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3.2.25. ∫

+ 2)dx

3.2.26. ∫

x − 2)dx

3.2.27. ∫

− 7)dx

− x − 2

+ 8x + 17

+ 3x − 1

3.2.28. ∫

(2x + 1)dx

. 3.2.29. ∫

(x − 4)dx

3.2.30. ∫

xdx

+ 2x + 1

− x + 1

− 8x + 20

3.3. Express the integrant as the sum of partial fractions and determine the resulting integral.

3.3.1. ∫

3x2 + 20x + 9

dx .

+ 4x +

3)(x +

3.3.3. ∫

43x − 67

dx .

− x − 12)(x − 1)

3.3.5. ∫

12x

dx .

+ 6x +

5)(x +

3.3.7. ∫

x2 + 8x − 4

dx .

+ 5x +

6)(x −

3.3.9. ∫

6x2 + 6x − 6

dx .

+ x − 2)(x + 1)

3.3.11. ∫

3x2 + 3x − 24

dx .

− x

−

2)(x −

3.3.13. ∫

3x2 − 15

dx .

+ 5x

+ 6)(x − 1)

3.3.15. ∫

dx .

+ 2x

− x − 2

3.3.17. ∫

2x2 + 41x − 91

dx .

+ 2x

− 3)(x + 2)

3.3.19. ∫

dx .

+ 8x

+ 15)(x +

3.3.21. ∫

6x4

dx .

− 1)(x + 2)

3.3.23. ∫

2x2 + 12x − 6

dx .

+ 8x + 15)(x +

3.3.2. ∫

dx .

− 2x + 3)(x − 2)

3.3.4. ∫

2x2 + 8x + 9

dx .

+ x − 2)(x + 3)

3.3.6. ∫

2x − 7

dx .

− 5x + 6)(x + 1)

3.3.8. ∫

5x + 17

dx .

+ 4x + 3)(x + 5)

3.3.10. ∫

37x − 85

dx .

+ 2x − 3)(x −

3.3.12. ∫

6x2 − 4x + 30

dx .

− 2x − 3)(x −

3.3.14. ∫

x2 − 19x + 6

dx .

+ 5x + 6)(x −

3.3.16. ∫

4x2 + 32x + 52

dx .

+ 6x + 5)(x +

3.3.18. ∫

dx .

+ 2x − 3)(x +

3.3.20. ∫

6x2

dx .

+ 3x + 2)(x −

3.3.22. ∫

2x2 − 26

dx .

+ 4x + 3)(x +

3.3.24. ∫

20x2

dx .

+ 2x − 3)(x −

111

3.3.25. ∫

x − 7

dx .

3.3.26. ∫

6x − 21

dx .

− 5x + 6)(x

+ 1)

+ x − 2)(x + 1)

2.3.27. ∫

2x4 − 3

dx .

3.3.28. ∫

7x2 − 17x

dx .

− 5x + 4)(x

+ 3)

−

2x − 3)(x − 2)

3.3.29. ∫

6x4 − 30x2 + 30

3.3.30. ∫

3x2 − 17x + 2

dx .

− 1)(x + 2)

5x + 6)(x − 1)

3.4. Express the integrant as the sum of partial fractions and determine the resulting integral.

3.4.1. ∫

x3 + 1

dx .

3.4.2. ∫

− 2x 2 − 2x + 1

dx .

− x

3.4.3. ∫

3x2 + 1

dx .

3.4.4. ∫

x + 2

dx .

− 1)(x −

− x

3.4.5. ∫

4x 4

+ 8x3 − 3x − 3

dx .

3.4.6. ∫

x + 2

dx .

+ 2x

+ x

3.4.7. ∫

2x 2

− 2x − 1

dx .

3.4.8. ∫

dx .

− x

−

2x + 1)(x + 1)

3.4.9. ∫

2x3 + 1

dx .

3.4.10. ∫

dx .

+ x

(x + 1)

3.4.11. ∫

2x2 − 5x + 1

dx .

3.4.12. ∫

x2 + x + 2

dx .

− 2x

+ x

3.4.13. ∫

3x 2 + 2

dx .

3.4.14. ∫

4x4 + 8x3 − 1

dx .

x(x + 1)

+ x)(x + 1)

3.4. 15. ∫

x + 2

dx .

3.4.16. ∫

dx .

− 2x

+ x

− x

3.4.17. ∫

x3 − 3

dx .

3.4.18. ∫

6x − 2x2 − 1

dx .

− 1)(x − 1)

−

+ x

3.4.19. ∫

x3 − 4x 2 − 1

dx .

3.4.20. ∫

4x 4 + 8x3 − 2

dx .

− 2x

+ x

x(x + 1)

3.4.21. ∫

x3 − 4x + 5

dx .

3.4.22. ∫

x 2 − 3x + 2

dx .

− 1)(x − 1)

+ x

112

3.4.23. ∫

x +

dx .

3.4.24. ∫

3x2 − 7x + 2

dx .

− x

− x + 1

− x)(x − 1)

3.4.25. ∫

dx .

3.4.26. ∫

2x3 + 4x + 3

dx .

+ x

− 1)(x + 1)

3.4.27. ∫

2x2 + 1

dx .

3.4.28. ∫

2x3

+ 5x2 − 1

dx .

− 2x

+ x

3x − x

−

+ 2x

3.4.29. ∫

dx .

3.4.30. ∫

+ 1

dx .

x(x − 1)

x(x + 1)

3.5. Express the integrant as the sum of partial fractions and determine the resulting integral.

3.5.1. ∫			3x + 13	dx .
	(x	2	+ 2x + 5)(x − 1)

3.5.3.∫ x2 − 6x + 8 dx .

x3 + 8

3.5.5. ∫		2x2 + 2x + 20		dx .
	(x	2	+ 2x + 5)(x − 1)

3.5.7.∫ 7x − 10 dx .

x3 + 8

3.5.9.∫ 4x − x2 − 12 dx .

x3 + 8

3.5.11. ∫		2x2 + 2x + 20		dx .
	(x	2	+ 2x + 5)(x − 1)

3.5.13.∫ 6 − 9x dx .

x3 + 8

3.5.15.∫ 4x + x + 10 dx .

x3 + 82

3.5.17. ∫		(x 2 + 4x + 20)dx				.
	2
	(x			− 4x + 13)(x + 1)
3.5.19. ∫				8			dx .
		(x	2	+ 6x +
					13)(x + 1)

3.5.2. ∫					12 − 6x
		(x		2	− 4x	+ 13)(x + 1)

3.5.4. ∫			4x + 2			dx .
	x		4 2
				+ 4x

3.5.6.∫ x2 + 3x + 2 dx .

x3 − 1

9(x − 1)dx 3.5.8. ∫ (x 2 − 4x + 13)(x + 1) .

3.5.10.∫ 3 − 9x dx .

x3 − 1


3.5.12. ∫			(4x − 10)dx			.
			2
	(x			− 2x + 10)x
3.5.14. ∫		(x		2 − 13x + 40)dx
	2			− 4x + 13)(x + 1)
	(x
3.5.16. ∫			6x		dx .
		x	3	− 1

3.5.18.∫ 3x + 2x + 1 dx .

x3 − 12

3.5.20. ∫		(4x 2 + 38)dx
3.5.20. ∫	(x	2
	(x	− 2x + 2)(x + 2)

dx .

113

3.5.21. ∫

19x − x

2 − 34

dx .

3.5.22. ∫

2x 2 + 7x

dx .

− 4x +

− 8)

13)(x + 1)

3.5.23. ∫

dx .

3.5.24. ∫

5x + 13

dx .

− 2x +

10)(x + 2)

+ 6x + 13)(x + 1)

3.5.25. ∫

x2 − 5x + 40

dx .

3.5.26. ∫

4x2 + 7x + 5

dx .

− 2x +

5)(x + 2)

−

1)(x

+ 2x +

3.5.27. ∫

x2 + 23

dx .

3.5.28. ∫

4x2 + 3x + 17

dx .

+ 2x +

5)(x + 1)

+ 2x + 5)(x − 1)

3.5.29. ∫

5x2 + 17x + 36

dx .

3.5.30. ∫

2x + 22

dx .

+ 6x + 13)(x + 1)

− 2x + 5)(x −

Topic 4. Integrals involving powers of trigonometric functions

This section shows how to integrate certain products of powers of the six trigonometric functions, sin x, cos x, tg x (tan x), ctg x (cot x), sec x, cosec x (csc x),

sin mx, cos nx and so on. Since tg x =	sin x	, ctg x =	cos x	, sec x =	1	and
	cos x		sin x		cos x

cosec x = sin1 x , any such product can be expressed in the form sinnx cosmx for some integers n and m, positive, zero, or negative. The methods in this section

thus show how to compute ∫ sinn x cosm x dx for certain convenient

combinations of n and m.

The technique described in this section — a particular substitution — reduces the integration of any rational function R of sin x and cos x such as

∫ R(sin x, cos x)dx to the integration of a rational function of t. The latter can be accomplished by partial fractions. The method depends on the fact that sin x and cos x can both be expressed as rational functions of tg 2x .

Consider – π < x < π and let t = tg 2x .

Literature: [1, section 6, ch. 6.5], [2, section 2, ch. 2.1], [4, section 7, § 22], [6, section 8], [7, section 10, § 12], [9, § 32].

114

Т.4	Main concepts

4.1. Integration of sin mx cos nx, sin mx sin nx and cos mx cos nx

In this case are given the formulas:

sin mx cos nx =		1	[sin(m + n)x + sin(m − n)x] ,
	2

sin mx sin nx =	1		[cos(m − n)x − cos(m + n)x] ,
	2

cos mx cos nx =		1	[cos(m − n)x + cos(m + n)x] .
	2

4.2. Integration of sinnx cosmx

Consider the integral ∫ sinn x cosm xdx , where n and m are nonnegative integers.

4.2.1.If m = 1 and n ≥ 1 then the integral becomes: ∫ sinn x cos xdx . The substitution t = sin x turns this integral into the easy integral ∫tndt.

4.2.2.If n = 1 and m ≥ 1 then the integral becomes: ∫ cosm xsin xdx .

The substitution t = cos x turns this integral into the easy integral −∫tndt. 4.2.3. If m = 0 and n ≥ 3 and n = 2k + 1 is odd positive integer then the given

integral becomes ∫sinn x dx .

Recall that d(cos x) = –sin x dx. Thus

∫sin2k +1x dx = ∫sin2k x sin xdx = −∫(1− cos 2 x)k d (cos x) . This integral is easier than is given.

4.2.4.Similarly if n = 0 and m ≥ 3 and m = 2k + 1 is odd positive integer:

∫cos2k +1x dx = ∫cos2k x cos xdx = ∫(1− sin 2 x)k d (sin x) .

4.2.5.If m = 0 and n ≥ 2 and n = 2k is even positive integer then the given integral becomes ∫sin2k x dx .

Replace sin2x by	1− cos2x	. Then the given integral becomes
	2

	1− cos 2 x k
∫		dx and is easier than
	2

4.2.6. If n = 0 and m ≥ 2 and m integral becomes ∫cos2k x dx .

is given.

= 2k is even positive integer then the given

115

Similarly replace cos2x by 1+ cos2x . 2

4.2.7. More generally to find ∫ sinn x cosm xdx , where m and n are non-

negative integers and n is odd, pair one sin x with dx to form sin x = − d(cosx) and use the identity sin2x = 1 − cos2x together with the substitution t = cosx. The new integrand will be a polynomial in t. A similar approach works on

∫ sinn x cosm xdx , if m is odd, as is illustrated by example 9.

If both n and m are even in ∫ sinn x cosm xdx the method of example 9 does

not apply. It	then helps	to use the identities sin2 x =	1− cos 2x	and
cos2 x = 1+ cos 2x .			2
cos2 x = 1+ cos 2x .
2
4.3. Integration of tgmx secnx
Recall that	d(tg x) = sec2	x dx and d(sec x) = sec x tg x dx. These formulas

facilitate the computation of ∫tgmx secnx dx m and n nonnegative integers, when

m is odd or n is even. When m is odd , form sec x tg x dx , when n is even, form sec2x dx.

4.4. Description of the method

The method depends on the fact that cos x and sin x can both be expressed

as rational functions of

= t

The substitution t = tg

= arctg t x = 2arctg t

dx =

2dt

thus leads to the formulas:

1+ t2

sin x =

cos x =

1− t 2

and

dx =

dt .

+ t 2

1+ t 2

+ t 2

This substitution transforms any integral of a rational function of cos x and sin x into an integral of a rational function of t. The resulting rational function can then be integrated by the method of partial fractions.

This method is used to integrate a function such as

∫ dx . a cos x + b sin x + c

Keep in mind that the substitution t = tg 2x is called upon only when easier ways, such as those in the preceding section, don’t work.

116

Summary of this section

The following table summarizes the techniques discussed and similar ones for other powers of trigonometric functions.

Table 2.2

№

Integrand

Technique

sin2x

Write sin2x as

1− cos2x

cos2x

Write cos2x as

1+ cos2x

sinnx (n odd)

Write sinnx dx = sinn - 1x(sin x dx) and

use t = cos x; hence 1 – t2 = sin2x

cosmx sinnx

Write cosmx sinnx dx = cosmx sinn - 1x(sin x dx) and

(n odd)

use t = cos x; hence 1 – t2 = sin2x

cosmx sinnx

Write cosmx sinnx dx = cosm - 1 x sinnx (cos x dx) and

(m odd)

use t = sin x; hence 1 – t2 = cos2x

cosmx sinnx

Replace cos2x by

1+ cos2x

and sin2x by

1− cos2x

(m and n

positive even

integers)

tgmx secnx

Write tgmx secnx dx as tgmx secn -2 x (sec2x dx) and

(n ≥ 2 even)

use t = tg x; hence 1 + t2 = sec2x

tgmx secnx

Write tgmx secnx dx as tgm - 1x secn - 1 x (tg x sec x dx)

(m odd)

and use t = sec x; hence t2 – 1 = tg2x.

tgnx (n ≥ 2)

Write tgnx= tgn - 2x tg2x = tgn – 2x sec2 x - tgn – 2x and

repeat

sec x

∫sec x dx = ln |sec x + tg x| + C

ctgmx cosecnx

Write ctgmx cosecnx as ctgmx cosecn - 2x (cosec2 x dx)

(n ≥ 2 even)

and use t = ctg x; hence 1 + t2 = cosec2x

ctgmx cosecnx

Write ctgmx cosecnx as ctgm - 1x cosecn - 1x (сtg x

(m odd)

cosec x dx) and use t = cosec x; hence t2 - 1= ctg2x

tgmx secnx(m

x + 1

and n even)

Replace sec x by tg

∫sin

n−1

n − 1

∫sin

n−2

xdx

sinn x

xdx

= − n cos xsin

x +

∫cos

n−1

n −1

∫ cos

n−2

cosn x

x dx =

sin x cos

x +

xdx

secn−2 x tg x

n − 2

∫sec

n−2

secn x (n ≥ 2)

∫secn x dx =

n − 1

x dx

n − 1

117

Т.4	Typical problems

Find the integrals 1. ∫sin 2x cos 6xdx .

Solution. Since sin 2x cos 6x = 12 (sin 8x − sin 4x) , then

∫sin 2x cos 6xdx = 12 ∫(sin 8x − sin 4x)dx = − 161 cos 8x + 18 cos 4x + C .

2.∫cos2 xdx .

Solution. Replace cos2x by													1+ cos2x , then
													2
		∫cos			2	xdx = ∫	1+ cos 2x						1							1							1				1
		∫cos				xdx = ∫		2		dx = ∫ 2 dx + ∫ 2 cos 2xdx =																	2 x +				4 sin 2x + C.
3.			I = ∫sin4 2xdx .										1− cos2x , then
Solution. Replace sin2x by													1− cos2x , then
													2
							I = ∫(sin				2		2			= ∫			1 − cos4x							2
							I = ∫(sin						2x) dx			= ∫							2			dx =
						∫(1− 2 cos 4x + cos							4x)dx =										2
			=	1								2						1	∫				− 2 cos 4x +				1+ cos8x
			=	4														4	∫		1		− 2 cos 4x +							2		dx =
				4														4												2
	1		3					1								1			3					1					1		1
=			∫			− 2 cos 4x +			cos8x dx =													x	− 2		sin 4x +							sin 8x		+ C=
=	4		∫	2		− 2 cos 4x +		2	cos8x dx =							4			2			x	− 2	4	sin 4x +				2		8	sin 8x		+ C=
	4			2				2								4			2					4					2		8
								=	3 x − 1 sin 4x +											1			sin 8x + C.
								=	3 x − 1 sin 4x +										64				sin 8x + C.
									8				8						64
4.			∫sin 5 xdx .
Solution. ∫sin 5 xdx = ∫sin 4 x sin xdx = ∫(1− cos2 x)2 sin xdx =
						= ∫(1− cos2 x)2 d(− cos x) =															cos x = t				= −∫(1− t 2 )2 dt =


	= −∫(1− 2t 2 + t 4 )dt = − t +													2	t 3 −		t 5		+ C = − cos x +								2	cos			3 x −		cos5 x		+ C .
	= −∫(1− 2t 2 + t 4 )dt = − t +														t 3 −				+ C = − cos x +								3	cos			3 x −				+ C .
													3			5											3					5
5.			∫ tg5 xdx .
Solution. Multiplying the integrand we get
																118

∫ tg

xdx

∫ tg x

(tg

dx = ∫ tg x (

− 1)

= ∫

tg x

−

cos

− 2 ∫

tg x

dx +∫ tg xdx = ∫

sin x

−2∫ tg xd tg x

+ ∫

sin x

dx =

cos

cos x

= −∫

d cos x

− tg

− ∫

d cos x

− tg

− ln

cos x

+ C .

cos x

cos

4 cos

6. ∫sin7 3xdx.

Solution.

Recall that d(cos 3x) = − 3sin 3x dx. Thus:

∫sin7 3xdx = ∫sin6 3x sin 3xdx = − 13 ∫(1− cos2 3x)3 d (cos3x).


Let t = cos3x, we obtain then							∫sinn 3xdx = −					1		∫(1− t2 )3 dt =
Let t = cos3x, we obtain then							∫sinn 3xdx = −					3		∫(1− t2 )3 dt =
	1	∫(1− 3t	2		4			6	)dt = −	1			3		3		5		1		6
= −				+ 3t		− t					t − t			+		t		−		t		−	3C	=
= −	3			+ 3t		− t				3	t − t			+	5	t		−	6	t		−	3C	=
	3									3					5				6

=− 13 cos3x + 13 cos3 3x − 15 cos5 3x + 181 cos6 3x + C.

7.∫sin5 x cos xdx .

Solution.

The substitution sin x = t transforms the given integral into

∫ sin5 x cos xdx = ∫ sin5 xd(sin x) = ∫ t5 dt =	t6	+ C =	sin6 x	+ C .

6		6

8. I = ∫sin4 xcos5 xdx.

Solution.

Let’s represent the product of cos x and dx as cos x dx = d(sin x), which suggests the substitution t = sin x:

I = ∫sin4 xcos4 xd(sin x) =

= ∫sin4 x(1 − sin2 x)2 d(sin x) = ∫t 4 (1 − t 2 )2 dt = ∫t 4 (1 − 2t 2 + t 4 )dt =

∫(t 4 − 2t6 + t8 )dt =	t5	−	2t7	+	t9	+ C = sin5 x	−	2sin7 x	+ sin9 x	+ C.
			7					7
5				9		5			9

9. I = ∫sin4 xcos2 xdx.

119

Solution. I = ∫(sin

xdx = ∫

1 − cos2x 2

1 + cos2x

x) cos

1 ∫ (1− 2 cos 2x + cos2

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

∫ (1− cos 2x

2x)dx =

sin 4x

sin3

− cos

2x + cos

−

+ C.

10. Find ∫ tgm x dx.

Solution. Keep in mind that tg′ x = sec2x

and that tg2x = sec2x − 1. The steps

are few: ∫ tgm x dx = ∫ tgm−2 x tg2 xdx = ∫ tgm−2 x (sec2 x − 1)dx =

= ∫ tgm−2 x sec2 xdx − ∫ tgm−2 xdx = t = tg x =∫tm−2dt − ∫ tgm−2 xdx =

=tgm−1 x − ∫ tgm−2 xdx. m − 1

Repeated application of this recursion eventually produces ∫ tg xdx or

∫ tg0 xdx. Both are easily computed: ∫ tg xdx = − ln

cos x

+ C;

∫ tg0 xdx = x + C.

11. Obtain a recursion formula for ∫secn xdx.

Solution. Write secnx as secn –2x sec2x and use integration by parts:

∫secn−2 x sec2 xdx =

u = secn−2

x du =

(

n − 2

)

secn−2

x tg xdx

dv = sec

xdx v = tg x

= secn−2 tg x − ∫ tg x (n − 2)secn−2 x tg xdx = secn−2 x tg x −

− (n − 2)∫secn−2 x(sec2 x − 1)dx = secn−2 x tg x − (n − 2)∫secn xdx +

+ (n − 2)∫secn−2 xdx. Collecting ∫secn x dx, we obtain

(n − 1)∫secn xdx = secn−2 x tg x + (n − 2)∫secn−2 xdx and therefore

∫secn xdx =

secn−2 x tg x

n − 2

∫secn−2 dx .

n − 1

12. ∫sec xdx.

cos x dx

Solution. ∫sec xdx = ∫

= ∫

cos x

cos

1 − sin

t = sin x;

= ∫

= 1 ln

1+ t

+ C.

dt = cos x dx.

− t

1− t

120

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