm.

3.2 Mensuration
3.2.1 Introduction—anything concerned with measuring, calculating and estimating lengths,
areas and volumes, as well as the construction of objects, comes under the Mensuration.
Therefore, units have an important role in Mensuration. Some standard units with their
conversion are listed below—
Length Area Volume Weight
1 Km 1000 m 1 m 10000 cm 1 m 1000 litres 1 tonne 1000 Kg
1 m 100 cm 1 cm 100 mm 1 litre 1000 ml 1 Kg 1000 g
1cm 10 mm 1 cm 1 ml
3.2.2 Triangle—a triangle is a polygon with three edges and three vertices. It is one of the basic
shapes in geometry. A triangle with vertices A, B and C is denoted by ∆ABC . Shape of a
triangle is shown below—
A
B C
Existence of a triangle—the triangle inequality states that the sum of the lengths of any two
sides of a triangle must be greater than or equal to the length of the third side. That sum can
equal the length of the third side only in the case of a degenerate triangle, one with collinear
vertices. It is not possible for that sum to be less than the length of the third side.
There are two important parameter related to a triangle are—
Area of Triangle— It depends upon the shape and size of the triangle.
Perimeter— It is equal to the sum of the sides of the triangle.
Perimeter = AB + BC + CA
Some important types of triangle are listed below— A
1. Right angle triangle— Square of any one side of triangle is equal
to the sum of squares of other sides.
Here, AB + BC = AC and triangle is right angled at B.
Perimeter = AB + BC + CA B C
Area = (AB)(BC)
Note— For any ∆ABC, ∠ + ∠ + ∠ = °

2. Isosceles triangle—A triangle having two equal sides is called as isosceles triangle.
In this diagram side AB and AC are equals. Hence, ∠B and ∠C are
equals. A
Perimeter = 2AB + BC
Area = (BC)√4AB − BC
B C
3. Equilateral triangle— A triangle having all sides equal is called as equilateral triangle.
In this diagram side AB, BC and AC all are equal. Hence, ∠A , ∠B and ∠C are
equals. A
Perimeter = 3AB
Area =
√
AB
B C
Quadrilateral— a quadrilateral is a polygon with four sides and four vertices or corners.
There are two types of quadrilateral— A
1. Planar quadrilateral—The interior angles of a simple (and planar)
quadrilateral ABCD add up to 360 degrees of arc,
that is— B D
C
2. Crossed quadrilateral— In a crossed quadrilateral, the four A
interior angles on either side of the crossing add up to 720°.
B C D
Parallelogram— A four-sided polygon with two pairs of parallel and equal sides. The
following is a parallelogram—
= ×
Rectangle— A rectangle is a parallelogram with 4 right angles. The following is a rectangle—
h
b

Square—A square is a rectangle with 4 equal sides. The following is square—
Rhombus— A rhombus is a parallelogram with 4 equal sides. The following is a rhombus—
Trapezoid— A trapezoid is a quadrilateral with only one pair of parallel sides. The following
are trapezoids—
1. Scalene trapezoid— A scalene trapezoid is a trapezoid with no equal sides. The
following is a scalene trapezoid—
2. Right-angled trapezoid— A right-angled is a trapezoid with two right angles. The
following is a right-angle trapezoid—
3. Isosceles trapezoid— In an isosceles trapezoid, non-parallel sides are equal.
The following is an isosceles trapezoid—

Circle—
Semicircle—
Sphere— Sphere is a locus of the points having fixed distance from a fixed point in three
dimensional planes.
Cone— It is the solid figure bounded by a base in a plane and by a surface (called the lateral
surface) formed by the locus of all straight line segments joining the apex to the perimeter of the
base.
Volume of the sphere =
4
3
πr
Surface area of the sphere = 4πr
A right circular cone and an oblique
circular cone
Right circular cone—

Cylinder—A cylinder is defined more broadly as any ruled surface spanned by a one-parameter
family of parallel lines. A cylinder whose cross section is an ellipse, parabola, or hyperbola is
called an elliptic cylinder, parabolic cylinder, or hyperbolic cylinder respectively.
A right circular cylinder
Cube—
Cuboids—
Right circular cylinder—
Lateral surface area = 2πrh
Total surface area = 2πrh + 2πr
Volume of the cone = πr h

EXERCISE
Question— Three sides of a triangle are AB=3 cm, BC=4cm and CA=5cm. Discuss about the
type of triangle. Also find its area.
Question— A triangle having two equal sides of length 12 cm and third side of length 9 cm.
Find the area of the triangle.
Question— Find the sides of an equilateral triangle having area 173 sq. cm.
Question— Find the volume of a cone having base radius 5cm and height 12cm.
Question— Find the volume of a cylinder having base radius and height both equals to 7cm.

References—
1. en.wikipedia.org/wiki/Trigonometry
3. https://www.khanacademy.org/math/trigonometry
4. www.sosmath.com
5. www.themathpage.com
6. www.clarku.edu
7. en.wikibooks.org/wiki/Trigonometry
8. www.cimt.plymouth.ac.uk
9. www.bbc.co.uk

Indefinite Integration
Course- Diploma
Semester-II
Unit- IV

Unit-IV Indefinite Integration
4. Indefinite Integration
4.1 Definition— A function ( ) is an anti-derivative or integral of a function ( ) if
( )
= ( )
For all in the domain of . The set of all anti-derivative of f is the indefinite integral of with
respect to , denoted by
( ) = ( )
The symbol ∫ is an integral sign. The function f is the integrand of the integral and x is the
variable of integration.
4.2 Algebra on integration—
1.∫[ ( ) + ( )] = ∫ ( ) + ∫ ( )
2. ∫ ( ) = ∫ ( ) , for constant.
4.3 Standard integrals—
1.∫ = + ( ≠ − )
2.∫ = | | +
3.∫ = − +
4. ∫ = +
5. ∫ = | | +
6. ∫ = | | +
7.∫ = +
8.∫ = − +
9. ∫ = +
10.∫ = − +
11.∫ = +
12.∫ = +

13.∫ = + − +
14.∫ = + +
15. ∫ = + − +
16. ∫ = +
17. ∫ = +
18. ∫ = +
19. ∫ = + √ + +
20. ∫ = + √ − +
21. ∫ √ + = √ + + + √ + +
22. ∫ √ − = √ − + + √ − +
23. ∫ √ − = √ − + +
Example— Solve the integral ∫ .
Solution— ∫ = + = +
Example— Solve the integral ∫ .
Solution— ∫ = + = + = +
Example— Solve the integral ∫ − + .
Solution— ∫ 4 − 3 +
√

= 4 ∫ cos − 3 ∫ + 2 ∫ √
= 4 sin − 3 + 2 +
4.4 Methods of integration
4.4.1 Integration by substitution—
If integrand is not in the standard form and can be transformed to integral form by a suitable
substitution then such process of integration is called as integration by substitution.
Take these steps to evaluate the integral ∫ { ( )} ′( ) , where and ′ are contineous
function—
Step 1: Substitute = ( ) and = ′( ) to obtain the integral ∫ ( ) .
Step 2: Integrate with respect to .
Step 3: Replace by ( ) in the result.
Example—Solve the following integral ∫ .
Solution— ∫
= ∫ 2
= ∫ .
= 2
⟹ = 2
⟹ =
= ∫ .
= +
= 2 +
Example— Evaluate ∫( + ) .
Solution— ∫(2 + 3)
=∫ Let 2 + 3 =
= ∫ 2 =
= + =
=
( )
+
Example— Integrate ∫ .
Solution— ∫
=∫ Let =

= ∫ =
= ∫ ( )
=
= tan + =
= tan +
4.4.2 Integration by method of parts—
If and are functions of then the integral of product of these functions is given by—
= −
This rule is called as ‘integration by method of parts’.
The choice of first and second integral is given by—
I— Inverse
L—Logarithmic
A—Algebraic
T—Trigonometric
E—Exponential
The term appearing first in this series have to take first integral (u).
Example—Integrate ∫ .
Solution— ∫ cos
=( ∫ cos ) − ∫ (∫ cos )
= − ∫ sin = sin + cos +
4.4.3 Integration of terms involving multiple trigonometric functions—
1. cos = . 2 sin cos = [sin( + ) + cos( − ) ]
2. sin sin = [cos( − ) − cos( + ) ]
3. cos cos = [cos( − ) − cos( + ) ]
3. sin = (1 − cos 2 )
4. cos = (1 + 2 )
5. sin = (3 sin − sin 3 )
6. cos = (3 cos + cos 3 )
Example— Evaluate ∫ .
Solution—∫ sin 3 cos 2
= ∫ 2 sin 3 cos 2
= ∫(sin 5 + sin )

= − 5 − + = (cos 5 + 5 cos ) +
Solution— ∫ sin
= ∫(3 sin − sin 3 )
= −3cos + cos 3 +
= cos + cos ) +
EXERCISE
Question— Integrate ∫ .
Question— Integrate ∫ sin 2 cos 3 .
Question— Integrate∫ sin 2 .
Question— Integrate∫ .
Question— Evaluate∫(3 − 5) .

References—
2. tutorial.math.lamar.edu
4. www.webmath.com
5. en.wikibooks.org/wiki/Calculus/Indefinite_integral

Definite Integration
Course- Diploma
Semester-II
Unit- V

Unit-V Definite Integration
5. Definite integration
5.1 Definition— If ( ) is continuous in the interval [ , ] and ( ) is an antiderivative of
( ), then
5.2 Elementary properties of definite integral—
1. ∫ ( ) = − ∫ ( )
2. ∫ ( ) = ∫ ( ) = ∫ ( )
3. ∫ ( ) =
4. ∫ ( ) = ∫ ( ) + ∫ ( ) , < <
5. ∫ ( + − ) = ∫ ( )
6. ∫ ( − ) = ∫ ( )
7. ∫ ( ) =
∫ ( )
8. ∫ ( ) =
∫ ( ) ( − ) = ( )
( − ) = − ( )
5.3 Standard properties of definite integral—
1. If ( ) ≥ is defined and continuous on[ , ], then ∫ ( ) ≥
2. If ( ) ≤ is defined and continuous on[ , ], then ∫ ( ) ≤
3. If ( ) ≤ ( ) is defined and continuous on[ , ], then ∫ ( ) ≤ ∫ ( )
( ) = ( ) − ( )

4. If ( ) and ( ) be two functions defined and continuous on [a, b] then—
( ) ( ) ≤ ( ) ( )
5. The mean value theorem—
Statement— If ( ) is continuous on [ , ], then at least one value of = in ( , )
such that—
( ) = ∫ ( )
Where ( ) is called as average or mean value of ( ) on [ , ].
Solution—∫
= = − =
Example— Integrate ∫
Solution—∫
= |− cos |
= − cos — cos 0
= 0 − (−1) = 1
Example— Integrate ∫
Solution— ∫
= ∫ − ∫
( )
(∫ )
= − ∫
= [ − ∫ ]
= [ − ]
= ( − ) − ( 1 − 1)
= ( − ) + 1 = 1
Example— Evaluate ∫
Solution—∫ 3 2
= ∫ 2 3 2
= ∫ ( 5 + )

= − 5 −
= − 5 − − − 0 − 0
= (0 − 0) − − − 1
=
EXERCISE
Question— Evaluate ∫ sin 4 cos 3 .
Question— Integrate ∫ (2 + 1) .
Question— Integrate ∫ cos .
Question— Integrate ∫ ( ) .

Reference—
3. tutorial.math.lamar.edu
4. en.wikipedia.org/wiki/Integral
6. www.sosmath.com

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