Section 6.2--Trigonometric Integrals and Substitutions

[Pages:20]Section 6.2 Trigonometric Integrals and Substitutions

Trigonometric Integrals and Substitutions

2010 Kiryl Tsishchanka

This section consists of two parts: Part I: Trigonometric Integrals. We will distinguish three main cases:

Case A: Integrals of type

sinm x cosn xdx

where m and n are nonnegative integers.

METHOD OF INTEGRATION: (i) If m is odd, then u = cos x. (ii) If n is odd, then u = sin x. (iii) If both m and n are even, then use the identities

sin2

x

=

1 (1

-

cos 2x),

cos2 x

=

1 (1 +

cos 2x)

2

2

or(and) sometimes

sin x cos x

=

1 2

sin 2x

EXAMPLE 1: Find sin3 xdx.

Solution: We have

cos x = u

sin3 xdx =

sin2 x ? sin xdx =

(1

-

cos2

x)

sin

xdx

=

d cos x = du

- sin xdx = du

sin xdx = -du

= - (1 - u2)du = -u + 1u3 + C = - cos x + 1 cos3 x + C

3

3

EXAMPLE 2: Find sin5 xdx.

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Section 6.2 Trigonometric Integrals and Substitutions

EXAMPLE 2: Find sin5 xdx.

2010 Kiryl Tsishchanka

Solution: We have

sin5 xdx = sin4 x ? sin xdx = (sin2 x)2 sin xdx = (1 - cos2 x)2 sin xdx

cos x = u

=

d cos x = du

=

-

(1 - u2)2du = -

(1 - 2u2 + u4)du

- sin xdx = du

sin xdx = -du

= -u + 2u3 - 1 u5 + C = - cos x + 2 cos3 x - 1 cos5 x + C

35

3

5

EXAMPLE 3: Find sin2 x cos3 xdx. Solution: We have

sin2 x cos3 xdx = sin2 x cos2 x ? cos xdx = sin2 x(1 - sin2 x) cos xdx

sin x = u

= d sin x = du =

cos xdx = du

u2(1 - u2)du =

(u2 - u4)du = 1u3 - 1 u5 + C = 1 sin3 x - 1 sin5 x + C

35

3

5

EXAMPLE 4: Find 3 sin x cos5 xdx.

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Section 6.2 Trigonometric Integrals and Substitutions

EXAMPLE 4: Find 3 sin x cos5 xdx.

2010 Kiryl Tsishchanka

Solution: We have 3 sin x cos5 xdx =

=

3 sin x cos4 x ? cos xdx = 3 sin x(cos2 x)2 ? cos xdx

sin x = u

3 sin x(1 - sin2 x)2 ? cos xdx = d sin x = du =

cos xdx = du

3 u(1

-

u2)2

du

= u1/3(1 - 2u2 + u4)du = (u1/3 - 2u2+1/3 + u4+1/3)du

= (u1/3 - 2u7/3 + u13/3)du = u1/3+1 - 2 u7/3+1 + u13/3+1 + C 1/3 + 1 7/3 + 1 13/3 + 1

= u4/3 - 2 u10/3 + u16/3 + C 4/3 10/3 16/3

=

3 4

u4/3

-

3 5

u10/3

+

3 16

u16/3

+C

= 3 sin4/3 x - 3 sin10/3 x + 3 sin16/3 x + C

4

5

16

EXAMPLE 5: Find sin2 x cos2 xdx.

Solution: We have

sin2 x cos2 xdx = (sin x cos x)2dx =

1

2

1

sin 2x dx =

sin2 2xdx = 1

1(1 - cos 4x)dx

2

4

42

4x = u

=

1 8

(1

-

cos 4x)dx

=

d(4x) = du 4dx = du

=

1 32

(1

-

cos u)du

=

1 32

(u

-

sin u)

+

C

=

1 32

(4x

-

sin

4x)

+

C

dx

=

1 4

du

EXAMPLE 6: Find cos4 xdx.

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Section 6.2 Trigonometric Integrals and Substitutions

EXAMPLE 6: Find cos4 xdx.

2010 Kiryl Tsishchanka

Solution: We have cos4 xdx = (cos2 x)2dx =

1 + cos 2x 2 dx = 1 (1 + cos 2x)2dx

2

4

= 1 (1 + 2 cos 2x + cos2 2x)dx = 1

4

4

1 + 2 cos 2x + 1 (1 + cos 4x) dx 2

=

1 4

3 2

+

2

cos

2x

+

1 2

cos

4x

dx

=

1 4

?

3 2

x

+

1 4

?

2

?

1 2

sin

2x

+

1 4

?

1 2

?

1 4

sin

4x

+

C

=

3 8

x

+

1 4

sin

2x

+

1 32

sin

4x

+

C

Case B: Integrals of type

tanm x secn xdx

Midterms where m and n are nonnegative integers.

METHOD OF INTEGRATION: (i) If m is odd, then u = sec x. (ii) If n is even, then u = tan x. (iii) In other cases the guidelines are not as clear-cut.

EXAMPLE 7: Find tan xdx.

Solution 1: We have

cos x = u

tan xdx =

sin cos

x x

dx

=

1 cos x

? sin xdx

=

d cos x = du - sin xdx = du

=

-

1 u

du

sin xdx = -du

= - ln |u| + C = - ln | cos x| + C = ln | cos x|-1 + C = ln | sec x| + C

Solution 2: We have

tan xdx =

sec x = u

tan x sec sec x

x

dx

=

d sec x = du

=

sec x tan xdx = du

1 u

du

=

ln

|u|

+

C

=

ln

|

sec

x|

+

C

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Section 6.2 Trigonometric Integrals and Substitutions

EXAMPLE 8: Find (a) csc xdx

(b) sec xdx

2010 Kiryl Tsishchanka

Solution 1: (a) We have

csc xdx =

=

dx = sin x

dx

sin

2

?

x 2

sec2 udu sec2 u sin u cos u

=

x 2

=

u

d

=

x 2

= du

=

1

2dx = du

2du = sin (2u)

2du = 2 sin u cos u

du sin u cos u

dx = 2du

tan u = v

sec2 udu

sec2 udu

sec2 udu

dv

sin u cos u =

cos2 u

sin u =

cos u

= d tan u = dv =

tan u

v

sec2 udu = dv

= ln |v| + C = ln | tan u| + C = ln tan

x 2

+C

In short,

csc xdx =

dx sin x

=

dx

sin

2

?

x 2

=

2 sin

dx

x 2

cos

x 2

=

sec2 x dx

2

2 sec2 x sin x cos x

2

2

2

=

sec2 x dx 2 x

2 tan 2

=

tan

x 2x

=v

d tan = dv

2

1 sec2 x dx = dv

2 sec2

x 2

2

dx

=

2dv

=

dv = ln |v| + C = ln tan x + C

v

2

This can be rewritten as ln |csc x - cot x| + C (see Appendix II).

(b) We have

sec xdx =

dx cos x

=

dx

sin

x

+

2

x

+

2

=

u

=

d

x

+

2

=

du

=

du sin u

dx = du

which is ln tan u + C by (a). Substituting x + for u, we get

2

2

x +

sec xdx = ln tan

2

2

+C

= ln

tan

x 2+4

+C

This can be rewritten as ln |sec x + tan x| + C (see Appendix II).

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Section 6.2 Trigonometric Integrals and Substitutions

2010 Kiryl Tsishchanka

Solution 2: (a) We have

csc xdx =

1 sin

x

dx

=

sin x sin2 x

dx

=

=-

(1

-

1 u)(1

+

u) du

=

-

cos x = u

1

sin x - cos2

x

dx

=

d cos x = du - sin xdx = du

=

-

sin xdx = -du

1

1 -

u2

du

1 2

1

1 -

u

+

1

1 +

u

du

=

-

1 2

1

1 -

u du

-

1 2

1

1 +

u

du

=

1 2

ln |1

-

u|

-

1 2

ln |1

+

u|

+

C

=

1 2

ln

1-u 1+u

+

C

=

1 2

ln

1 - cos x 1 + cos x

+C

This can be rewritten as ln |csc x - cot x| + C (see Appendix II).

(b) We have sec xdx =

1 cos

x

dx

=

cos x cos2 x

dx

=

sin x = u

1

cos x - sin2

dx x

=

d sin x = du

=

cos xdx = du

1

1 -

u2

du

=

(1

-

1 u)(1

+

u)

du

=

1 2

1

1 -

u

+

1

1 +

u

du

=

1 2

1

1 -

u

du

+

1 2

1

1 +

u

du

= -1 ln |1 - u| + 1 ln |1 + u| + C = 1 ln

2

2

2

1+u 1-u

+ C = 1 ln 2

1 + sin x 1 - sin x

+C

This can be rewritten as ln |sec x + tan x| + C (see Appendix II).

Solution 3: (a) We have

csc xdx =

csc

x

csc csc

x x

- -

cot cot

x dx x

=

csc2 x csc

- x

csc x cot - cot x

x

dx

= 1 du = ln |u| + C = ln | csc x - cot x| + C u

csc x - cot x = u

= d(csc x - cot x) = du

(- csc x cot x + csc2 x)dx = du

(b) We have

sec x + tan x = u

sec xdx =

sec x sec x + tan xdx = sec x + tan x

sec2 x + sec x tan xdx = d(sec x + tan x) = du

sec x + tan x

(sec x tan x + sec2 x)dx = du

= 1 du = ln |u| + C = ln | sec x + tan x| + C u

EXAMPLE 9: Find tan8 x sec4 xdx.

6

Section 6.2 Trigonometric Integrals and Substitutions

2010 Kiryl Tsishchanka

EXAMPLE 9: Find tan8 x sec4 xdx.

Solution: We have tan8 x sec4 xdx =

= u8(1 + u2)du =

tan x = u

tan8 x sec2 x ? sec2 xdx = tan8 x(1 + tan2 x) sec2 xdx = d tan x = du

sec2 xdx = du

(u8 + u10)du = 1 u9 + 1 u11 + C = 1 tan9 x + 1 tan11 x + C

9 11

9

11

EXAMPLE 10: Find tan3 xdx.

Solution 1: We have tan3 xdx = tan2 x ? tan xdx = (sec2 x - 1) ? tan xdx

sec x = u

(sec2 x - 1) ? sec x tan xdx

=

=

d sec x = du

(u2 - 1)du =

sec x

u

sec x tan xdx = du

=

u

-

1 u

du

=

u2 2

-

ln |u|

+

C

=

1 2

sec2

x

-

ln | sec

x|

+

C

Solution 2: We have tan3 xdx = tan2 x ? tan xdx = (sec2 x - 1) ? tan xdx = (sec2 x tan x - tan x)dx

= sec2 x tan xdx - tan xdx = [by Example 7] = sec2 x tan xdx - ln | sec x|

To evaluate sec2 x tan xdx, we either

sec2 x tan xdx =

sec x = u

sec x ? sec x tan xdx = d sec x = du =

sec x tan xdx = du

udu = u2 + C = 1 sec2 x + C

2

2

or

tan x = u

sec2 x tan xdx = d tan x = du = udu = u2 + C = 1 tan2 x + C

2

2

sec2 xdx = du

Therefore

tan3 xdx = 1 sec2 x - ln | sec x| + C or 1 tan2 x - ln | sec x| + C

2

2

Other possible answers are

tan3 xdx = 1 sec2 x + ln | cos x| + C or 1 tan2 x + ln | cos x| + C

2

2

7

Section 6.2 Trigonometric Integrals and Substitutions

EXAMPLE 11: Find sec3 xdx.

2010 Kiryl Tsishchanka

Solution: We have

sec3 xdx =

sec x = u

sec2 xdx = dv

sec x ? sec2 xdx = d(sec x) = du

tan x = v = sec x tan x -

sec x tan xdx = du

sec x tan2 xdx

= sec x tan x - sec x(sec2 x - 1)dx = sec x tan x - (sec3 x - sec x)dx

= sec x tan x - sec3 xdx + sec xdx

It follows that

2 sec3 xdx = sec x tan x + sec xdx

so

sec3 xdx = 1 sec x tan x + sec xdx = [by Example 8] = 1(sec x tan x + ln | sec x + tan x|) + C

2

2

Case C: Integrals of type sin x sin xdx, sin x cos xdx or cos x cos xdx

METHOD OF INTEGRATION: Use the identities

sin A sin B = 1 [cos(A - B) - cos(A + B)] 2

cos

A

cos

B

=

1 2

[cos(A

+

B)

+

cos(A

-

B)]

sin A cos B

=

1 2

[sin(A

+

B)

+

sin(A

-

B)]

EXAMPLE 12: Find cos 3x cos 2xdx.

8

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