This document presents reduction formulas for integrals of sinnx and cosnx (where n is greater than or equal to 2). It derives the reduction formulas by repeatedly applying integration by parts. For sinnx, the reduction formula expresses In (the integral of sinnx) in terms of In-1 and In-2. For cosnx, the reduction formula expresses In in terms of In-1 and In-2. The document provides detailed step-by-step working to arrive at each reduction formula.
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Reduction forumla
1. 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
2. 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.
Dr. Nirav Vyas Reduction Formula
3. 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.
Reduction formula is generally obtained by repeated
application of integration by parts.
Dr. Nirav Vyas Reduction Formula
4. 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.
Reduction formula is generally obtained by repeated
application of integration by parts.
Reduction formula are used as a tool to find area and
volume.
Dr. Nirav Vyas Reduction Formula
5. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
Dr. Nirav Vyas Reduction Formula
6. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
x in two
parts sinn
x = (sinn−1
x)(sin x)
Dr. Nirav Vyas Reduction Formula
7. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx
Dr. Nirav Vyas Reduction Formula
8. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
x in two
parts sinn
x = (sinn−1
x)(sin x)
Let In = sinn
xdx
= (sinn−1
x)(sin x)dx
Dr. Nirav Vyas Reduction Formula
9. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
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
Dr. Nirav Vyas Reduction Formula
10. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
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
= − sinn−1
x cos x + (n − 1) sinn−2
x cos2
xdx
Dr. Nirav Vyas Reduction Formula
11. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
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
= − sinn−1
x cos x + (n − 1) sinn−2
x cos2
xdx
= − sinn−1
x cos x + (n − 1) sinn−2
x(1 − sin2
x)dx
Dr. Nirav Vyas Reduction Formula
12. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
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
= − sinn−1
x cos x + (n − 1) sinn−2
x cos2
xdx
= − sinn−1
x cos x + (n − 1) sinn−2
x(1 − sin2
x)dx
= − sinn−1
x cos x + (n − 1) sinn−2
xdx − (n − 1) sinn
xdx
Dr. Nirav Vyas Reduction Formula
13. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
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
= − sinn−1
x cos x + (n − 1) sinn−2
x cos2
xdx
= − sinn−1
x cos x + (n − 1) sinn−2
x(1 − sin2
x)dx
= − sinn−1
x cos x + (n − 1) sinn−2
xdx − (n − 1) sinn
xdx
∴ In = − sinn−1
x cos x + (n − 1)In−2 − (n − 1)In
Dr. Nirav Vyas Reduction Formula
14. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
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
= − sinn−1
x cos x + (n − 1) sinn−2
x cos2
xdx
= − sinn−1
x cos x + (n − 1) sinn−2
x(1 − sin2
x)dx
= − sinn−1
x cos x + (n − 1) sinn−2
xdx − (n − 1) sinn
xdx
∴ In = − sinn−1
x cos x + (n − 1)In−2 − (n − 1)In
∴ In + (n − 1)In = − sinn−1
x cos x + (n − 1)In−2
Dr. Nirav Vyas Reduction Formula
15. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
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
= − sinn−1
x cos x + (n − 1) sinn−2
x cos2
xdx
= − sinn−1
x cos x + (n − 1) sinn−2
x(1 − sin2
x)dx
= − sinn−1
x cos x + (n − 1) sinn−2
xdx − (n − 1) sinn
xdx
∴ In = − sinn−1
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
Dr. Nirav Vyas Reduction Formula
16. Reduction formula - sinn
x, cosn
x
1. Reduction Formula for sinn
xdx, n ≥ 2
In order to apply integration by parts, we split sinn
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
= − sinn−1
x cos x + (n − 1) sinn−2
x cos2
xdx
= − sinn−1
x cos x + (n − 1) sinn−2
x(1 − sin2
x)dx
= − sinn−1
x cos x + (n − 1) sinn−2
xdx − (n − 1) sinn
xdx
∴ In = − sinn−1
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)
Dr. Nirav Vyas Reduction Formula
17. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
Dr. Nirav Vyas Reduction Formula
18. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
x in two
parts cosn
x = (cosn−1
x)(cos x)
Dr. Nirav Vyas Reduction Formula
19. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx
Dr. Nirav Vyas Reduction Formula
20. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
x in two
parts cosn
x = (cosn−1
x)(cos x)
Let In = cosn
xdx
= (cosn−1
x)(cos x)dx
Dr. Nirav Vyas Reduction Formula
21. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
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
Dr. Nirav Vyas Reduction Formula
22. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
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
= cosn−1
x sin x + (n − 1) cosn−2
x sin2
xdx
Dr. Nirav Vyas Reduction Formula
23. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
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
= cosn−1
x sin x + (n − 1) cosn−2
x sin2
xdx
= cosn−1
x sin x + (n − 1) cosn−2
x(1 − cos2
x)dx
Dr. Nirav Vyas Reduction Formula
24. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
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
= cosn−1
x sin x + (n − 1) cosn−2
x sin2
xdx
= cosn−1
x sin x + (n − 1) cosn−2
x(1 − cos2
x)dx
= cosn−1
x sin x + (n − 1) cosn−2
xdx − (n − 1) cosn
xdx
Dr. Nirav Vyas Reduction Formula
25. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
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
= cosn−1
x sin x + (n − 1) cosn−2
x sin2
xdx
= cosn−1
x sin x + (n − 1) cosn−2
x(1 − cos2
x)dx
= cosn−1
x sin x + (n − 1) cosn−2
xdx − (n − 1) cosn
xdx
∴ In = cosn−1
x sin x + (n − 1)In−2 − (n − 1)In
Dr. Nirav Vyas Reduction Formula
26. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
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
= cosn−1
x sin x + (n − 1) cosn−2
x sin2
xdx
= cosn−1
x sin x + (n − 1) cosn−2
x(1 − cos2
x)dx
= cosn−1
x sin x + (n − 1) cosn−2
xdx − (n − 1) cosn
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
Dr. Nirav Vyas Reduction Formula
27. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
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
= cosn−1
x sin x + (n − 1) cosn−2
x sin2
xdx
= cosn−1
x sin x + (n − 1) cosn−2
x(1 − cos2
x)dx
= cosn−1
x sin x + (n − 1) cosn−2
xdx − (n − 1) cosn
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
Dr. Nirav Vyas Reduction Formula
28. Reduction formula - sinn
x, cosn
x
2. Reduction Formula for cosn
xdx, n ≥ 2
In order to apply integration by parts, we split cosn
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
= cosn−1
x sin x + (n − 1) cosn−2
x sin2
xdx
= cosn−1
x sin x + (n − 1) cosn−2
x(1 − cos2
x)dx
= cosn−1
x sin x + (n − 1) cosn−2
xdx − (n − 1) cosn
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)
Dr. Nirav Vyas Reduction Formula
29. Reduction formula - sinn
x, cosn
x
The formulae (1) and (2) are called Reduction formulae
because the exponent is reduced from n to n − 2.
Dr. Nirav Vyas Reduction Formula
30. Reduction formula - sinn
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.
Dr. Nirav Vyas Reduction Formula
31. Reduction formula - sinn
x, cosn
x
Let In =
π/2
0
sinn
xdx
Dr. Nirav Vyas Reduction Formula
32. Reduction formula - sinn
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
Dr. Nirav Vyas Reduction Formula
33. Reduction formula - sinn
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
∴
π/2
0
sinn
xdx = {−
1
n
sinn−1
x cos x}
π/2
0 +
n − 1
n
π/2
0
sinn−2
xdx
Dr. Nirav Vyas Reduction Formula
34. Reduction formula - sinn
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
∴
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
Dr. Nirav Vyas Reduction Formula
35. Reduction formula - sinn
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
∴
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)
Dr. Nirav Vyas Reduction Formula
36. Reduction formula - sinn
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
∴
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,
Dr. Nirav Vyas Reduction Formula
37. Reduction formula - sinn
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
∴
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,
∴ In−2 =
n − 3
n − 2
In−4 — (4)
Dr. Nirav Vyas Reduction Formula
38. Reduction formula - sinn
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
∴
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,
∴ In−2 =
n − 3
n − 2
In−4 — (4)
from (3) and (4) In =
n − 1
n
n − 3
n − 2
In−4
Dr. Nirav Vyas Reduction Formula
39. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Dr. Nirav Vyas Reduction Formula
40. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 1: n is an even integer
Dr. Nirav Vyas Reduction Formula
41. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 1: n is an even integer
Continuing the above process we find
Dr. Nirav Vyas Reduction Formula
42. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 1: n is an even integer
Continuing the above process we find
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
1
2
I0
Dr. Nirav Vyas Reduction Formula
43. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 1: n is an even integer
Continuing the above process we find
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
Dr. Nirav Vyas Reduction Formula
44. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 1: n is an even integer
Continuing the above process we find
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)
Dr. Nirav Vyas Reduction Formula
45. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
Dr. Nirav Vyas Reduction Formula
46. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Dr. Nirav Vyas Reduction Formula
47. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 2: n is an odd integer
Dr. Nirav Vyas Reduction Formula
48. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 2: n is an odd integer
Continuing the above process we find
Dr. Nirav Vyas Reduction Formula
49. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 2: n is an odd integer
Continuing the above process we find
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
2
3
I1
Dr. Nirav Vyas Reduction Formula
50. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 2: n is an odd integer
Continuing the above process we find
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
Dr. Nirav Vyas Reduction Formula
51. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 2: n is an odd integer
Continuing the above process we find
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)
Dr. Nirav Vyas Reduction Formula
52. Reduction formula - sinn
x, cosn
x
Proceeding in the same way we get,
In =
n − 1
n
n − 3
n − 2
n − 5
n − 4
In−6
Case 2: n is an odd integer
Continuing the above process we find
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)
Dr. Nirav Vyas Reduction Formula
53. Reduction formula - sinn
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
Dr. Nirav Vyas Reduction Formula
54. Reduction formula - sinn
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
the reason is
a
0
f(x)dx =
a
0
f(a − x)dx
Dr. Nirav Vyas Reduction Formula
55. Reduction formula - sinn
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
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
Dr. Nirav Vyas Reduction Formula
56. Reduction formula - sinn
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
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
Case 1: n is an even integer
Dr. Nirav Vyas Reduction Formula
57. Reduction formula - sinn
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
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
Case 1: n is an even integer
∴
π/2
0
cosn
xdx =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
3
4
.
1
2
.
π
2
— (5)
Dr. Nirav Vyas Reduction Formula
58. Reduction formula - sinn
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
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
Case 1: n is an even integer
∴
π/2
0
cosn
xdx =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
3
4
.
1
2
.
π
2
— (5)
Case 2: n is an odd integer
Dr. Nirav Vyas Reduction Formula
59. Reduction formula - sinn
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
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
Case 1: n is an even integer
∴
π/2
0
cosn
xdx =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
3
4
.
1
2
.
π
2
— (5)
Case 2: n is an odd integer
∴
π/2
0
cosn
xdx =
n − 1
n
n − 3
n − 2
n − 5
n − 4
. . .
4
5
.
2
3
.1 — (6)
Dr. Nirav Vyas Reduction Formula
60. Reduction formula - sinn
x, cosn
x - Example 1
Evaluate
π
0
(1 + cos x)4
dx
Dr. Nirav Vyas Reduction Formula
61. Reduction formula - sinn
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
Dr. Nirav Vyas Reduction Formula
62. Reduction formula - sinn
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
Dr. Nirav Vyas Reduction Formula
63. Reduction formula - sinn
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
Dr. Nirav Vyas Reduction Formula
64. Reduction formula - sinn
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
I = 16
π/2
0
cos8
t.2dt
Dr. Nirav Vyas Reduction Formula
65. Reduction formula - sinn
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
I = 16
π/2
0
cos8
t.2dt
= 32
π/2
0
cos8
tdt
Dr. Nirav Vyas Reduction Formula
66. Reduction formula - sinn
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
I = 16
π/2
0
cos8
t.2dt
= 32
π/2
0
cos8
tdt
using formula (5) with n = 8
Dr. Nirav Vyas Reduction Formula
67. Reduction formula - sinn
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
I = 16
π/2
0
cos8
t.2dt
= 32
π/2
0
cos8
tdt
using formula (5) with n = 8
I = 32.
7
8
.
5
6
.
3
4
.
1
2
.
π
2
Dr. Nirav Vyas Reduction Formula
68. Reduction formula - sinn
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
I = 16
π/2
0
cos8
t.2dt
= 32
π/2
0
cos8
tdt
using formula (5) with n = 8
I = 32.
7
8
.
5
6
.
3
4
.
1
2
.
π
2
=
35π
8
Dr. Nirav Vyas Reduction Formula
69. Reduction formula - sinn
x, cosn
x - Example 2
Evaluate
2π
−2π
sin6
xdx
Dr. Nirav Vyas Reduction Formula
70. Reduction formula - sinn
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
Dr. Nirav Vyas Reduction Formula
71. Reduction formula - sinn
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
∴
2π
−2π
sin6
xdx = 2
2π
0
sin6
xdx
Dr. Nirav Vyas Reduction Formula
72. Reduction formula - sinn
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
∴
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
Dr. Nirav Vyas Reduction Formula
73. Reduction formula - sinn
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
∴
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}
Dr. Nirav Vyas Reduction Formula
74. Reduction formula - sinn
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
∴
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
Dr. Nirav Vyas Reduction Formula
75. Reduction formula - sinn
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
∴
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
= 4{
π/2
0
sin6
xdx +
π/2
0
sin6
(π − x)dx}
Dr. Nirav Vyas Reduction Formula
76. Reduction formula - sinn
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
∴
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
= 4{
π/2
0
sin6
xdx +
π/2
0
sin6
(π − x)dx}
= 8
π/2
0
sin6
xdx
Dr. Nirav Vyas Reduction Formula
77. Reduction formula - sinn
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
∴
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
= 4{
π/2
0
sin6
xdx +
π/2
0
sin6
(π − x)dx}
= 8
π/2
0
sin6
xdx
using formula (5) with n = 6
Dr. Nirav Vyas Reduction Formula
78. Reduction formula - sinn
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
∴
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
= 4{
π/2
0
sin6
xdx +
π/2
0
sin6
(π − x)dx}
= 8
π/2
0
sin6
xdx
using formula (5) with n = 6
I = 8.
5
6
.
3
4
.
1
2
.
π
2
Dr. Nirav Vyas Reduction Formula
79. Reduction formula - sinn
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
∴
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
= 4{
π/2
0
sin6
xdx +
π/2
0
sin6
(π − x)dx}
= 8
π/2
0
sin6
xdx
using formula (5) with n = 6
I = 8.
5
6
.
3
4
.
1
2
.
π
2
=
5π
4
Dr. Nirav Vyas Reduction Formula
80. Reduction formula - sinn
x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
Dr. Nirav Vyas Reduction Formula
81. Reduction formula - sinn
x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
Sol. Integration by parts, gives
Dr. Nirav Vyas Reduction Formula
82. Reduction formula - sinn
x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
Sol. Integration by parts, gives
1
0
x6
sin−1
xdx = {(sin−1
x)
x7
7
}1
0 −
1
0
1
√
1 − x2
.
x7
7
Dr. Nirav Vyas Reduction Formula
83. Reduction formula - sinn
x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
Sol. Integration by parts, gives
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
Dr. Nirav Vyas Reduction Formula
84. Reduction formula - sinn
x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
Sol. Integration by parts, gives
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
Dr. Nirav Vyas Reduction Formula
85. Reduction formula - sinn
x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
Sol. Integration by parts, gives
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
when x = 0 → t = 0 and x = 1 → t = π/2
Dr. Nirav Vyas Reduction Formula
86. Reduction formula - sinn
x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
Sol. Integration by parts, gives
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
when x = 0 → t = 0 and x = 1 → t = π/2
∴ I =
π
14
−
1
7
π/2
0
sin7
t
cos t
costdt
Dr. Nirav Vyas Reduction Formula
87. Reduction formula - sinn
x, cosn
x - Example 3
Evaluate
1
0
x6
sin−1
xdx
Sol. Integration by parts, gives
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
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...)
Dr. Nirav Vyas Reduction Formula
88. Reduction formula - sinn
x, cosn
x - Exercise
Evaluate the following
1
π/4
0
sin7
2xdx Ans. 8
35
Dr. Nirav Vyas Reduction Formula
89. Reduction formula - sinn
x, cosn
x - Exercise
Evaluate the following
1
π/4
0
sin7
2xdx Ans. 8
35
2
π
0
(1 − cos x)3
dx Ans. 5π
2
Dr. Nirav Vyas Reduction Formula
90. Reduction formula - sinn
x, cosn
x - Exercise
Evaluate the following
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
Dr. Nirav Vyas Reduction Formula
91. Reduction formula - sinn
x, cosn
x - Exercise
Evaluate the following
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
Dr. Nirav Vyas Reduction Formula
92. Reduction formula - sinn
x, cosn
x - Exercise
Evaluate the following
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
Dr. Nirav Vyas Reduction Formula
93. Reduction formula - sinn
x, cosn
x - Exercise
Evaluate the following
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
Dr. Nirav Vyas Reduction Formula
94. 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
Dr. Nirav Vyas Reduction Formula
95. 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
Let Im,n = sinm
x cosn
xdx
Dr. Nirav Vyas Reduction Formula
96. 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
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)
Dr. Nirav Vyas Reduction Formula
97. 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
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)
Let Im,n = (sinm−1
x)(cosn
x sin x)dx
Dr. Nirav Vyas Reduction Formula
98. 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
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)
Let Im,n = (sinm−1
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
Dr. Nirav Vyas Reduction Formula
99. 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
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)
Let Im,n = (sinm−1
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
= (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
sinm−2
x cosn+2
xdx
Dr. Nirav Vyas Reduction Formula
100. 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
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)
Let Im,n = (sinm−1
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
= (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
sinm−2
x cosn+2
xdx
= (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
sinm−2
x cosn
x(1 − sin2
x)dx
Dr. Nirav Vyas Reduction Formula
101. 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
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)
Let Im,n = (sinm−1
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
= (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
sinm−2
x cosn+2
xdx
= (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
sinm−2
x cosn
x(1 − sin2
x)dx
= (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
sinm−2
x cosn
xdx −
m−1
n+1
sinm
x cosn
xdx
Dr. Nirav Vyas Reduction Formula
102. 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
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)
Let Im,n = (sinm−1
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
= (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
sinm−2
x cosn+2
xdx
= (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
sinm−2
x cosn
x(1 − sin2
x)dx
= (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
sinm−2
x cosn
xdx −
m−1
n+1
sinm
x cosn
xdx
Dr. Nirav Vyas Reduction Formula
103. Reduction formula - sinm
x cosn
xdx
= (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
sinm−2
x cosn
xdx −
m−1
n+1
sinm
x cosn
xdx
Dr. Nirav Vyas Reduction Formula
104. Reduction formula - sinm
x cosn
xdx
= (− sinm−1 x cosn+1 x
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
Dr. Nirav Vyas Reduction Formula
105. Reduction formula - sinm
x cosn
xdx
= (− sinm−1 x cosn+1 x
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
∴ Im,n + m−1
n+1
Im,n = (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
Im−2,n
Dr. Nirav Vyas Reduction Formula
106. Reduction formula - sinm
x cosn
xdx
= (− sinm−1 x cosn+1 x
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
∴ Im,n + m−1
n+1
Im,n = (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
Im−2,n
∴ m+n
n+1
Im,n = (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
Im−2,n
Dr. Nirav Vyas Reduction Formula
107. Reduction formula - sinm
x cosn
xdx
= (− sinm−1 x cosn+1 x
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
∴ Im,n + m−1
n+1
Im,n = (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
Im−2,n
∴ m+n
n+1
Im,n = (− sinm−1 x cosn+1 x
n+1
) + m−1
n+1
Im−2,n
∴ Im,n = (− sinm−1 x cosn+1 x
m+n
) + m−1
m+n
Im−2,n —(1)
Dr. Nirav Vyas Reduction Formula
108. Reduction formula - sinm
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
Dr. Nirav Vyas Reduction Formula
109. Reduction formula - sinm
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
Im,n = (cosn−1 x sinm+1 x
m+n
) + n−1
m+n
Im,n−2 —(2)
Dr. Nirav Vyas Reduction Formula
110. Reduction formula - sinm
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
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.
Dr. Nirav Vyas Reduction Formula
111. Reduction formula - sinm
x cosn
xdx
Let
π/2
0
sinm
x cosn
xdx
Dr. Nirav Vyas Reduction Formula
112. Reduction formula - sinm
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
Dr. Nirav Vyas Reduction Formula
113. Reduction formula - sinm
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
Im,n = m−1
m+n
Im−2,n —(3)
Dr. Nirav Vyas Reduction Formula
114. Reduction formula - sinm
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
Im,n = m−1
m+n
Im−2,n —(3)
and
Im,n = n−1
m+n
Im,n−2 —(4)
Dr. Nirav Vyas Reduction Formula
115. Reduction formula - sinm
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
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.
Dr. Nirav Vyas Reduction Formula
116. Reduction formula - sinm
x cosn
xdx
we will be using Im,n = n−1
m+n
Im,n−2 —(4)
Dr. Nirav Vyas Reduction Formula
117. Reduction formula - sinm
x cosn
xdx
we will be using Im,n = n−1
m+n
Im,n−2 —(4)
Case 1: Let m and n be both positive integers
Dr. Nirav Vyas Reduction Formula
118. Reduction formula - sinm
x cosn
xdx
we will be using Im,n = n−1
m+n
Im,n−2 —(4)
Case 1: Let m and n be both positive integers
we have Im,n = n−1
m+n
Im,n−2 — (i)
Dr. Nirav Vyas Reduction Formula
119. Reduction formula - sinm
x cosn
xdx
we will be using Im,n = n−1
m+n
Im,n−2 —(4)
Case 1: Let m and n be both positive integers
we have Im,n = n−1
m+n
Im,n−2 — (i)
by successive application of above formula , replacing n by
n − 2 we get
Dr. Nirav Vyas Reduction Formula
120. Reduction formula - sinm
x cosn
xdx
we will be using Im,n = n−1
m+n
Im,n−2 —(4)
Case 1: Let m and n be both positive integers
we have Im,n = n−1
m+n
Im,n−2 — (i)
by successive application of above formula , replacing n by
n − 2 we get
Im,n−2 = n−3
m+n−2
Im,n−4 — (ii)
Dr. Nirav Vyas Reduction Formula
121. Reduction formula - sinm
x cosn
xdx
we will be using Im,n = n−1
m+n
Im,n−2 —(4)
Case 1: Let m and n be both positive integers
we have Im,n = n−1
m+n
Im,n−2 — (i)
by successive application of above formula , replacing n by
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) )
Dr. Nirav Vyas Reduction Formula
122. Reduction formula - sinm
x cosn
xdx
we will be using Im,n = n−1
m+n
Im,n−2 —(4)
Case 1: Let m and n be both positive integers
we have Im,n = n−1
m+n
Im,n−2 — (i)
by successive application of above formula , replacing n by
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) )
∴ Im,n = n−1
m+n
. n−3
m+n−2
. n−5
m+n−4
Im,n−6
Dr. Nirav Vyas Reduction Formula
123. Reduction formula - sinm
x cosn
xdx
we will be using Im,n = n−1
m+n
Im,n−2 —(4)
Case 1: Let m and n be both positive integers
we have Im,n = n−1
m+n
Im,n−2 — (i)
by successive application of above formula , replacing n by
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) )
∴ 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)
Dr. Nirav Vyas Reduction Formula
124. Reduction formula - sinm
x cosn
xdx
(Case 1 contd...)
Dr. Nirav Vyas Reduction Formula
125. Reduction formula - sinm
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)
Dr. Nirav Vyas Reduction Formula
126. Reduction formula - sinm
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)
Dr. Nirav Vyas Reduction Formula
127. Reduction formula - sinm
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
Dr. Nirav Vyas Reduction Formula
128. Reduction formula - sinm
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
∴ 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
Dr. Nirav Vyas Reduction Formula
129. Reduction formula - sinm
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
∴ 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)
Dr. Nirav Vyas Reduction Formula
130. Reduction formula - sinm
x cosn
xdx
Case 2: Let m be even and n be odd.
Dr. Nirav Vyas Reduction Formula
131. Reduction formula - sinm
x cosn
xdx
Case 2: Let m be even and n be odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
Dr. Nirav Vyas Reduction Formula
132. Reduction formula - sinm
x cosn
xdx
Case 2: Let m be even and n be odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
get
Dr. Nirav Vyas Reduction Formula
133. Reduction formula - sinm
x cosn
xdx
Case 2: Let m be even and n be odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
get
∴ Im,n = m−1
m+n
. m−3
m+n−2
. m−5
m+n−4
. . . 1
n+2
I0,n
Dr. Nirav Vyas Reduction Formula
134. Reduction formula - sinm
x cosn
xdx
Case 2: Let m be even and n be odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
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)
Dr. Nirav Vyas Reduction Formula
135. Reduction formula - sinm
x cosn
xdx
Case 2: Let m be even and n be odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
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
Dr. Nirav Vyas Reduction Formula
136. Reduction formula - sinm
x cosn
xdx
Case 2: Let m be even and n be odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
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
Im,n =
[(m − 1)(m − 3) . . . 3.1] [(n − 1)(n − 3) . . . 4.2]
(m + n)(m + n − 2) . . . 5.3
.1 —
(7)
Dr. Nirav Vyas Reduction Formula
137. Reduction formula - sinm
x cosn
xdx
Case 3: Let n be even and m be odd.
Dr. Nirav Vyas Reduction Formula
138. Reduction formula - sinm
x cosn
xdx
Case 3: Let n be even and m be odd.
Then Im,n =
π/2
0
sinm
x cosn
xdx
Dr. Nirav Vyas Reduction Formula
139. Reduction formula - sinm
x cosn
xdx
Case 3: Let n be even and m be odd.
Then Im,n =
π/2
0
sinm
x cosn
xdx
=
π/2
0
sinm π
2
− x cosn π
2
− x dx
Dr. Nirav Vyas Reduction Formula
140. Reduction formula - sinm
x cosn
xdx
Case 3: Let n be even and m be odd.
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
Dr. Nirav Vyas Reduction Formula
141. Reduction formula - sinm
x cosn
xdx
Case 3: Let n be even and m be odd.
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
Dr. Nirav Vyas Reduction Formula
142. Reduction formula - sinm
x cosn
xdx
Case 3: Let n be even and m be odd.
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
Dr. Nirav Vyas Reduction Formula
143. Reduction formula - sinm
x cosn
xdx
Case 3: Let n be even and m be odd.
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
Dr. Nirav Vyas Reduction Formula
144. Reduction formula - sinm
x cosn
xdx
Case 3: Let n be even and m be odd.
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)
Dr. Nirav Vyas Reduction Formula
145. Reduction formula - sinm
x cosn
xdx
Case 4: Let m and n both are odd.
Dr. Nirav Vyas Reduction Formula
146. Reduction formula - sinm
x cosn
xdx
Case 4: Let m and n both are odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
Dr. Nirav Vyas Reduction Formula
147. Reduction formula - sinm
x cosn
xdx
Case 4: Let m and n both are odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
get
Dr. Nirav Vyas Reduction Formula
148. Reduction formula - sinm
x cosn
xdx
Case 4: Let m and n both are odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
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
Dr. Nirav Vyas Reduction Formula
149. Reduction formula - sinm
x cosn
xdx
Case 4: Let m and n both are odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
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
Dr. Nirav Vyas Reduction Formula
150. Reduction formula - sinm
x cosn
xdx
Case 4: Let m and n both are odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
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
substituting the value of I1,n is previous formula, we get
Dr. Nirav Vyas Reduction Formula
151. Reduction formula - sinm
x cosn
xdx
Case 4: Let m and n both are odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
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
substituting the value of I1,n is previous formula, we get
∴ 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
Dr. Nirav Vyas Reduction Formula
152. Reduction formula - sinm
x cosn
xdx
Case 4: Let m and n both are odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
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
substituting the value of I1,n is previous formula, we get
∴ 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
Dr. Nirav Vyas Reduction Formula
153. Reduction formula - sinm
x cosn
xdx
Case 4: Let m and n both are odd.
we will be using Im,n = m−1
m+n
Im−2,n —(3)
from (3) and by successive application of above formula , we
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
substituting the value of I1,n is previous formula, we get
∴ 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)
Dr. Nirav Vyas Reduction Formula
154. Reduction formula - sinm
x cosn
xdx
All the above four cases can be written as
Dr. Nirav Vyas Reduction Formula
155. Reduction formula - sinm
x cosn
xdx
All the above four cases can be written as
Im,n =
π/2
0
sinm
x cosn
xdx
Dr. Nirav Vyas Reduction Formula
156. Reduction formula - sinm
x cosn
xdx
All the above four cases can be written as
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)
Dr. Nirav Vyas Reduction Formula
157. Reduction formula - sinm
x cosn
xdx
All the above four cases can be written as
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
Dr. Nirav Vyas Reduction Formula
158. Reduction formula - sinm
x cosn
xdx
All the above four cases can be written as
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
Dr. Nirav Vyas Reduction Formula
159. Reduction formula - sinm
x cosn
xdx - Example 1
Evaluate
π
0
sin2
x(1 + cos x)4
dx
Dr. Nirav Vyas Reduction Formula
160. Reduction formula - sinm
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
Dr. Nirav Vyas Reduction Formula
161. Reduction formula - sinm
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
Dr. Nirav Vyas Reduction Formula
162. Reduction formula - sinm
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
Dr. Nirav Vyas Reduction Formula
163. Reduction formula - sinm
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
Dr. Nirav Vyas Reduction Formula
164. Reduction formula - sinm
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
= 26 π/2
0
sin2
t cos10
t.2dt
Dr. Nirav Vyas Reduction Formula
165. Reduction formula - sinm
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
= 26 π/2
0
sin2
t cos10
t.2dt
using formula (10) with m = 2, n = 10
Dr. Nirav Vyas Reduction Formula
166. Reduction formula - sinm
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
= 26 π/2
0
sin2
t cos10
t.2dt
using formula (10) with m = 2, n = 10
= 27 [1] [9.7.5.3.1]
12.10.8.6.4.2
.
π
2
Dr. Nirav Vyas Reduction Formula
167. Reduction formula - sinm
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
= 26 π/2
0
sin2
t cos10
t.2dt
using formula (10) with m = 2, n = 10
= 27 [1] [9.7.5.3.1]
12.10.8.6.4.2
.
π
2
=
21π
16
Dr. Nirav Vyas Reduction Formula
168. Reduction formula - sinm
x cosn
xdx - Example 2
Evaluate
π/6
0
cos6
3x sin2
6xdx
Dr. Nirav Vyas Reduction Formula
169. Reduction formula - sinm
x cosn
xdx - Example 2
Evaluate
π/6
0
cos6
3x sin2
6xdx
Sol.
π/6
0
cos6
3x sin2
6xdx
Dr. Nirav Vyas Reduction Formula
170. Reduction formula - sinm
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,
Dr. Nirav Vyas Reduction Formula
171. Reduction formula - sinm
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
Dr. Nirav Vyas Reduction Formula
172. Reduction formula - sinm
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)
Dr. Nirav Vyas Reduction Formula
173. Reduction formula - sinm
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
Dr. Nirav Vyas Reduction Formula
174. Reduction formula - sinm
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
Dr. Nirav Vyas Reduction Formula
175. Reduction formula - sinm
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
using formula (10) with m = 2, n = 8
Dr. Nirav Vyas Reduction Formula
176. Reduction formula - sinm
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
using formula (10) with m = 2, n = 8
= 4
3
[1] [7.5.3.1]
10.8.6.4.2
.
π
2
Dr. Nirav Vyas Reduction Formula
177. Reduction formula - sinm
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
using formula (10) with m = 2, n = 8
= 4
3
[1] [7.5.3.1]
10.8.6.4.2
.
π
2
=
7π
384
Dr. Nirav Vyas Reduction Formula
178. Reduction formula - sinm
x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
Dr. Nirav Vyas Reduction Formula
179. Reduction formula - sinm
x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
Sol. Let 2x = sin t → dx = cos tdt/2
Dr. Nirav Vyas Reduction Formula
180. Reduction formula - sinm
x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
Sol. Let 2x = sin t → dx = cos tdt/2
when x = 0, t = 0 and x = 1/2 => sin t = 1 => t = π/2
Dr. Nirav Vyas Reduction Formula
181. Reduction formula - sinm
x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
Sol. Let 2x = sin t → dx = cos tdt/2
when x = 0, t = 0 and x = 1/2 => sin t = 1 => t = π/2
∴
1/2
0
x3
√
1 − 4x2dx
Dr. Nirav Vyas Reduction Formula
182. Reduction formula - sinm
x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
Sol. Let 2x = sin t → dx = cos tdt/2
when x = 0, t = 0 and x = 1/2 => sin t = 1 => t = π/2
∴
1/2
0
x3
√
1 − 4x2dx
=
π/2
0
(1/2 sin t)3
cos t.(cos tdt/2)
Dr. Nirav Vyas Reduction Formula
183. Reduction formula - sinm
x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
Sol. Let 2x = sin t → dx = cos tdt/2
when x = 0, t = 0 and x = 1/2 => sin t = 1 => t = π/2
∴
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
Dr. Nirav Vyas Reduction Formula
184. Reduction formula - sinm
x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
Sol. Let 2x = sin t → dx = cos tdt/2
when x = 0, t = 0 and x = 1/2 => sin t = 1 => t = π/2
∴
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
using formula (10) with m = 3, n = 2
Dr. Nirav Vyas Reduction Formula
185. Reduction formula - sinm
x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
Sol. Let 2x = sin t → dx = cos tdt/2
when x = 0, t = 0 and x = 1/2 => sin t = 1 => t = π/2
∴
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
using formula (10) with m = 3, n = 2
= 1
16
[2] [1]
5.3.1
.1
Dr. Nirav Vyas Reduction Formula
186. Reduction formula - sinm
x cosn
xdx - Example 3
Evaluate
1/2
0
x3
√
1 − 4x2dx
Sol. Let 2x = sin t → dx = cos tdt/2
when x = 0, t = 0 and x = 1/2 => sin t = 1 => t = π/2
∴
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
using formula (10) with m = 3, n = 2
= 1
16
[2] [1]
5.3.1
.1
=
1
120
Dr. Nirav Vyas Reduction Formula
187. Reduction formula - sinm
x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx
Dr. Nirav Vyas Reduction Formula
188. Reduction formula - sinm
x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx
Sol. Let x = tan t → dx = sec2
tdt
Dr. Nirav Vyas Reduction Formula
189. Reduction formula - sinm
x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx
Sol. Let x = tan t → dx = sec2
tdt
when x = 0, t = 0 and x → ∞ => t = π/2
Dr. Nirav Vyas Reduction Formula
190. Reduction formula - sinm
x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx
Sol. Let x = tan t → dx = sec2
tdt
when x = 0, t = 0 and x → ∞ => t = π/2
∴
∞
0
x4
(1 + x2)4
dx
Dr. Nirav Vyas Reduction Formula
191. Reduction formula - sinm
x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx
Sol. Let x = tan t → dx = sec2
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
Dr. Nirav Vyas Reduction Formula
192. Reduction formula - sinm
x cosn
xdx - Example 4
Evaluate
∞
0
x4
(1 + x2)4
dx
Sol. Let x = tan t → dx = sec2
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...)
Dr. Nirav Vyas Reduction Formula
193. Reduction formula - sinm
x cosn
xdx - Exercise
Evaluate the following
1
π/6
0
cos4
3x sin2
6xdx Ans. 5π
192
Dr. Nirav Vyas Reduction Formula
194. Reduction formula - sinm
x cosn
xdx - Exercise
Evaluate the following
1
π/6
0
cos4
3x sin2
6xdx Ans. 5π
192
2
π/4
0
cos3
2x sin4
4x Ans. 128
1155
Dr. Nirav Vyas Reduction Formula
195. Reduction formula - sinm
x cosn
xdx - Exercise
Evaluate the following
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
Dr. Nirav Vyas Reduction Formula
196. Reduction formula - sinm
x cosn
xdx - Exercise
Evaluate the following
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
Dr. Nirav Vyas Reduction Formula
197. Reduction formula - sinm
x cosn
xdx - Exercise
Evaluate the following
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
5
2
0
x3
√
2x − x2dx Ans. 7π
8
Dr. Nirav Vyas Reduction Formula
198. Reduction formula - sinm
x cosn
xdx - Exercise
Evaluate the following
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
5
2
0
x3
√
2x − x2dx Ans. 7π
8
6
∞
0
x2
(1 + x2)7/2
dx Ans. 2
15
Dr. Nirav Vyas Reduction Formula
199. Reduction formula - tann
xdx, cotn
xdx
4. Reduction Formula for tann
xdx, n ≥ 2
Dr. Nirav Vyas Reduction Formula
200. Reduction formula - tann
xdx, cotn
xdx
4. Reduction Formula for tann
xdx, n ≥ 2
In order to apply integration by parts, we split tann
x in two
parts tann
x = (tann−2
x)(tan2
x)
Dr. Nirav Vyas Reduction Formula
201. Reduction formula - tann
xdx, cotn
xdx
4. Reduction Formula for tann
xdx, n ≥ 2
In order to apply integration by parts, we split tann
x in two
parts tann
x = (tann−2
x)(tan2
x)
Let In = tann
xdx
Dr. Nirav Vyas Reduction Formula
202. Reduction formula - tann
xdx, cotn
xdx
4. Reduction Formula for tann
xdx, n ≥ 2
In order to apply integration by parts, we split tann
x in two
parts tann
x = (tann−2
x)(tan2
x)
Let In = tann
xdx
= (tann−2
x)(tan2
x)dx
Dr. Nirav Vyas Reduction Formula
203. Reduction formula - tann
xdx, cotn
xdx
4. Reduction Formula for tann
xdx, n ≥ 2
In order to apply integration by parts, we split tann
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
Dr. Nirav Vyas Reduction Formula
204. Reduction formula - tann
xdx, cotn
xdx
4. Reduction Formula for tann
xdx, n ≥ 2
In order to apply integration by parts, we split tann
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
Dr. Nirav Vyas Reduction Formula
205. Reduction formula - tann
xdx, cotn
xdx
4. Reduction Formula for tann
xdx, n ≥ 2
In order to apply integration by parts, we split tann
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
Dr. Nirav Vyas Reduction Formula
206. Reduction formula - tann
xdx, cotn
xdx
4. Reduction Formula for tann
xdx, n ≥ 2
In order to apply integration by parts, we split tann
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
∴ we have following reduction formula for tann
xdx
Dr. Nirav Vyas Reduction Formula
207. Reduction formula - tann
xdx, cotn
xdx
4. Reduction Formula for tann
xdx, n ≥ 2
In order to apply integration by parts, we split tann
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
∴ we have following reduction formula for tann
xdx
In = tann−1 x
n−1
+ In−2 — (1)
Dr. Nirav Vyas Reduction Formula
208. Reduction formula - tann
xdx, cotn
xdx
5. Reduction Formula for cotn
xdx, n ≥ 2
Dr. Nirav Vyas Reduction Formula
209. Reduction formula - tann
xdx, cotn
xdx
5. Reduction Formula for cotn
xdx, n ≥ 2
In order to apply integration by parts, we split cotn
x in two
parts cotn
x = (cotn−2
x)(cot2
x)
Dr. Nirav Vyas Reduction Formula
210. Reduction formula - tann
xdx, cotn
xdx
5. Reduction Formula for cotn
xdx, n ≥ 2
In order to apply integration by parts, we split cotn
x in two
parts cotn
x = (cotn−2
x)(cot2
x)
Let In = cotn
xdx
Dr. Nirav Vyas Reduction Formula
211. Reduction formula - tann
xdx, cotn
xdx
5. Reduction Formula for cotn
xdx, n ≥ 2
In order to apply integration by parts, we split cotn
x in two
parts cotn
x = (cotn−2
x)(cot2
x)
Let In = cotn
xdx
= (cotn−2
x)(cot2
x)dx
Dr. Nirav Vyas Reduction Formula
212. Reduction formula - tann
xdx, cotn
xdx
5. Reduction Formula for cotn
xdx, n ≥ 2
In order to apply integration by parts, we split cotn
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
Dr. Nirav Vyas Reduction Formula
213. Reduction formula - tann
xdx, cotn
xdx
5. Reduction Formula for cotn
xdx, n ≥ 2
In order to apply integration by parts, we split cotn
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
Dr. Nirav Vyas Reduction Formula
214. Reduction formula - tann
xdx, cotn
xdx
5. Reduction Formula for cotn
xdx, n ≥ 2
In order to apply integration by parts, we split cotn
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
Dr. Nirav Vyas Reduction Formula
215. Reduction formula - tann
xdx, cotn
xdx
5. Reduction Formula for cotn
xdx, n ≥ 2
In order to apply integration by parts, we split cotn
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
∴ we have following reduction formula for cotn
xdx
Dr. Nirav Vyas Reduction Formula
216. Reduction formula - tann
xdx, cotn
xdx
5. Reduction Formula for cotn
xdx, n ≥ 2
In order to apply integration by parts, we split cotn
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
∴ we have following reduction formula for cotn
xdx
In = − cotn−1 x
n−1
+ In−2 — (2)
Dr. Nirav Vyas Reduction Formula
217. Reduction formula - tann
xdx, cotn
xdx
Now let us consider corresponding definite integrals
Let In =
π/4
0
tann
xdx, n ≥ 2
Dr. Nirav Vyas Reduction Formula
218. Reduction formula - tann
xdx, cotn
xdx
Now let us consider corresponding definite integrals
Let In =
π/4
0
tann
xdx, n ≥ 2
from eq. (1), we have
Dr. Nirav Vyas Reduction Formula
219. Reduction formula - tann
xdx, cotn
xdx
Now let us consider corresponding definite integrals
Let In =
π/4
0
tann
xdx, n ≥ 2
from eq. (1), we have
In =
π/4
0
tann
xdx
Dr. Nirav Vyas Reduction Formula
220. Reduction formula - tann
xdx, cotn
xdx
Now let us consider corresponding definite integrals
Let In =
π/4
0
tann
xdx, n ≥ 2
from eq. (1), we have
In =
π/4
0
tann
xdx
= tann−1 x
n−1
π/4
0
−
π/4
0
tann−2
xdx
Dr. Nirav Vyas Reduction Formula
221. Reduction formula - tann
xdx, cotn
xdx
Now let us consider corresponding definite integrals
Let In =
π/4
0
tann
xdx, n ≥ 2
from eq. (1), we have
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)
Dr. Nirav Vyas Reduction Formula
222. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Dr. Nirav Vyas Reduction Formula
223. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
θdθ
Dr. Nirav Vyas Reduction Formula
224. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
θdθ
x = 0, θ = 0 and x = 1, θ = π/4
Dr. Nirav Vyas Reduction Formula
225. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
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 =
Dr. Nirav Vyas Reduction Formula
226. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
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
Dr. Nirav Vyas Reduction Formula
227. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
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
Dr. Nirav Vyas Reduction Formula
228. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
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
Dr. Nirav Vyas Reduction Formula
229. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
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
I6 = 1
5
− I4
Dr. Nirav Vyas Reduction Formula
230. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
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
I6 = 1
5
− I4
I6 = 1
5
− 1
3
− I2
Dr. Nirav Vyas Reduction Formula
231. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
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
I6 = 1
5
− I4
I6 = 1
5
− 1
3
− I2
I6 = 1
5
− 1
3
− (1 − I0)
Dr. Nirav Vyas Reduction Formula
232. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
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
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
Dr. Nirav Vyas Reduction Formula
233. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
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
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
Dr. Nirav Vyas Reduction Formula
234. Reduction formula - tann
xdx - Example 1
Evaluate
1
0
x6
1 + x2
dx
Let x = tanθ, ∴ dx = sec2
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
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
Dr. Nirav Vyas Reduction Formula
235. Reduction formula - tann
xdx, cotn
xdx - Exercise
1
π/4
0
tan4
xdx
Dr. Nirav Vyas Reduction Formula
236. Reduction formula - tann
xdx, cotn
xdx - Exercise
1
π/4
0
tan4
xdx
2
π/4
0
tan5
xdx
Dr. Nirav Vyas Reduction Formula
237. Reduction formula - tann
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
Dr. Nirav Vyas Reduction Formula