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

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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

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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.

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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

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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.

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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

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Winter 2018 7 / 7

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