Reduction forumla

Reduction Formula
Dr. Nirav Vyas
Department of Science & Humanities
Atmiya University
Yogidham, Kalavad road
Rajkot - 360005 . Gujarat
Email: nbvyas@aits.edu.in
Dr. Nirav Vyas Reduction Formula

Introduction
A Reduction Formula for an integral is a formula which
connects integral linearly with another integral of the same
type, but of the lower degree.

Introduction
Reduction formula is generally obtained by repeated
application of integration by parts.

Introduction
Reduction formula is generally obtained by repeated
application of integration by parts.
Reduction formula are used as a tool to ﬁnd area and
volume.

Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2

x, cosn
x
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
x in two
parts sinn
x = (sinn−1
x)(sin x)

x, cosn
x
xdx, n ≥ 2
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx

x, cosn
x
xdx, n ≥ 2
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx
= (sinn−1
x)(sin x)dx

x, cosn
x
xdx, n ≥ 2
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx
= (sinn−1
x)(sin x)dx
= (sinn−1
x)(− cos x) − {(n − 1) sinn−2
x cos x}{− cos x}dx

x, cosn
x
xdx, n ≥ 2
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx
= (sinn−1
x)(sin x)dx
= (sinn−1
x)(− cos x) − {(n − 1) sinn−2
= − sinn−1
x cos x + (n − 1) sinn−2
x cos2
xdx

x, cosn
x
xdx, n ≥ 2
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx
= (sinn−1
x)(sin x)dx
= (sinn−1
x)(− cos x) − {(n − 1) sinn−2
= − sinn−1
x cos2
xdx
= − sinn−1
x(1 − sin2
x)dx

x, cosn
x
xdx, n ≥ 2
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx
= (sinn−1
x)(sin x)dx
= (sinn−1
x)(− cos x) − {(n − 1) sinn−2
= − sinn−1
x cos2
xdx
= − sinn−1
x(1 − sin2
x)dx
= − sinn−1
xdx − (n − 1) sinn
xdx

x, cosn
x
xdx, n ≥ 2
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx
= (sinn−1
x)(sin x)dx
= (sinn−1
x)(− cos x) − {(n − 1) sinn−2
= − sinn−1
x cos2
xdx
= − sinn−1
x(1 − sin2
x)dx
= − sinn−1
xdx
∴ In = − sinn−1
x cos x + (n − 1)In−2 − (n − 1)In

x, cosn
x
xdx, n ≥ 2
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx
= (sinn−1
x)(sin x)dx
= (sinn−1
x)(− cos x) − {(n − 1) sinn−2
= − sinn−1
x cos2
xdx
= − sinn−1
x(1 − sin2
x)dx
= − sinn−1
xdx
x cos x + (n − 1)In−2 − (n − 1)In
∴ In + (n − 1)In = − sinn−1
x cos x + (n − 1)In−2

x, cosn
x
xdx, n ≥ 2
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx
= (sinn−1
x)(sin x)dx
= (sinn−1
x)(− cos x) − {(n − 1) sinn−2
= − sinn−1
x cos2
xdx
= − sinn−1
x(1 − sin2
x)dx
= − sinn−1
xdx
x cos x + (n − 1)In−2 − (n − 1)In
∴ In + (n − 1)In = − sinn−1
x cos x + (n − 1)In−2
∴ nIn = − sinn−1
x cos x + (n − 1)In−2

x, cosn
x
xdx, n ≥ 2
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx
= (sinn−1
x)(sin x)dx
= (sinn−1
x)(− cos x) − {(n − 1) sinn−2
= − sinn−1
x cos2
xdx
= − sinn−1
x(1 − sin2
x)dx
= − sinn−1
xdx
x cos x + (n − 1)In−2 − (n − 1)In
∴ In + (n − 1)In = − sinn−1
x cos x + (n − 1)In−2
∴ nIn = − sinn−1
x cos x + (n − 1)In−2
∴ In = −
1
n
sinn−1
x cos x +
n − 1
n
In−2 —(1)

x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2

x, cosn
x
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
x in two
parts cosn
x = (cosn−1
x)(cos x)

x, cosn
x
xdx, n ≥ 2
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx

x, cosn
x
xdx, n ≥ 2
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx
= (cosn−1
x)(cos x)dx

x, cosn
x
xdx, n ≥ 2
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx
= (cosn−1
x)(cos x)dx
= (cosn−1
x)(sin x) − {(n − 1) cosn−2
x(− sin x)}{sin x}dx

x, cosn
x
xdx, n ≥ 2
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx
= (cosn−1
x)(cos x)dx
= (cosn−1
x)(sin x) − {(n − 1) cosn−2
= cosn−1
x sin x + (n − 1) cosn−2
x sin2
xdx

x, cosn
x
xdx, n ≥ 2
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx
= (cosn−1
x)(cos x)dx
= (cosn−1
x)(sin x) − {(n − 1) cosn−2
= cosn−1
x sin2
xdx
= cosn−1
x(1 − cos2
x)dx

x, cosn
x
xdx, n ≥ 2
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx
= (cosn−1
x)(cos x)dx
= (cosn−1
x)(sin x) − {(n − 1) cosn−2
= cosn−1
x sin2
xdx
= cosn−1
x(1 − cos2
x)dx
= cosn−1
xdx − (n − 1) cosn
xdx

x, cosn
x
xdx, n ≥ 2
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx
= (cosn−1
x)(cos x)dx
= (cosn−1
x)(sin x) − {(n − 1) cosn−2
= cosn−1
x sin2
xdx
= cosn−1
x(1 − cos2
x)dx
= cosn−1
xdx
∴ In = cosn−1
x sin x + (n − 1)In−2 − (n − 1)In

x, cosn
x
xdx, n ≥ 2
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx
= (cosn−1
x)(cos x)dx
= (cosn−1
x)(sin x) − {(n − 1) cosn−2
= cosn−1
x sin2
xdx
= cosn−1
x(1 − cos2
x)dx
= cosn−1
xdx
∴ In = cosn−1
x sin x + (n − 1)In−2 − (n − 1)In
∴ In + (n − 1)In = cosn−1
x sin x + (n − 1)In−2

x, cosn
x
xdx, n ≥ 2
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx
= (cosn−1
x)(cos x)dx
= (cosn−1
x)(sin x) − {(n − 1) cosn−2
= cosn−1
x sin2
xdx
= cosn−1
x(1 − cos2
x)dx
= cosn−1
xdx
∴ In = cosn−1
x sin x + (n − 1)In−2 − (n − 1)In
∴ In + (n − 1)In = cosn−1
x sin x + (n − 1)In−2
∴ nIn = cosn−1
x sin x + (n − 1)In−2

x, cosn
x
xdx, n ≥ 2
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx
= (cosn−1
x)(cos x)dx
= (cosn−1
x)(sin x) − {(n − 1) cosn−2
= cosn−1
x sin2
xdx
= cosn−1
x(1 − cos2
x)dx
= cosn−1
xdx
∴ In = cosn−1
x sin x + (n − 1)In−2 − (n − 1)In
∴ In + (n − 1)In = cosn−1
x sin x + (n − 1)In−2
∴ nIn = cosn−1
x sin x + (n − 1)In−2
∴ In =
1
n
cosn−1
x sin x +
n − 1
n
In−2 — (2)

x, cosn
x
The formulae (1) and (2) are called Reduction formulae
because the exponent is reduced from n to n − 2.

x, cosn
x
The formulae (1) and (2) are called Reduction formulae
because the exponent is reduced from n to n − 2.
Now we will consider some deﬁnite integral and apply above
reduction formulae to obtain some generalized result.

x, cosn
x
Let In =
π/2
0
sinn
xdx

x, cosn
x
Let In =
π/2
0
sinn
xdx
from (1), we have In = −
1
n
sinn−1
x cos x +
n − 1
n
In−2

x, cosn
x
Let In =
π/2
0
sinn
xdx
1
n
sinn−1
x cos x +
n − 1
n
In−2
∴
π/2
0
sinn
xdx = {−
1
n
sinn−1
x cos x}
π/2
0 +
n − 1
n
π/2
0
sinn−2
xdx

x, cosn
x
Let In =
π/2
0
sinn
xdx
1
n
sinn−1
x cos x +
n − 1
n
In−2
∴
π/2
0
sinn
xdx = {−
1
n
sinn−1
x cos x}
π/2
0 +
n − 1
n
π/2
0
sinn−2
xdx
∴
π/2
0
sinn
xdx = (0 − 0) +
n − 1
n
In−2

x, cosn
x
Let In =
π/2
0
sinn
xdx
1
n
sinn−1
x cos x +
n − 1
n
In−2
∴
π/2
0
sinn
xdx = {−
1
n
sinn−1
x cos x}
π/2
0 +
n − 1
n
π/2
0
sinn−2
xdx
∴
π/2
0
sinn
xdx = (0 − 0) +
n − 1
n
In−2
∴ In =
n − 1
n
In−2 — (3)

x, cosn
x
Let In =
π/2
0
sinn
xdx
1
n
sinn−1
x cos x +
n − 1
n
In−2
∴
π/2
0
sinn
xdx = {−
1
n
sinn−1
x cos x}
π/2
0 +
n − 1
n
π/2
0
sinn−2
xdx
∴
π/2
0
sinn
xdx = (0 − 0) +
n − 1
n
In−2
∴ In =
n − 1
n
In−2 — (3)
Now if we apply successive use of eq. (3) by replacing n by
n − 2 in (3), we get,

x, cosn
x
Let In =
π/2
0
sinn
xdx
1
n
sinn−1
x cos x +
n − 1
n
In−2
∴
π/2
0
sinn
xdx = {−
1
n
sinn−1
x cos x}
π/2
0 +
n − 1
n
π/2
0
sinn−2
xdx
∴
π/2
0
sinn
xdx = (0 − 0) +
n − 1
n
In−2
∴ In =
n − 1
n
In−2 — (3)
n − 2 in (3), we get,
∴ In−2 =
n − 3
n − 2
In−4 — (4)

x, cosn
x
Let In =
π/2
0
sinn
xdx
1
n
sinn−1
x cos x +
n − 1
n
In−2
∴
π/2
0
sinn
xdx = {−
1
n
sinn−1
x cos x}
π/2
0 +
n − 1
n
π/2
0
sinn−2
xdx
∴
π/2
0
sinn
xdx = (0 − 0) +
n − 1
n
In−2
∴ In =
n − 1
n
In−2 — (3)
n − 2 in (3), we get,
∴ In−2 =
n − 3
n − 2
In−4 — (4)
from (3) and (4) In =
n − 1
n
n − 3
n − 2
In−4

x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6

x, cosn
x
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 1: n is an even integer

x, cosn
x
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Continuing the above process we ﬁnd

x, cosn
x
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
1
2
I0

x, cosn
x
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
1
2
I0
where I0 =
π
2
0
sin0
xdx =
π
2
0
dx = π
2

x, cosn
x
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
1
2
I0
where I0 =
π
2
0
sin0
xdx =
π
2
0
dx = π
2
∴
π/2
0
sinn
xdx =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
3
4
.
1
2
.
π
2
— (5)

x, cosn
x

x, cosn
x
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 2: n is an odd integer

x, cosn
x
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6

x, cosn
x
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
2
3
I1

x, cosn
x
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
2
3
I1
where I1 =
π
2
0
sin xdx = {− cos x}
π
2
0 dx = 1

x, cosn
x
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
2
3
I1
where I1 =
π
2
0
sin xdx = {− cos x}
π
2
0 dx = 1
∴
π/2
0
sinn
xdx =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
4
5
.
2
3
.1 — (6)

x, cosn
x
We can verify that reduction formula for
π/2
0
cosn
xdx is
exactly same as that for
π/2
0
sinn
xdx

x, cosn
x
π/2
0
cosn
xdx is
π/2
0
sinn
xdx
the reason is
a
0
f(x)dx =
a
0
f(a − x)dx

x, cosn
x
π/2
0
cosn
xdx is
π/2
0
sinn
xdx
the reason is
a
0
f(x)dx =
a
0
f(a − x)dx
∴
π/2
0
sinn
xdx =
π/2
0
sinn π
2
− x dx =
π/2
0
cosn
xdx

x, cosn
x
π/2
0
cosn
xdx is
π/2
0
sinn
xdx
the reason is
a
0
f(x)dx =
a
0
f(a − x)dx
∴
π/2
0
sinn
xdx =
π/2
0
sinn π
2
− x dx =
π/2
0
cosn
xdx
∴
π/2
0
cosn
xdx =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
3
4
.
1
2
.
π
2
— (5)

x, cosn
x
π/2
0
cosn
xdx is
π/2
0
sinn
xdx
the reason is
a
0
f(x)dx =
a
0
f(a − x)dx
∴
π/2
0
sinn
xdx =
π/2
0
sinn π
2
− x dx =
π/2
0
cosn
xdx
∴
π/2
0
cosn
xdx =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
3
4
.
1
2
.
π
2
— (5)
∴
π/2
0
cosn
xdx =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
4
5
.
2
3
.1 — (6)

x, cosn
x - Example 1
Evaluate
π
0
(1 + cos x)4
dx

x, cosn
x - Example 1
Evaluate
π
0
(1 + cos x)4
dx
Sol.
π
0
(1 + cos x)4
dx =
π
0
(2 cos2 x
2
)4
dx

x, cosn
x - Example 1
Evaluate
π
0
(1 + cos x)4
dx
Sol.
π
0
(1 + cos x)4
dx =
π
0
(2 cos2 x
2
)4
dx
= 24 π
0
cos8 x
2
dx

x, cosn
x - Example 1
Evaluate
π
0
(1 + cos x)4
dx
Sol.
π
0
(1 + cos x)4
dx =
π
0
(2 cos2 x
2
)4
dx
= 24 π
0
cos8 x
2
dx
Taking x/2 = t, ∴ dx = 2dt

x, cosn
x - Example 1
Evaluate
π
0
(1 + cos x)4
dx
Sol.
π
0
(1 + cos x)4
dx =
π
0
(2 cos2 x
2
)4
dx
= 24 π
0
cos8 x
2
dx
I = 16
π/2
0
cos8
t.2dt

x, cosn
x - Example 1
Evaluate
π
0
(1 + cos x)4
dx
Sol.
π
0
(1 + cos x)4
dx =
π
0
(2 cos2 x
2
)4
dx
= 24 π
0
cos8 x
2
dx
I = 16
π/2
0
cos8
t.2dt
= 32
π/2
0
cos8
tdt

x, cosn
x - Example 1
Evaluate
π
0
(1 + cos x)4
dx
Sol.
π
0
(1 + cos x)4
dx =
π
0
(2 cos2 x
2
)4
dx
= 24 π
0
cos8 x
2
dx
I = 16
π/2
0
cos8
t.2dt
= 32
π/2
0
cos8
tdt
using formula (5) with n = 8

x, cosn
x - Example 1
Evaluate
π
0
(1 + cos x)4
dx
Sol.
π
0
(1 + cos x)4
dx =
π
0
(2 cos2 x
2
)4
dx
= 24 π
0
cos8 x
2
dx
I = 16
π/2
0
cos8
t.2dt
= 32
π/2
0
cos8
tdt
I = 32.
7
8
.
5
6
.
3
4
.
1
2
.
π
2

x, cosn
x - Example 1
Evaluate
π
0
(1 + cos x)4
dx
Sol.
π
0
(1 + cos x)4
dx =
π
0
(2 cos2 x
2
)4
dx
= 24 π
0
cos8 x
2
dx
I = 16
π/2
0
cos8
t.2dt
= 32
π/2
0
cos8
tdt
I = 32.
7
8
.
5
6
.
3
4
.
1
2
.
π
2
=
35π
8

x, cosn
x - Example 2
Evaluate
2π
−2π
sin6
xdx

x, cosn
x - Example 2
Evaluate
2π
−2π
sin6
xdx
Sol. Since
a
−a
f(x)dx = 2
a
0
f(x)dx if f(x) is even

x, cosn
x - Example 2
Evaluate
2π
−2π
sin6
xdx
Sol. Since
a
−a
f(x)dx = 2
a
0
∴
2π
−2π
sin6
xdx = 2
2π
0
sin6
xdx

x, cosn
x - Example 2
Evaluate
2π
−2π
sin6
xdx
Sol. Since
a
−a
f(x)dx = 2
a
0
∴
2π
−2π
sin6
xdx = 2
2π
0
sin6
xdx
Now using the
2a
0
f(x)dx =
a
0
f(x)dx +
a
0
f(2a − x)dx

x, cosn
x - Example 2
Evaluate
2π
−2π
sin6
xdx
Sol. Since
a
−a
f(x)dx = 2
a
0
∴
2π
−2π
sin6
xdx = 2
2π
0
sin6
xdx
Now using the
2a
0
f(x)dx =
a
0
f(x)dx +
a
0
f(2a − x)dx
∴ I = 2{
π
0
sin6
xdx +
π
0
sin6
(2π − x)dx}

x, cosn
x - Example 2
Evaluate
2π
−2π
sin6
xdx
Sol. Since
a
−a
f(x)dx = 2
a
0
∴
2π
−2π
sin6
xdx = 2
2π
0
sin6
xdx
Now using the
2a
0
f(x)dx =
a
0
f(x)dx +
a
0
f(2a − x)dx
∴ I = 2{
π
0
sin6
xdx +
π
0
sin6
(2π − x)dx}
= 4
π
0
sin6
xdx Apply same result again

x, cosn
x - Example 2
Evaluate
2π
−2π
sin6
xdx
Sol. Since
a
−a
f(x)dx = 2
a
0
∴
2π
−2π
sin6
xdx = 2
2π
0
sin6
xdx
Now using the
2a
0
f(x)dx =
a
0
f(x)dx +
a
0
f(2a − x)dx
∴ I = 2{
π
0
sin6
xdx +
π
0
sin6
(2π − x)dx}
= 4
π
0
sin6
= 4{
π/2
0
sin6
xdx +
π/2
0
sin6
(π − x)dx}

x, cosn
x - Example 2
Evaluate
2π
−2π
sin6
xdx
Sol. Since
a
−a
f(x)dx = 2
a
0
∴
2π
−2π
sin6
xdx = 2
2π
0
sin6
xdx
Now using the
2a
0
f(x)dx =
a
0
f(x)dx +
a
0
f(2a − x)dx
∴ I = 2{
π
0
sin6
xdx +
π
0
sin6
(2π − x)dx}
= 4
π
0
sin6
= 4{
π/2
0
sin6
xdx +
π/2
0
sin6
(π − x)dx}
= 8
π/2
0
sin6
xdx

x, cosn
x - Example 2
Evaluate
2π
−2π
sin6
xdx
Sol. Since
a
−a
f(x)dx = 2
a
0
∴
2π
−2π
sin6
xdx = 2
2π
0
sin6
xdx
Now using the
2a
0
f(x)dx =
a
0
f(x)dx +
a
0
f(2a − x)dx
∴ I = 2{
π
0
sin6
xdx +
π
0
sin6
(2π − x)dx}
= 4
π
0
sin6
= 4{
π/2
0
sin6
xdx +
π/2
0
sin6
(π − x)dx}
= 8
π/2
0
sin6
xdx
I = 8.
5
6
.
3
4
.
1
2
.
π
2

x, cosn
x - Example 2
Evaluate
2π
−2π
sin6
xdx
Sol. Since
a
−a
f(x)dx = 2
a
0
∴
2π
−2π
sin6
xdx = 2
2π
0
sin6
xdx
Now using the
2a
0
f(x)dx =
a
0
f(x)dx +
a
0
f(2a − x)dx
∴ I = 2{
π
0
sin6
xdx +
π
0
sin6
(2π − x)dx}
= 4
π
0
sin6
= 4{
π/2
0
sin6
xdx +
π/2
0
sin6
(π − x)dx}
= 8
π/2
0
sin6
xdx
I = 8.
5
6
.
3
4
.
1
2
.
π
2
=
5π
4

x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx

x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
Sol. Integration by parts, gives

x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
1
0
x6
sin−1
xdx = {(sin−1
x)
x7
7
}1
0 −
1
0
1
√
1 − x2
.
x7
7

x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
1
0
x6
sin−1
xdx = {(sin−1
x)
x7
7
}1
0 −
1
0
1
√
1 − x2
.
x7
7
=
π
2
.
1
7
−
1
7
1
0
x7
√
1 − x2
dx

x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
1
0
x6
sin−1
xdx = {(sin−1
x)
x7
7
}1
0 −
1
0
1
√
1 − x2
.
x7
7
=
π
2
.
1
7
−
1
7
1
0
x7
√
1 − x2
dx
put x = sin t, ∴ dx = cos tdt

x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
1
0
x6
sin−1
xdx = {(sin−1
x)
x7
7
}1
0 −
1
0
1
√
1 − x2
.
x7
7
=
π
2
.
1
7
−
1
7
1
0
x7
√
1 − x2
dx
when x = 0 → t = 0 and x = 1 → t = π/2

x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
1
0
x6
sin−1
xdx = {(sin−1
x)
x7
7
}1
0 −
1
0
1
√
1 − x2
.
x7
7
=
π
2
.
1
7
−
1
7
1
0
x7
√
1 − x2
dx
when x = 0 → t = 0 and x = 1 → t = π/2
∴ I =
π
14
−
1
7
π/2
0
sin7
t
cos t
costdt

x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
1
0
x6
sin−1
xdx = {(sin−1
x)
x7
7
}1
0 −
1
0
1
√
1 − x2
.
x7
7
=
π
2
.
1
7
−
1
7
1
0
x7
√
1 − x2
dx
when x = 0 → t = 0 and x = 1 → t = π/2
∴ I =
π
14
−
1
7
π/2
0
sin7
t
cos t
costdt
(complete the remaining solution...)

x, cosn
x - Exercise
Evaluate the following
1
π/4
0
sin7
2xdx Ans. 8
35

x, cosn
x - Exercise
1
π/4
0
sin7
2xdx Ans. 8
35
2
π
0
(1 − cos x)3
dx Ans. 5π
2

x, cosn
x - Exercise
1
π/4
0
sin7
2xdx Ans. 8
35
2
π
0
(1 − cos x)3
dx Ans. 5π
2
3
1
0
x7
1 − x4
dx Ans. 1
3

x, cosn
x - Exercise
1
π/4
0
sin7
2xdx Ans. 8
35
2
π
0
(1 − cos x)3
dx Ans. 5π
2
3
1
0
x7
1 − x4
dx Ans. 1
3
4
1
0
x5
sin−1
xdx Ans. 11π
192

x, cosn
x - Exercise
1
π/4
0
sin7
2xdx Ans. 8
35
2
π
0
(1 − cos x)3
dx Ans. 5π
2
3
1
0
x7
1 − x4
dx Ans. 1
3
4
1
0
x5
sin−1
xdx Ans. 11π
192
5
1
0
x2
cos−1
xdx Ans. 2
9

x, cosn
x - Exercise
1
π/4
0
sin7
2xdx Ans. 8
35
2
π
0
(1 − cos x)3
dx Ans. 5π
2
3
1
0
x7
1 − x4
dx Ans. 1
3
4
1
0
x5
sin−1
xdx Ans. 11π
192
5
1
0
x2
cos−1
xdx Ans. 2
9
6
π/2
−π/2
sin7
xdx Ans. 0

Reduction formula - sinm
x cosn
xdx
3. Reduction Formula for sinm
x cosn
xdx where m and n
are positive integers with m ≥ 2, n ≥ 2

x cosn
xdx
x cosn
xdx where m and n
Let Im,n = sinm
x cosn
xdx

x cosn
xdx
x cosn
xdx where m and n
Let Im,n = sinm
x cosn
xdx
In order to apply integration by parts, we split sinm
x cosn
x
in two parts sinm
x cosn
x = (sinm−1
x)(cosn
x sin x)

x cosn
xdx
x cosn
xdx where m and n
Let Im,n = sinm
x cosn
xdx
x cosn
x
in two parts sinm
x cosn
x = (sinm−1
x)(cosn
x sin x)
Let Im,n = (sinm−1
x)(cosn
x sin x)dx

x cosn
xdx
x cosn
xdx where m and n
Let Im,n = sinm
x cosn
xdx
x cosn
x
in two parts sinm
x cosn
x = (sinm−1
x)(cosn
x sin x)
x)(cosn
x sin x)dx
= (sinm−1
x)(− cosn+1 x
n+1
) − {(m − 1) sinm−2
x cos x}{− cosn+1 x
n+1
}dx

x cosn
xdx
x cosn
xdx where m and n
Let Im,n = sinm
x cosn
xdx
x cosn
x
in two parts sinm
x cosn
x = (sinm−1
x)(cosn
x sin x)
x)(cosn
x sin x)dx
= (sinm−1
x)(− cosn+1 x
n+1
) − {(m − 1) sinm−2
n+1
}dx
= (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
sinm−2
x cosn+2
xdx

x cosn
xdx
x cosn
xdx where m and n
Let Im,n = sinm
x cosn
xdx
x cosn
x
in two parts sinm
x cosn
x = (sinm−1
x)(cosn
x sin x)
x)(cosn
x sin x)dx
= (sinm−1
x)(− cosn+1 x
n+1
) − {(m − 1) sinm−2
n+1
}dx
n+1
) + m−1
n+1
sinm−2
x cosn+2
xdx
n+1
) + m−1
n+1
sinm−2
x cosn
x(1 − sin2
x)dx

x cosn
xdx
x cosn
xdx where m and n
Let Im,n = sinm
x cosn
xdx
x cosn
x
in two parts sinm
x cosn
x = (sinm−1
x)(cosn
x sin x)
x)(cosn
x sin x)dx
= (sinm−1
x)(− cosn+1 x
n+1
) − {(m − 1) sinm−2
n+1
}dx
n+1
) + m−1
n+1
sinm−2
x cosn+2
xdx
n+1
) + m−1
n+1
sinm−2
x cosn
x(1 − sin2
x)dx
n+1
) + m−1
n+1
sinm−2
x cosn
xdx −
m−1
n+1
sinm
x cosn
xdx

x cosn
xdx
n+1
) + m−1
n+1
sinm−2
x cosn
xdx −
m−1
n+1
sinm
x cosn
xdx

x cosn
xdx
n+1
) + m−1
n+1
sinm−2
x cosn
xdx −
m−1
n+1
sinm
x cosn
xdx
∴ Im,n = (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
Im−2,n − m−1
n+1
Im,n

x cosn
xdx
n+1
) + m−1
n+1
sinm−2
x cosn
xdx −
m−1
n+1
sinm
x cosn
xdx
n+1
) + m−1
n+1
Im−2,n − m−1
n+1
Im,n
∴ Im,n + m−1
n+1
Im,n = (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
Im−2,n

x cosn
xdx
n+1
) + m−1
n+1
sinm−2
x cosn
xdx −
m−1
n+1
sinm
x cosn
xdx
n+1
) + m−1
n+1
Im−2,n − m−1
n+1
Im,n
∴ Im,n + m−1
n+1
n+1
) + m−1
n+1
Im−2,n
∴ m+n
n+1
n+1
) + m−1
n+1
Im−2,n

x cosn
xdx
n+1
) + m−1
n+1
sinm−2
x cosn
xdx −
m−1
n+1
sinm
x cosn
xdx
n+1
) + m−1
n+1
Im−2,n − m−1
n+1
Im,n
∴ Im,n + m−1
n+1
n+1
) + m−1
n+1
Im−2,n
∴ m+n
n+1
n+1
) + m−1
n+1
Im−2,n
m+n
) + m−1
m+n
Im−2,n —(1)

x cosn
xdx
Similarly in sinm
x cosn
xdx, if we reduce the powers of
cosine instead of sine we get the another reduction formula

x cosn
xdx
Similarly in sinm
x cosn
Im,n = (cosn−1 x sinm+1 x
m+n
) + n−1
m+n
Im,n−2 —(2)

x cosn
xdx
Similarly in sinm
x cosn
Im,n = (cosn−1 x sinm+1 x
m+n
) + n−1
m+n
Im,n−2 —(2)
Now we will consider some deﬁnite integral and apply above
reduction formulae to obtain some generalized result.

x cosn
xdx
Let
π/2
0
sinm
x cosn
xdx

x cosn
xdx
Let
π/2
0
sinm
x cosn
xdx
Substituting the limits of integration 0 and π/2 in (1) and
(2) we have

x cosn
xdx
Let
π/2
0
sinm
x cosn
xdx
(2) we have
Im,n = m−1
m+n
Im−2,n —(3)

x cosn
xdx
Let
π/2
0
sinm
x cosn
xdx
(2) we have
Im,n = m−1
m+n
Im−2,n —(3)
and
Im,n = n−1
m+n
Im,n−2 —(4)

x cosn
xdx
Let
π/2
0
sinm
x cosn
xdx
(2) we have
Im,n = m−1
m+n
Im−2,n —(3)
and
Im,n = n−1
m+n
Im,n−2 —(4)
we shall consider following four diﬀerent cases.

x cosn
xdx
we will be using Im,n = n−1
m+n
Im,n−2 —(4)

x cosn
xdx
m+n
Im,n−2 —(4)
Case 1: Let m and n be both positive integers

x cosn
xdx
m+n
Im,n−2 —(4)
we have Im,n = n−1
m+n
Im,n−2 — (i)

x cosn
xdx
m+n
Im,n−2 —(4)
m+n
Im,n−2 — (i)
by successive application of above formula , replacing n by
n − 2 we get

x cosn
xdx
m+n
Im,n−2 —(4)
m+n
Im,n−2 — (i)
n − 2 we get
Im,n−2 = n−3
m+n−2
Im,n−4 — (ii)

x cosn
xdx
m+n
Im,n−2 —(4)
m+n
Im,n−2 — (i)
n − 2 we get
Im,n−2 = n−3
m+n−2
Im,n−4 — (ii)
∴ Im,n = n−1
m+n
. n−3
m+n−2
Im,n−4 (from (i) and (ii) )

x cosn
xdx
m+n
Im,n−2 —(4)
m+n
Im,n−2 — (i)
n − 2 we get
Im,n−2 = n−3
m+n−2
Im,n−4 — (ii)
∴ Im,n = n−1
m+n
. n−3
m+n−2
∴ Im,n = n−1
m+n
. n−3
m+n−2
. n−5
m+n−4
Im,n−6

x cosn
xdx
m+n
Im,n−2 —(4)
m+n
Im,n−2 — (i)
n − 2 we get
Im,n−2 = n−3
m+n−2
Im,n−4 — (ii)
∴ Im,n = n−1
m+n
. n−3
m+n−2
∴ Im,n = n−1
m+n
. n−3
m+n−2
. n−5
m+n−4
Im,n−6
∴ Im,n = n−1
m+n
. n−3
m+n−2
. n−5
m+n−4
. . . 1
m+2
Im,0 — (5)

x cosn
xdx
(Case 1 contd...)

x cosn
xdx
(Case 1 contd...)
∴ Im,n = n−1
m+n
. n−3
m+n−2
. n−5
m+n−4
. . . 1
m+2
Im,0 — (5)

x cosn
xdx
(Case 1 contd...)
∴ Im,n = n−1
m+n
. n−3
m+n−2
. n−5
m+n−4
. . . 1
m+2
Im,0 — (5)
Now, Im,0 =
π/2
0
sinm
xdx = m−1
m
.m−3
m−2
. . . 1
2
.π
2
(as n is even)

x cosn
xdx
(Case 1 contd...)
∴ Im,n = n−1
m+n
. n−3
m+n−2
. n−5
m+n−4
. . . 1
m+2
Im,0 — (5)
Now, Im,0 =
π/2
0
sinm
xdx = m−1
m
.m−3
m−2
. . . 1
2
.π
2
(as n is even)
substituting the value of Im,0 in eq. (5), we get

x cosn
xdx
(Case 1 contd...)
∴ Im,n = n−1
m+n
. n−3
m+n−2
. n−5
m+n−4
. . . 1
m+2
Im,0 — (5)
Now, Im,0 =
π/2
0
sinm
xdx = m−1
m
.m−3
m−2
. . . 1
2
.π
2
(as n is even)
∴ Im,n = n−1
m+n
. n−3
m+n−2
. n−5
m+n−4
. . . 1
m+2
. m−1
m
.m−3
m−2
. . . 1
2
.π
2

x cosn
xdx
(Case 1 contd...)
∴ Im,n = n−1
m+n
. n−3
m+n−2
. n−5
m+n−4
. . . 1
m+2
Im,0 — (5)
Now, Im,0 =
π/2
0
sinm
xdx = m−1
m
.m−3
m−2
. . . 1
2
.π
2
(as n is even)
∴ Im,n = n−1
m+n
. n−3
m+n−2
. n−5
m+n−4
. . . 1
m+2
. m−1
m
.m−3
m−2
. . . 1
2
.π
2
∴ Im,n =
[(m − 1)(m − 3) . . . 3.1] [(n − 1)(n − 3) . . . 3.1]
(m + n)(m + n − 2) . . . 4.2
.π
2
—
(6)

x cosn
xdx
Case 2: Let m be even and n be odd.

x cosn
xdx
we will be using Im,n = m−1
m+n
Im−2,n —(3)

x cosn
xdx
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
get

x cosn
xdx
m+n
Im−2,n —(3)
get
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 1
n+2
I0,n

x cosn
xdx
m+n
Im−2,n —(3)
get
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 1
n+2
I0,n
I0,n =
π/2
0
cosn
xdx = n−1
n
n−3
n−2
. . . 2
3
.1 (as n is odd)

x cosn
xdx
m+n
Im−2,n —(3)
get
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 1
n+2
I0,n
I0,n =
π/2
0
cosn
xdx = n−1
n
n−3
n−2
. . . 2
3
.1 (as n is odd)
substituting the value of I0,n is previous formula, we get

x cosn
xdx
m+n
Im−2,n —(3)
get
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 1
n+2
I0,n
I0,n =
π/2
0
cosn
xdx = n−1
n
n−3
n−2
. . . 2
3
.1 (as n is odd)
Im,n =
[(m − 1)(m − 3) . . . 3.1] [(n − 1)(n − 3) . . . 4.2]
(m + n)(m + n − 2) . . . 5.3
.1 —
(7)

x cosn
xdx
Case 3: Let n be even and m be odd.

x cosn
xdx
Then Im,n =
π/2
0
sinm
x cosn
xdx

x cosn
xdx
Then Im,n =
π/2
0
sinm
x cosn
xdx
=
π/2
0
sinm π
2
− x cosn π
2
− x dx

x cosn
xdx
Then Im,n =
π/2
0
sinm
x cosn
xdx
=
π/2
0
sinm π
2
− x cosn π
2
− x dx
a
0
f(x)dx =
a
0
f(a − x)dx

x cosn
xdx
Then Im,n =
π/2
0
sinm
x cosn
xdx
=
π/2
0
sinm π
2
− x cosn π
2
− x dx
a
0
f(x)dx =
a
0
f(a − x)dx
=
π/2
0
cosm
x sinn
xdx = In,m

x cosn
xdx
Then Im,n =
π/2
0
sinm
x cosn
xdx
=
π/2
0
sinm π
2
− x cosn π
2
− x dx
a
0
f(x)dx =
a
0
f(a − x)dx
=
π/2
0
cosm
x sinn
xdx = In,m
∴ Im,n = In,m

x cosn
xdx
Then Im,n =
π/2
0
sinm
x cosn
xdx
=
π/2
0
sinm π
2
− x cosn π
2
− x dx
a
0
f(x)dx =
a
0
f(a − x)dx
=
π/2
0
cosm
x sinn
xdx = In,m
∴ Im,n = In,m
Therefore we can use the formula (7) which is

x cosn
xdx
Then Im,n =
π/2
0
sinm
x cosn
xdx
=
π/2
0
sinm π
2
− x cosn π
2
− x dx
a
0
f(x)dx =
a
0
f(a − x)dx
=
π/2
0
cosm
x sinn
xdx = In,m
∴ Im,n = In,m
Therefore we can use the formula (7) which is
Im,n =
[(n − 1)(n − 3) . . . 3.1] [(m − 1)(m − 3) . . . 4.2]
(m + n)(m + n − 2) . . . 5.3
.1 —-
(8)

x cosn
xdx
Case 4: Let m and n both are odd.

x cosn
xdx
m+n
Im−2,n —(3)

x cosn
xdx
m+n
Im−2,n —(3)
get

x cosn
xdx
m+n
Im−2,n —(3)
get
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 4
n+5
. 2
n+3
I1,n

x cosn
xdx
m+n
Im−2,n —(3)
get
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 4
n+5
. 2
n+3
I1,n
I1,n =
π/2
0
sin x cosn
xdx = −
cosn+1
x
n + 1
π/2
0
=
1
n + 1

x cosn
xdx
m+n
Im−2,n —(3)
get
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 4
n+5
. 2
n+3
I1,n
I1,n =
π/2
0
sin x cosn
xdx = −
cosn+1
x
n + 1
π/2
0
=
1
n + 1
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 4
n+5
. 2
n+3
. 1
n+1

x cosn
xdx
m+n
Im−2,n —(3)
get
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 4
n+5
. 2
n+3
I1,n
I1,n =
π/2
0
sin x cosn
xdx = −
cosn+1
x
n + 1
π/2
0
=
1
n + 1
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 4
n+5
. 2
n+3
. 1
n+1
In order to get conversion from we multiply numerator and
denominator by (n − 1)(n − 3)..4.2 we ﬁnd

x cosn
xdx
m+n
Im−2,n —(3)
get
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 4
n+5
. 2
n+3
I1,n
I1,n =
π/2
0
sin x cosn
xdx = −
cosn+1
x
n + 1
π/2
0
=
1
n + 1
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 4
n+5
. 2
n+3
. 1
n+1
In order to get conversion from we multiply numerator and
denominator by (n − 1)(n − 3)..4.2 we ﬁnd
Im,n =
[(m − 1)(m − 3) . . . 4.2] [(n − 1)(n − 3) . . . 4.2]
(m + n)(m + n − 2) . . . (n + 3)(n + 1)(n − 1)(n − 3) . . . 4.2
—- (9)

x cosn
xdx
All the above four cases can be written as

x cosn
xdx
Im,n =
π/2
0
sinm
x cosn
xdx

x cosn
xdx
Im,n =
π/2
0
sinm
x cosn
xdx
=
[(m − 1)(m − 3) . . . .2or1] [(n − 1)(n − 3) . . . .2or1]
(m + n)(m + n − 2)(m + n − 4) . . . .2or1
.k —- (10)

x cosn
xdx
Im,n =
π/2
0
sinm
x cosn
xdx
=
[(m − 1)(m − 3) . . . .2or1] [(n − 1)(n − 3) . . . .2or1]
(m + n)(m + n − 2)(m + n − 4) . . . .2or1
.k —- (10)
where k =
π
2
if both m and n are even

x cosn
xdx
Im,n =
π/2
0
sinm
x cosn
xdx
=
[(m − 1)(m − 3) . . . .2or1] [(n − 1)(n − 3) . . . .2or1]
(m + n)(m + n − 2)(m + n − 4) . . . .2or1
.k —- (10)
where k =
π
2
if both m and n are even
or k = 1 for all other cases

x cosn
xdx - Example 1
Evaluate
π
0
sin2
x(1 + cos x)4
dx

x cosn
xdx - Example 1
Evaluate
π
0
sin2
x(1 + cos x)4
dx
Sol.
π
0
sin2
x(1 + cos x)4
dx

x cosn
xdx - Example 1
Evaluate
π
0
sin2
x(1 + cos x)4
dx
Sol.
π
0
sin2
x(1 + cos x)4
dx
=
π
0
2 cos x
2
sin x
2
2
2 cos2 x
2
4
dx

x cosn
xdx - Example 1
Evaluate
π
0
sin2
x(1 + cos x)4
dx
Sol.
π
0
sin2
x(1 + cos x)4
dx
=
π
0
2 cos x
2
sin x
2
2
2 cos2 x
2
4
dx
= 26 π
0
sin2 x
2
cos10 x
2
dx

x cosn
xdx - Example 1
Evaluate
π
0
sin2
x(1 + cos x)4
dx
Sol.
π
0
sin2
x(1 + cos x)4
dx
=
π
0
2 cos x
2
sin x
2
2
2 cos2 x
2
4
dx
= 26 π
0
sin2 x
2
cos10 x
2
dx
Let x/2 = t → dx = 2dt

x cosn
xdx - Example 1
Evaluate
π
0
sin2
x(1 + cos x)4
dx
Sol.
π
0
sin2
x(1 + cos x)4
dx
=
π
0
2 cos x
2
sin x
2
2
2 cos2 x
2
4
dx
= 26 π
0
sin2 x
2
cos10 x
2
dx
= 26 π/2
0
sin2
t cos10
t.2dt

x cosn
xdx - Example 1
Evaluate
π
0
sin2
x(1 + cos x)4
dx
Sol.
π
0
sin2
x(1 + cos x)4
dx
=
π
0
2 cos x
2
sin x
2
2
2 cos2 x
2
4
dx
= 26 π
0
sin2 x
2
cos10 x
2
dx
= 26 π/2
0
sin2
t cos10
t.2dt
using formula (10) with m = 2, n = 10

x cosn
xdx - Example 1
Evaluate
π
0
sin2
x(1 + cos x)4
dx
Sol.
π
0
sin2
x(1 + cos x)4
dx
=
π
0
2 cos x
2
sin x
2
2
2 cos2 x
2
4
dx
= 26 π
0
sin2 x
2
cos10 x
2
dx
= 26 π/2
0
sin2
t cos10
t.2dt
= 27 [1] [9.7.5.3.1]
12.10.8.6.4.2
.
π
2

x cosn
xdx - Example 1
Evaluate
π
0
sin2
x(1 + cos x)4
dx
Sol.
π
0
sin2
x(1 + cos x)4
dx
=
π
0
2 cos x
2
sin x
2
2
2 cos2 x
2
4
dx
= 26 π
0
sin2 x
2
cos10 x
2
dx
= 26 π/2
0
sin2
t cos10
t.2dt
= 27 [1] [9.7.5.3.1]
12.10.8.6.4.2
.
π
2
=
21π
16

x cosn
xdx - Example 2
Evaluate
π/6
0
cos6
3x sin2
6xdx

x cosn
xdx - Example 2
Evaluate
π/6
0
cos6
3x sin2
6xdx
Sol.
π/6
0
cos6
3x sin2
6xdx

x cosn
xdx - Example 2
Evaluate
π/6
0
cos6
3x sin2
6xdx
Sol.
π/6
0
cos6
3x sin2
6xdx
let 3x = t => dx = dt/3,

x cosn
xdx - Example 2
Evaluate
π/6
0
cos6
3x sin2
6xdx
Sol.
π/6
0
cos6
3x sin2
6xdx
let 3x = t => dx = dt/3,
when x = 0 → t = 0 and x = π/6 → t = π/2

x cosn
xdx - Example 2
Evaluate
π/6
0
cos6
3x sin2
6xdx
Sol.
π/6
0
cos6
3x sin2
6xdx
let 3x = t => dx = dt/3,
when x = 0 → t = 0 and x = π/6 → t = π/2
=
π/2
0
cos6
t sin2
2t(dt/3)

x cosn
xdx - Example 2
Evaluate
π/6
0
cos6
3x sin2
6xdx
Sol.
π/6
0
cos6
3x sin2
6xdx
let 3x = t => dx = dt/3,
when x = 0 → t = 0 and x = π/6 → t = π/2
=
π/2
0
cos6
t sin2
2t(dt/3)
= 1
3
π/2
0
cos6
t(2 sin t cos t)2
dt

x cosn
xdx - Example 2
Evaluate
π/6
0
cos6
3x sin2
6xdx
Sol.
π/6
0
cos6
3x sin2
6xdx
let 3x = t => dx = dt/3,
when x = 0 → t = 0 and x = π/6 → t = π/2
=
π/2
0
cos6
t sin2
2t(dt/3)
= 1
3
π/2
0
cos6
t(2 sin t cos t)2
dt
= 4
3
π/2
0
sin2
t cos8
tdt

x cosn
xdx - Example 2
Evaluate
π/6
0
cos6
3x sin2
6xdx
Sol.
π/6
0
cos6
3x sin2
6xdx
let 3x = t => dx = dt/3,
when x = 0 → t = 0 and x = π/6 → t = π/2
=
π/2
0
cos6
t sin2
2t(dt/3)
= 1
3
π/2
0
cos6
t(2 sin t cos t)2
dt
= 4
3
π/2
0
sin2
t cos8
tdt
= 4
3
[1] [7.5.3.1]
10.8.6.4.2
.
π
2

x cosn
xdx - Example 2
Evaluate
π/6
0
cos6
3x sin2
6xdx
Sol.
π/6
0
cos6
3x sin2
6xdx
let 3x = t => dx = dt/3,
when x = 0 → t = 0 and x = π/6 → t = π/2
=
π/2
0
cos6
t sin2
2t(dt/3)
= 1
3
π/2
0
cos6
t(2 sin t cos t)2
dt
= 4
3
π/2
0
sin2
t cos8
tdt
= 4
3
[1] [7.5.3.1]
10.8.6.4.2
.
π
2
=
7π
384

x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx

x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
Sol. Let 2x = sin t → dx = cos tdt/2

x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
when x = 0, t = 0 and x = 1/2 => sin t = 1 => t = π/2

x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
∴
1/2
0
x3
√
1 − 4x2dx

x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
∴
1/2
0
x3
√
1 − 4x2dx
=
π/2
0
(1/2 sin t)3
cos t.(cos tdt/2)

x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
∴
1/2
0
x3
√
1 − 4x2dx
=
π/2
0
(1/2 sin t)3
cos t.(cos tdt/2)
= 1
16
π/2
0
sin3
t cos2
tdt

x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
∴
1/2
0
x3
√
1 − 4x2dx
=
π/2
0
(1/2 sin t)3
cos t.(cos tdt/2)
= 1
16
π/2
0
sin3
t cos2
tdt
= 1
16
[2] [1]
5.3.1
.1

x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
∴
1/2
0
x3
√
1 − 4x2dx
=
π/2
0
(1/2 sin t)3
cos t.(cos tdt/2)
= 1
16
π/2
0
sin3
t cos2
tdt
= 1
16
[2] [1]
5.3.1
.1
=
1
120

x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx

x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx
Sol. Let x = tan t → dx = sec2
tdt

x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx
tdt
when x = 0, t = 0 and x → ∞ => t = π/2

x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx
tdt
when x = 0, t = 0 and x → ∞ => t = π/2
∴
∞
0
x4
(1 + x2)4
dx

x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx
tdt
when x = 0, t = 0 and x → ∞ => t = π/2
∴
∞
0
x4
(1 + x2)4
dx
=
π/2
0
tan4
t
sec8 t
. sec2
tdt

x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx
tdt
when x = 0, t = 0 and x → ∞ => t = π/2
∴
∞
0
x4
(1 + x2)4
dx
=
π/2
0
tan4
t
sec8 t
. sec2
tdt
(complete the remaining solution...)

x cosn
xdx - Exercise
1
π/6
0
cos4
3x sin2
6xdx Ans. 5π
192

x cosn
xdx - Exercise
1
π/6
0
cos4
3x sin2
6xdx Ans. 5π
192
2
π/4
0
cos3
2x sin4
4x Ans. 128
1155

x cosn
xdx - Exercise
1
π/6
0
cos4
3x sin2
6xdx Ans. 5π
192
2
π/4
0
cos3
2x sin4
4x Ans. 128
1155
3
a
0
x3
(a2
− x2
)3/2
dx Ans. 2a7
35

x cosn
xdx - Exercise
1
π/6
0
cos4
3x sin2
6xdx Ans. 5π
192
2
π/4
0
cos3
2x sin4
4x Ans. 128
1155
3
a
0
x3
(a2
− x2
)3/2
dx Ans. 2a7
35
4
2a
0
x2
√
2ax − x2dx Ans. 5π
8
a4

x cosn
xdx - Exercise
1
π/6
0
cos4
3x sin2
6xdx Ans. 5π
192
2
π/4
0
cos3
2x sin4
4x Ans. 128
1155
3
a
0
x3
(a2
− x2
)3/2
dx Ans. 2a7
35
4
2a
0
x2
√
8
a4
5
2
0
x3
√
2x − x2dx Ans. 7π
8

x cosn
xdx - Exercise
1
π/6
0
cos4
3x sin2
6xdx Ans. 5π
192
2
π/4
0
cos3
2x sin4
4x Ans. 128
1155
3
a
0
x3
(a2
− x2
)3/2
dx Ans. 2a7
35
4
2a
0
x2
√
8
a4
5
2
0
x3
√
2x − x2dx Ans. 7π
8
6
∞
0
x2
(1 + x2)7/2
dx Ans. 2
15

Reduction formula - tann
xdx, cotn
xdx
4. Reduction Formula for tann
xdx, n ≥ 2

xdx, cotn
xdx
xdx, n ≥ 2
In order to apply integration by parts, we split tann
x in two
parts tann
x = (tann−2
x)(tan2
x)

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts tann
x = (tann−2
x)(tan2
x)
Let In = tann
xdx

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts tann
x = (tann−2
x)(tan2
x)
Let In = tann
xdx
= (tann−2
x)(tan2
x)dx

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts tann
x = (tann−2
x)(tan2
x)
Let In = tann
xdx
= (tann−2
x)(tan2
x)dx
= (tann−2
x)(sec2
x − 1)dx

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts tann
x = (tann−2
x)(tan2
x)
Let In = tann
xdx
= (tann−2
x)(tan2
x)dx
= (tann−2
x)(sec2
x − 1)dx
= tann−2
x sec2
xdx − tann−2
xdx

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts tann
x = (tann−2
x)(tan2
x)
Let In = tann
xdx
= (tann−2
x)(tan2
x)dx
= (tann−2
x)(sec2
x − 1)dx
= tann−2
x sec2
xdx − tann−2
xdx
Now the integral tann−2
x sec2
xdx of the form
[f(x)]n
f (x)dx hence it is equal to tann−1 x
n−1

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts tann
x = (tann−2
x)(tan2
x)
Let In = tann
xdx
= (tann−2
x)(tan2
x)dx
= (tann−2
x)(sec2
x − 1)dx
= tann−2
x sec2
xdx − tann−2
xdx
x sec2
xdx of the form
[f(x)]n
n−1
∴ we have following reduction formula for tann
xdx

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts tann
x = (tann−2
x)(tan2
x)
Let In = tann
xdx
= (tann−2
x)(tan2
x)dx
= (tann−2
x)(sec2
x − 1)dx
= tann−2
x sec2
xdx − tann−2
xdx
x sec2
xdx of the form
[f(x)]n
n−1
∴ we have following reduction formula for tann
xdx
In = tann−1 x
n−1
+ In−2 — (1)

xdx, cotn
xdx
5. Reduction Formula for cotn
xdx, n ≥ 2

xdx, cotn
xdx
xdx, n ≥ 2
In order to apply integration by parts, we split cotn
x in two
parts cotn
x = (cotn−2
x)(cot2
x)

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts cotn
x = (cotn−2
x)(cot2
x)
Let In = cotn
xdx

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts cotn
x = (cotn−2
x)(cot2
x)
Let In = cotn
xdx
= (cotn−2
x)(cot2
x)dx

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts cotn
x = (cotn−2
x)(cot2
x)
Let In = cotn
xdx
= (cotn−2
x)(cot2
x)dx
= (cotn−2
x)(cosec2
x − 1)dx

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts cotn
x = (cotn−2
x)(cot2
x)
Let In = cotn
xdx
= (cotn−2
x)(cot2
x)dx
= (cotn−2
x)(cosec2
x − 1)dx
= cotn−2
x cosec2
xdx − cotn−2
xdx

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts cotn
x = (cotn−2
x)(cot2
x)
Let In = cotn
xdx
= (cotn−2
x)(cot2
x)dx
= (cotn−2
x)(cosec2
x − 1)dx
= cotn−2
x cosec2
xdx − cotn−2
xdx
Now the integral cotn−2
x cosec2
xdx of the form
[f(x)]n
f (x)dx hence it is equal to − cotn−1 x
n−1

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts cotn
x = (cotn−2
x)(cot2
x)
Let In = cotn
xdx
= (cotn−2
x)(cot2
x)dx
= (cotn−2
x)(cosec2
x − 1)dx
= cotn−2
x cosec2
xdx − cotn−2
xdx
x cosec2
xdx of the form
[f(x)]n
n−1
∴ we have following reduction formula for cotn
xdx

xdx, cotn
xdx
xdx, n ≥ 2
x in two
parts cotn
x = (cotn−2
x)(cot2
x)
Let In = cotn
xdx
= (cotn−2
x)(cot2
x)dx
= (cotn−2
x)(cosec2
x − 1)dx
= cotn−2
x cosec2
xdx − cotn−2
xdx
x cosec2
xdx of the form
[f(x)]n
n−1
∴ we have following reduction formula for cotn
xdx
In = − cotn−1 x
n−1
+ In−2 — (2)

xdx, cotn
xdx
Now let us consider corresponding deﬁnite integrals
Let In =
π/4
0
tann
xdx, n ≥ 2

xdx, cotn
xdx
Let In =
π/4
0
tann
xdx, n ≥ 2
from eq. (1), we have

xdx, cotn
xdx
Let In =
π/4
0
tann
xdx, n ≥ 2
In =
π/4
0
tann
xdx

xdx, cotn
xdx
Let In =
π/4
0
tann
xdx, n ≥ 2
In =
π/4
0
tann
xdx
= tann−1 x
n−1
π/4
0
−
π/4
0
tann−2
xdx

xdx, cotn
xdx
Let In =
π/4
0
tann
xdx, n ≥ 2
In =
π/4
0
tann
xdx
= tann−1 x
n−1
π/4
0
−
π/4
0
tann−2
xdx
∴ In =
1
n − 1
− In−2 — (3)

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
θdθ

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
θdθ
x = 0, θ = 0 and x = 1, θ = π/4

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
θdθ
x = 0, θ = 0 and x = 1, θ = π/4
∴
1
0
x6
1 + x2
dx =
π/4
0
tan6 θ
1+tan2 θ
sec2
θdθ =

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
θdθ
x = 0, θ = 0 and x = 1, θ = π/4
∴
1
0
x6
1 + x2
dx =
π/4
0
tan6 θ
1+tan2 θ
sec2
θdθ =
π/4
0
tan6
θdθ

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
θdθ
x = 0, θ = 0 and x = 1, θ = π/4
∴
1
0
x6
1 + x2
dx =
π/4
0
tan6 θ
1+tan2 θ
sec2
θdθ =
π/4
0
tan6
θdθ
we have In = 1
n−1
− In−2

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
θdθ
x = 0, θ = 0 and x = 1, θ = π/4
∴
1
0
x6
1 + x2
dx =
π/4
0
tan6 θ
1+tan2 θ
sec2
θdθ =
π/4
0
tan6
θdθ
we have In = 1
n−1
− In−2
using above reduction formula by succession, we have

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
θdθ
x = 0, θ = 0 and x = 1, θ = π/4
∴
1
0
x6
1 + x2
dx =
π/4
0
tan6 θ
1+tan2 θ
sec2
θdθ =
π/4
0
tan6
θdθ
we have In = 1
n−1
− In−2
I6 = 1
5
− I4

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
θdθ
x = 0, θ = 0 and x = 1, θ = π/4
∴
1
0
x6
1 + x2
dx =
π/4
0
tan6 θ
1+tan2 θ
sec2
θdθ =
π/4
0
tan6
θdθ
we have In = 1
n−1
− In−2
I6 = 1
5
− I4
I6 = 1
5
− 1
3
− I2

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
θdθ
x = 0, θ = 0 and x = 1, θ = π/4
∴
1
0
x6
1 + x2
dx =
π/4
0
tan6 θ
1+tan2 θ
sec2
θdθ =
π/4
0
tan6
θdθ
we have In = 1
n−1
− In−2
I6 = 1
5
− I4
I6 = 1
5
− 1
3
− I2
I6 = 1
5
− 1
3
− (1 − I0)

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
θdθ
x = 0, θ = 0 and x = 1, θ = π/4
∴
1
0
x6
1 + x2
dx =
π/4
0
tan6 θ
1+tan2 θ
sec2
θdθ =
π/4
0
tan6
θdθ
we have In = 1
n−1
− In−2
I6 = 1
5
− I4
I6 = 1
5
− 1
3
− I2
I6 = 1
5
− 1
3
− (1 − I0) but I0 =
π/4
0
tan0
θdθ =
π/4
0
dθ = π
4

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
θdθ
x = 0, θ = 0 and x = 1, θ = π/4
∴
1
0
x6
1 + x2
dx =
π/4
0
tan6 θ
1+tan2 θ
sec2
θdθ =
π/4
0
tan6
θdθ
we have In = 1
n−1
− In−2
I6 = 1
5
− I4
I6 = 1
5
− 1
3
− I2
I6 = 1
5
− 1
3
− (1 − I0) but I0 =
π/4
0
tan0
θdθ =
π/4
0
dθ = π
4
∴ I6 = 1
5
− 1
3
+ 1 − π
4

xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
θdθ
x = 0, θ = 0 and x = 1, θ = π/4
∴
1
0
x6
1 + x2
dx =
π/4
0
tan6 θ
1+tan2 θ
sec2
θdθ =
π/4
0
tan6
θdθ
we have In = 1
n−1
− In−2
I6 = 1
5
− I4
I6 = 1
5
− 1
3
− I2
I6 = 1
5
− 1
3
− (1 − I0) but I0 =
π/4
0
tan0
θdθ =
π/4
0
dθ = π
4
∴ I6 = 1
5
− 1
3
+ 1 − π
4
∴
1
0
x6
1 + x2
dx =
π/4
0
tan6
θdθ = 13
15
− π
4

xdx, cotn
xdx - Exercise
1
π/4
0
tan4
xdx

xdx, cotn
xdx - Exercise
1
π/4
0
tan4
xdx
2
π/4
0
tan5
xdx

xdx, cotn
xdx - Exercise
1
π/4
0
tan4
xdx
2
π/4
0
tan5
xdx
3
π/2
π/4
cot5
xdx
Try to ﬁnd reduction formula for
1 secn
xdx
2
π/4
0
secn
xdx

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