Integration: Reduction Formulas
[Pages:7]Integration: Reduction Formulas
Any positive integer power of sin x can be integrated by using a reduction formula.
Example
Prove that for any integer n 2, Z sinn x dx =
1 sinn 1 x cos x + n 1 Z sinn 2 x dx.
n
n
Solution. We use integration by parts. Let u = sinn 1 x and dv = sin x dx. Then, du = (n 1) sinn 2 x cos x dx and we can use v =
So,
Z
Z
sinn x dx =
sinn 1 x sin x dx
Z
= sinn 1 x( cos x)
( cos x)(n 1) sinn 2 x cos x dx
Z
= -sinn 1 x cos x + (n 1) sinn 2 x cos2 x dx
Z
= -sinn 1 x cos x + (n 1) sinn 2 x (1 sin2 x) dx
Z
= -sinn 1 x cos x + (n 1) sinn 2 x sinn x dx
Z
Z
= -sinn 1 x cos x + (n 1) sinn 2 x dx (n 1) sinn x dx
Z
Z
Re-arranging, we get n sinn x dx = sinn 1 x cos x + (n 1) sinn 2 x dx.
Z Dividing both sides by n, we get sinn x dx =
1 sinn 1 x cos x + n
1 Z sinn 2 x dx.
n
n
cos x.
University Calculus II (University of Calgary)
Winter 2018 1 / 7
Example
Use the reduction formula to find the integrals of sin2 x, sin3 x, sin4 x.
Solution. We recall the reduction formula proved above. For n 2,
Z sinn x dx =
1 sinn 1 x cos x + n
1 Z sinn 2 x dx.
n
n
WZith n = 2, sin2 x dx
=
1
1Z
- sin x cos x +
1 dx
2
2
1
1
= - sin x cos x + x + C .
2
2
WZith n = 3, sin3 x dx
=
1 -
sin2 x
cos x
+
2
Z
sin x dx
3
3
=
1 -
sin2 x
cos x
2 cos x + C .
3
3
With n = 4,
Z sin4 x dx
=
1 -
sin3 x
cos x
+
3
Z
sin2 x dx
=
4 1 -
sin3 x
cos x
+
4 3
1
1
sin x cos x + x + C .
4
42
2
=
1 -
sin3 x
cos x
3
3
sin x cos x + x + C .
4
8
8
University Calculus II (University of Calgary)
Winter 2018 2 / 7
Example
Express sin3 x cos4 x as a sum of constant multiples of sin x. Hence, or otherwise, find the integral of sin3 x cos4 x.
Solution. Since cos2 x = 1 sin2 x, cos6 x = (cos2 x)4 = (1 sin2 x)2 = 1 2 sin2 x + sin4 x.
So, sin3 x cos6 x = sin3 1 2 sin2 x + sin4 x = sin3 x Z
For the sake of simplicity, we will denote sinn x dx by In.
2 sin5 x + sin7 x.
The reduction formula reads In =
1 sinn n
1x
cos x
+
n
n
1 In
2.
By using the reduction formula, we get
I1 = -cos x + C
I3 I5 I7
= = =
1 -
sin2 x
cos x
31 -
sin4 x
cos x
51 -
sin6 x
cos x
7
2 cos x + C
34 sin2 x cos x 165 sin4 x cos x 35
8 cos x + C
185 sin2 x cos x 35
16 cos x + C
35
Integrating both sides of the identity
sin3 x cos4 x = sin3 x 2 sin5 x + sin7 x,
we get Z
sin3 x cos4 x dx = I3
2I5 + I7 =
1 sin6 x cos x + 8 sin4 x cos x
7
35
1 sin2 x cos x 35
2 cos x + C .
35
University Calculus II (University of Calgary)
Winter 2018 3 / 7
Remarks.
One can use integration by parts to derive a reduction formula for integrals of powers of cosine:
Z cosn x dx = 1 cosn 1 x sin x + n 1 Z cosn 2 x dx.
n
n
One can integrate all positive integer powers of cos x.
By using the identity sin2 = 1 cos2 x, one can express sinm x cosn x as a sum of constant multiples of powers of cos x if m is even.
In view of this and our previous examples, we can integrate sinm x cosn x as long as m and/or n is even.
If both m and n are odd, then we need a dierent approach. It turns out that either of the substitutions u = sin x and u = cos x will work.
Example
If we use u = sin x, then du = cos x dx, and since cos4 x = (1 sin2 x)2 = (1 u2)2,
Z
Z
Z
Z
sin3 x cos5 xdx = sin3 cos4 x dx = u3(1 u2)2du = (u3 2u5 + u7)du = ...
If we use u = cos x, then du = sin x dx, and since cos2 x = 1 sin2 x = 1 u2,
Z
Z
Z
Z
sin3 x cos5 xdx = ( sin2 x) cos5 x( sin x)dx =
(1 u2)u5du = ( u5 + u7)du = ...
Convince yourself this: If m is odd, then u = cos x will work (even if n is even). If n is odd, then u = sin x will work (even if m is even).
All functions of the form sinm x cosn x can be integrated.
University Calculus II (University of Calgary)
Winter 2018 4 / 7
Z Let's consider integrals of the form secm x tann x dx.
Example
Prove that for any integer n 2,
Z tann x dx =
1
tann 1 x
n1
Z
Z tann 2 x dx.
Use this and the fact that tan x dx = ln | sec x| + C to find the integrals of tan2 x, tan3 x, tan4 x and tan5 x.
Using the identity tan2 x = sec2 x 1,
Z
Z
Z
Z
Z
tann x dx = tann 2 xtan2x dx = tann 2 x(sec2x 1) dx = tann 2 xsec2x dx
tann 2 x dx.
Using the substitution u = tan x, we have du = sec2 x dx. So,
Z
Z
tann 2 x sec2 x dx = un 2du =
1 un 1 + C =
1
tann 1 x + C .
n1
n1
The reduction formula is proved. Z
With n = 2, we have tan2 x dx = tan x
Z 1dx = tan x
x + C.
Z With n = 3, we have
tan3 x dx = 1 tan2 x
2
Z With n = 4, we have
tan4 x dx = 1 tan3 x
3
Z With n = 5, we have
tan5 x dx = 1 tan4 x
4
Z tan x dx = 1 tan2 x 2
Z tan2 x dx = 1 tan3 x 3
Z tan3 x dx = 1 tan4 x 4
ln | sec x| + C . tan x + x + C . 1 tan2 x + ln | sec x| + C . 2
University Calculus II (University of Calgary)
Winter 2018 5 / 7
Remarks. One can integrate all positive integer powers of tan x. One can derive a reduction formula for sec x by integration by parts. Z Using the reduction formula and the fact sec x dx = ln | sec x + tan x| + C , we can integrate all positive integer powers of sec x. Similar strategies used for sinm x cosn x can be formulated to integrate all functions of the form secm x tann x.
University Calculus II (University of Calgary)
Winter 2018 6 / 7
Further Examples and Exercises
Prove the reduction formula for integrals of powers of cos x:
Z cosn x dx = 1 cosn 1 x sin x + n 1 Z cosn 2 x dx.
n
n
Use it to find the integrals of cos2 x, cos3 x, cos4 x, cos5 x, cos6 x.
Express sin4 x cos6 x as a sum of constant multiples of cos x. Hence, or otherwise, find the integral of sin4 x cos6 x.
Use integration by parts to find a reduction formula for integrals of positive integer powers of sec x.
Find the following integrals.
Z
Z
Z
sin5 x cos2 x dx, cos4 x sin2 x dx, sin4 x cos4 x dx
Find the following integrals.
Z
Z
Z
Z
Z
tan3 x dx, sec5 x dx, tan4 x dx, sec3 x tan2 x dx, sec4 x tan3 x dx
University Calculus II (University of Calgary)
Winter 2018 7 / 7
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