Unit - 2
Reduction formula for integrals
Reduction formula for: 
What is reduction formula?
A formula of the form,
= g(x) + 
Where k<n, is called a reduction formula.
(a). Reduction formula for 
Let, Iⁿ = 
=
sinxdx , if n>1
Taking 
¹x as first function and sinx as second function, integrate by parts,
Iⁿ = -
¹x cosx – (n -1)
cosx(-cosx)dx
= -
¹x cosx + (n - 1)
cos²xdx
= -
¹x cosx + (n - 1)
( 1 - sin²x) dx
= -
¹x cosx + (n - 1)
- 
dx
= -
¹x cosx + (n - 1) [ Iⁿ ‾ ² - Iⁿ]
By solving above
Iⁿ =
+ 
Iⁿ‾²
This is the reduction formula for 
Example: Evaluate 
by using reduction formula.
Solution: we know that,
+ 
Iⁿ‾², here Iⁿ‾² = 
Now using this formula,
=
= 
= 
= 
(b). Reduction formula for 
Similarly we can get the formula for
= 
+ 
Iⁿ‾²
We can use the following formulas to get the result easily,
= 
. 
. 
, only if n is odd and n ≥3
=
. 
. 
. 
, only if n is even and n ≥2
We apply same formula for cos x.
Example2: evaluate 
.
Solution: we will apply reduction formula to evaluate,
So, here n = 5, which is odd, we will get ,
(C) Reduction formula for: 
Where
Example1: Evaluate 
by using reduction formula.
Solution: here, m = 4, n = 6,
Now we will apply above reduction formula,
Example2: Evaluate 
by using reduction formula.
Solution: Here we can see that, m = 3 and n = 5,
Now we will apply reduction formula,
(D) Reduction formula for 
and 
Let
Thus, the required reduction formula is-
And for 
, the reduction formula will be-
Example: Evaluate 
by using reduction formula.
Sol.
Put n = 6, 4, and 2 successively in the reduction formula for 
, we obtain-
Thus
Example: If 
then prove that 
Sol.
The reduction formula for 
is-
On changing n to n+1, we get-
Or
Hence proved
Key takeaways-
- Iⁿ =
+ 
Iⁿ‾² 
= 
+ 
Iⁿ‾²
Improper integrals
(1) Let f is function defined on [a, ∞) and it is integrable on [a, t] for all t >a, then
If 
exists, then we define the improper integral of f over [a, ∞) as follows-
(2) Let f is function defined on (-∞,b] and it is integrable on [t, b] for all t >b, then
If 
exists, then we define the improper integral of f over (-∞, b] as follows-
(3) Let f is function defined on (-∞, ∞] and it is integrable on [a, b] for every closed and bounded interval [a, b] which is the subset of R., then
If 
and 
exist for some c belongs to R, then we define the improper integral of f over (-∞,∞) as follows-
= 
+ 
(4) Let f is function defined on (a,∞) and 
exists for all t>a, then
If 
exists, then we define the improper integral of f over (a ∞) as follows-
(5) Let f is function defined on (-∞, b) and 
exists for all t<b, then
If 
exists, then we define the improper integral of f over (-∞, b) as follows-
Improper integrals over finite intervals-
(1) Let f is function defined on (a, b] and 
exists for all t ∈ (a,b), then
If 
exists, then we define the improper integral of f over (a, b] as follows-
(2) Let f is function defined on [a, b) and 
exists for all t ∈ (a,b), then
If 
exists, then we define the improper integral of f over [a, b) as follows-
(3) Let f is function defined on [a, c) and (c, b]. If
and 
exist
Then we define the improper integral of f over [a, b] as follows-
When we apply limits in indefinite integrals are called definite integrals.
If an expression is written as 
, here ‘b’ is called upper limit and ‘a’ is called lower limit.
If f is an increasing or decreasing function on interval [a, b], then
Where 
Properties-
1. The definite integral applies only if a<b, but it would be appropriate to include the case a = b and a>b as well, in that case-
If a = b, then
And if a>b, then
2. Integral of a constant function-
3. Constant multiple property-
4. Interval union property-
If a < c < b, then
5. Inequality-
If c and d are constants such that 
for all x in [a, b], then
c (b – a)
Note- if a function f: [a,b] →R is continuous, then the function ‘f’ is always Integrable.
Example-1: Evaluate
.
Sol. Here we notice that f:x→cos x is a decreasing function on [a, b],
Therefore, by the definition of the definite integrals-
Then
Now,
Here 
Thus
Example-2: Evaluate 
Sol. Here 
is an increasing function on [1, 2]
So that,
…. (1)
We know that-
And 
Then equation (1) becomes-
Note- we can find the definite integral directly as-
Example-3: Evaluate-
Sol. 
Improper integrals
(1) Let f is function defined on [a,∞) and it is integrable on [a, t] for all t >a, then
If 
exists, then we define the improper integral of f over [a,∞) as follows-
(2) Let f is function defined on (-∞,b] and it is integrable on [t, b] for all t >b, then
If 
exists, then we define the improper integral of f over (-∞, b] as follows-
(3)Let f is function defined on (-∞, ∞] and it is integrable on [a, b] for every closed and bounded interval [a, b] which is the subset of R., then
If 
and 
exist for some c belongs to R, then we define the improper integral of f over (-∞,∞) as follows-
= 
+ 
(4) Let f is function defined on (a,∞) and 
exists for all t>a, then
If 
exists, then we define the improper integral of f over (a, ∞) as follows-
(5) Let f is function defined on (-∞, b) and 
exists for all t<b, then
If 
exists, then we define the improper integral of f over (-∞, b) as follows-
Improper integrals over finite intervals-
(1) Let f is function defined on (a, b] and 
exists for all t ∈ (a,b), then
If 
exists, then we define the improper integral of f over (a, b] as follows-
(2) Let f is function defined on [a, b) and 
exists for all t ∈ (a,b), then
If 
exists, then we define the improper integral of f over [a, b) as follows-
(3) Let f is function defined on [a, c) and (c, b]. If
and 
exist
Then we define the improper integral of f over [a, b] as follows-
The beta and gamma functions are defined as-
And
These integrals are also known as first and second Eulerian integrals.
Note- Beta function is symmetrical with respect to m and n.
Some important results-
Ex.1: Evaluate
dx
Solution
dx = 
dx
= Γ (5/2)
= Γ (3/2+ 1)
= 3/2 Γ (3/2)
= 3/2. ½ Γ (½)
= 3/2. ½ π
= ¾ π
Ex. 2: Find γ(-½)
Solution: (-½) + 1= ½
Γ (-1/2) =Γ (-½ + 1) / (-½)
= - 2 Γ (1/2)
= - 2 π
Ex. 3: Show that 
Solution:
= 
= 
= 
) .......................
= 
= 
Ex. 4: Evaluate 
dx.
Solution: Let 
dx
X | 0 |  |
t | 0 |  |
Put 
or 
;dx =2t dt
dt
dt
Ex. 5: Evaluate 
dx.
Solution: Let 
dx.
x | 0 |  |
t | 0 |  |
Put 
or 
; 4x dx = dt
dx
Evaluation of beta function 𝛃 (m, n)-
Here we have-
Or
Again, integrate by parts, we get-
Repeating the process above, integrating by parts we get-
Or
Evaluation of gamma function-
Integrating by parts, we take 
as first function-
We get-
Replace n by n+1,
Relation between beta and gamma function-
We know that-
………… (1)
…………………..(2)
Multiply equation (1) by 
, we get-
Integrate both sides with respect to x within limits x = 0 to x = 
, we get-
But
By putting λ= 1 + y and n = m + n
We get by using this result in (2)-
So that-
Definition: Beta function
Properties of Beta function:
Example (1): Evaluate I = 
Solution:
= 2 π/3
Example (2): Evaluate: I = 02 x2 / (2 – x). Dx
Solution:
Letting x = 2y, we get
I = (8/2) 01 y2 (1 – y)-1/2dy
= (8/2). B (3, 1/2)= 642 /15
Beta Function More Problems
Relation between Beta and Gamma functions:
Example(1): Evaluate: I = 0a x4 (a2 – x2). Dx
Solution: Letting x2 = a2y, we get
I = (a6 / 2) 01 y3/2 (1 – y )1/2dy
= (a6 / 2).B (5/2, 3/2)
= a6 /3 2
Example (2): Evaluate: I = 02 x (8 – x3). Dx
Solution: Let x3 = 8y
I = (8/3) 01 y-1/3 (1 – y) 1/3. Dy
= (8/3) B (2/3, 4/3)
= 16 π / (93)
Example(3): Prove that
Solution: Let
Put
or 
, 
Example(4): Evaluate
Solution: Let
Put 
or 
,
,
When
,
;
,
Also
Example(5): Show that
Solution:
=
(
0<p<1)
(By above result) 
RULE I: If      |
Q1) Evaluate: (i) 
Sol) Let  Differentiating by DUIS rule 
 
 
 
 
 
 
 
   Separating vectors and integrating, we have
         Putting     
(ii)  Solution: Let  Differentiating by DUIS rule 
 
  Put   
 
  Integrating, we get
  
 
  Putting     
|
DIFFERENTIATION UNDER INTEGRAL SIGN RULE – II
(Leibnitz’s Rule)
RULE II: If     
|
Q)Prove that: (i) If  
Sol) We have given that
 Then we have to prove that  Differentiating by DUIS rule (Leibnitz’s Rule), we have 
 
 
  Again differentiating by DUIS rule (Leibnitz’s Rule), we have   
 
 
     This is the desired result.
(ii)If   Solution: We have given that
 Then we have to prove that  Differentiating by DUIS rule (Leibnitz’s Rule), we have        Again differentiating by DUIS rule (Leibnitz’s Rule), we have 
 
 
  Again differentiating by DUIS rule (Leibnitz’s Rule), we have 
 
 
 
|
ERROR FUNCTION
Introduction to error function and its properties.
Error Function:
- Error function is denoted by
and defined as
2. Complementary Error function is denoted by 
and defined as
3. Properties of Error function
(i) 
(ii) 
(iii) 
(iv) 
i.e. error function is odd function.
(v) 
Q) Prove that
(i) 
is an odd function and hence deduce 
.
Sol)
We have to prove that
is an odd function i.e.
We know that
- - - - - - - - - - - - - - - - (1)
Replace 
by
, we have
Put 
, then 
and
Hence the error function is odd function.
Now we have to prove that
Consider
This is the desired result.
(ii)
.
Solution:
We have to prove that
Consider
This is the desired result.
(iii) 
Solution:
We have to prove that
Consider
Put 
, then 
and
This is the desired result.
(iv) 
Solution:
We have to prove that
We know that
Area in Cartesian coordinates-
Example-1: Find the area enclosed by two curves using double integration.
y = 2 – x and y² = 2 (2 – x)
Sol. Let,
y = 2 – x ………………..(1)
And y² = 2 (2 – x) ………………..(2)
On solving eq. (1) and (2)
We get the intersection points (2,0) and (0,2) ,
We know that,
Area = 
Here we will find the area as below,
Area = 
Which gives,
= ( - 4 + 4 /2 ) + 8 / 3 = 2 / 3.
Example-2: Find the area between the parabola y ² = 4ax and another parabola x² = 4ay.
Sol. Let,
y² = 4ax ………………..(1)
And
x² = 4ay…………………..(2)
Then if we solve these equations , we get the values of points where these two curves intersect
x varies from y²/4a to 
and y varies from o to 4a,
Now using the conceot of double integral,
Area = 
Area in polar coordinates-
Example-3: Find the area lying inside the cardioid r = a(1+cosθ) and outside the circle r = a, by using double integration.
Sol. We have,
r = a(1+cosθ) …………………….(1)
And
r = a ……………………………….(2)
On solving these equations by eliminating r , we get
a(1+cosθ) = a
(1+cosθ) = 1
Cosθ = 0
Here a θ varies from – π/2 to π/2
Limit of r will be a and 1+cosθ)
Which is the required area.
Example-4: Find the are lying inside a cardioid r = 1 + cosθ and ouside the parabola r(1 + cosθ) = 1.
Sol. Let,
r = 1 + cosθ ……………………..(1)
r(1 + cosθ) = 1……………………..(2)
Solving these equestions , we get
(1 + cosθ )(1 + cosθ ) = 1
(1 + cosθ )² = 1
1 + cosθ = 1
Cosθ = 0
θ = ±π / 2
So that, limits of r are,
1 + cosθ and 1 / 1 + cosθ
The area can be founded as below,
Double integrals as volumes
Suppose we have a curve y = f(x) is revolved about an axis , then a solid is generated , now we need to find out the volume of the solid generated ,
The formula for volume of the solid generated about x-axis,
Example-1: Calculate the volume generated by the revolution of a cardioid,
r = a ( 1 – cosθ) about its axis
Sol. Here, r = a ( 1 – cosθ)
Volume =
=
Which is the volume of generated by cardioid.
Example-2: Find the volume generated by revolving a circle x ² + y² = 4 about the line x= 3.
Sol. We know that,
Volume = 
Here , PQ = 3 – x,
= 
The volume is 24π².
Example-3: Calculate the volume under the surface z = 3 + x² - 2y over the region R defined as 0≤ x ≤ 1 and -x ≤ y ≤ x
Sol. The is a double integral of z = 3 + x² - 2y over the region R .
Volume will be,
Areas and volumes of revolutions
The volume of revolution (V) is obtained by rotating area A through one revolution about the x-axis is given by-
Suppose the curve x = f(y) is rotated 
about y-axis between the limits y = c and y = d, then the volume generated V, is given by-
Example-1: Find the area enclosed by the curves 
and if the area is rotated 
about the x-axis then determine the volume of the solid of revolution.
Sol. We know that, at the point of intersection the coordinates of the curve are equal. So that first we will find the point of intersection-
We get,
x = 0 and x = 2
The curve of the given equations will look like as follows-
Then,
The area of the shaded region will be-
A = 
So that the area will be 8/3 square unit.
The volume will be
= (volume produced by revolving 
– (volume produced by revolving 
= 
Method of cylindrical shells-
Let f(x) be a continuous and positive function. Define R as the region bounded above by the graph f(x), below by the x-axis, on the left by the line x = a and on the right x = b, then the volume of the solid of revolution formed by revolving R around the y-axis is given by
Example-2: Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 1/x over the interval [1 , 3].
Sol. The graph of the function f(x) = 1/x will look like-
The volume of the solid of revolution generated by revolving R(violet region) about the y-axis over the interval [1 , 3]
Then the volume of the solid will be-
Example-3: Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 2x - x² over the interval [0 , 2].
Sol. The graph of the function f(x) = 2x - x² will be-
The volume of the solid is given by-
= 
Area of surfaces of revolution
Suppose we have a curve y = f(x) and if we revolve this curve around x – axis then the area formed is known as surface area of revolution.
Let f(x) be a non-negative smooth function over the interval[ [ a , b] , then the surface area of the surface of revolution formed by revolving the graph of f(x) around x-axis is given by,
S = 
Example-1: Find the surface of the surface generated by revolving the graph f(x) = 
over the interval [1,4] around x-axis.
Sol. The graph of the function will look like
f(x) = 
then f’(x) = 1/2 (
(f’(x))² = 1/4x
Then,
S = 
S = 
S = 
,
Let u = 
, then du = dx, when x = 1 , u = 5/4 and when x = 4 , u = 17/4,
This will give,
= 
On solving the integral we get,
= 30.84 (approx.)
Example-1: find the surface of the surface generated by revolving the graph g(y) = 
over the interval [0,2] around y-axis.
Sol.
Hint - use formula
S = 
Ans. = 24.118
CENTER OF MASS AND GRAVITY-
Centre of mass-
We have-
Mass = volume 
Example: Find the mass of a plate formed by the coordinate planes and the plane-
The variable density 
Sol.
Here we the mass will be-
The limits of y are from y = 0 to y = b(1 – x/a) and limits of x are from 0 to a.
Therefore the required mass will be-
Which is the required answer.
Centre of gravity-
The centre of gravity of a lamina is the point where it balances perfectly which means the lamina’s centre of mass.
If 
and 
are the coordinates of the centroid C of area A, then-
And
Example: Determine the coordinates of the centroid of the area lying between the curve y = 5x - x² and the x-axis.
Sol.
Here y = 5x - x²when y = 0, x = 0 or x = 5
Therefore the curve cuts the x-axis at 0 and 5 as in the figure-
Now
= 2.5
Therefore the centroid of the area lies at (2.5, 2.5)
Key takeaways-
- The formula for volume of the solid generated about x-axis,
2. Centre of mass-
We have-
Mass = volume 
References:
1. G.B. Thomas and R.L. Finney, “Calculus and Analytic Geometry”, Pearson, 2002.
2. T. Veerarajan, “Engineering Mathematics”, McGraw-Hill, New Delhi, 2008.
3. B. V. Ramana, “Higher Engineering Mathematics”, McGraw Hill, New Delhi, 2010.
4. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
5. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
6. HK Dass
