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Национальный авиационный университет

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DDR 2 p.64-131.pdf

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1 x 4

T3. Exercises for class and homework

In Exercises 1 to 9 complete the square to find the integral.


1.				dx			.			2.				dx			.			3.					dx	.
1.			x	2			.			2.	x	2				4	.			3.		x	2	8x 20		.
			x			6x					x	4x				4						x		8x 20
4.			(x 2)dx						.	5.	(2		x 3)dx					.		6.					dx	.
4.			2						.	5.	2							.		6.		2x		2	5x 7	.
			x 4x 3								x		2x 10									2x			5x 7
7.		(3x 4)dx						.		8.				e x dx					.	9.			cos xdx				.
7.	3x			2	x 4			.		8.	e	2x		2e	x	6			.	9.	sin	2					.
	3x				x 4						e			2e		6					sin		x 2 sin x 2

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


10.				dx									.
10.		(x 2)(x 3)											.
		(x 2)(x 3)
12.			x 2 4x							dx .
12.		x	2						6	dx .
		x		5x					6
14.			(2x 3)dx												.
14.		(x 1)			2										.
		(x 1)					(x 3)
16.				dx												.
16.				2							4)					.
		(x 1)(x									4)
18.				1 x3										dx .
18.	(x		2	4x 5)								2		dx .
	(x			4x 5)
20.			dx					.
20.		x	7			5		.
		x		x
22.		(x3 6)dx											.
22.		x	4	6x				2	8				.
		x		6x					8

11.			x3dx				.
11.	(x 2)(x 4)						.
	(x 2)(x 4)
13.	(x2 10x 7)dx									.
13.	(x 1)(x 2)(x 5)									.
	(x 1)(x 2)(x 5)
15.			dx					.
15.	x	4	2x	3	x		2	.
	x		2x		x
17.	1 2x 5x 2 2x								3	dx .
17.		(x 1)		4	(x	2	1)			dx .
		(x 1)			(x		1)

19. x3 x 2 x 1 dx .

21. 1 dxx3 .

23.				xdx						.
	(1	x	2	)	3	(4	x	2	)

Answers

1.		1	ln		x		C .	2		1		C .	3			1	arctg			x 4	C . 4.		1		ln		x2 4x 3		C .	5.
1.		1	ln		x		C .	2		1		C .	3			1	arctg			x 4	C . 4.		1		ln		x2 4x 3		C .	5.
		6		x		6				x	2					2		2				2
	x2 2x 10					1 arctg		x 1	C .			6.	1			x 1			C .		7.	1				(3x 4)8		C .		8.
ln														ln										ln
ln														ln										ln
						3		3					9		2x 7							7					x 1

111

	1			arctg						ex 1						C .				9. arctg(sin x 1) C . 10. ln																						x 3				C . 11.				1	(x 2)2								4		ln		x 2


5												5																														x 2								2					3
+ 32 ln									x 4						C .					12.				x		3	ln		x 1				60 ln					x 6			C .				13.		ln		(x 1)(x 2) /(x 5)																C .
																										3

3																										7									7
3						3							x 3							1						7									7		x 1						1			1	C								1							x +1
14.									ln															C		. 15.								2ln																16.							ln										+

				4									x 1						2(x 1)																			x					x			x 1									2							4 + x2
				4									x 1						2(x 1)																																				2
										x
+			1	arctg						x			+C			.		17.							1						1			arctg x C .									18.			1 ln \| x2				4x 3 \|					9 arctg(x 2)
4										2												3(x				1)3				x 1															2										2
					3x 5												C.			19.			x2		x ln							x 1				arctg x C .									20.			1				1				1		ln				x2				C .
					3x 5																		x2									x 1																1				1				1						x2
		2(x2 4x 3)																						2								1 x2																	4x4			2x2			2					1 x2
		2(x2 4x 3)																						2								1 x2																	4x4			2x2			2					1 x2
21.									1		ln				(x		1)2					1			arctg			2x 1						C . 22.					3	arctg				x		ln	x2 4						3 2	arctg						x 2				C .
21.							6				ln			x2											arctg									C . 22.					2	arctg				2		ln						2		arctg						2				C .
							6							x2		x 1								3						3									2					2			2 x2					2								2
23.				1				ln					x2			1					1										1					C.

														2								2									2				2
				54										2		4			18(x			2		1)			12(x				2			1)	2
				54							x					4			18(x					1)			12(x							1)

T3. Individual test problems

3.1. Complete the square to find the integral.

3.1.1.						dx			.			3.1.2.						dx				.					3.1.3.							dx			.
3.1.1.	4x			2	5x 4				.			3.1.2.	x	2								.					3.1.3.		2x				2	7x 3			.
	4x				5x 4								x				4x 10												2x					7x 3
3.1.4.					dx			.				3.1.5.						dx							.		3.1.6.							dx			.
3.1.4.	2x			2	x		6	.				3.1.5.	9x			2		6x		2					.		3.1.6.		4x				2	4x	3		.
	2x				x		6						9x					6x		2									4x					4x	3
3.1.7.						dx				.		3.1.8.						dx			.						3.1.9.							dx				.
3.1.7.	2x			2	11x 2					.		3.1.8.	2x				2				.						3.1.9.	3x				2						.
	2x				11x 2								2x					x 2										3x						12x 3
3.1.10.					dx		.					3.1.11.						dx					.				3.1.12.							dx							.
3.1.10.		2x			2	3x	.					3.1.11.		x		2		5x					.				3.1.12.				2x			3 4x		2					.
		2x				3x								x				5x		6											2x			3 4x
3.1.13.						dx				.		3.1.14.						dx						.			3.1.15.							dx		.
3.1.13.		3x			2	8x 3				.		3.1.14.		8		2x x				2				.			3.1.15.			5x				6 x	2	.
		3x				8x 3								8		2x x														5x				6 x
3.1.16.						dx				.		3.1.17.						dx								.	3.1.18.							dx								.
3.1.16.		x		2	4x 25					.		3.1.17.			x	2		4x 15								.	3.1.18.				3x			2		6						.
		x			4x 25										x			4x 15													3x			9x		6
3.1.19.						dx					.	3.1.20.					dx		.								3.1.21.							dx					.
3.1.19.		2x			2	2x 5					.	3.1.20.			x		2		.								3.1.21.				x		2	6x					.
		2x				2x 5									x			3x													x			6x	11
3.1.22.						dx			.			3.1.23.						dx								.	3.1.24.							dx						.
3.1.22.		x	2		8x		19		.			3.1.23.			x		2	8x 11								.	3.1.24.				x		2	8x						.
		x			8x		19								x			8x 11													x			8x	21

112

3.1.25.				dx			.		3.1.26.						dx			.
		5x	2	x	2							2x		2	6x		3

3.1.28.				dx				.	3.1.29.						dx		.
		1 2x x				2						x	2	3x		6

3.2. Complete the square to find the integral.
3.2.1.		(x 1)dx			.				2.2.2.		(x		6)dx			.
		2									2		2x 1
	x 3x 4									3x


3.2.4.				xdx		.
3.2.4.	x	2		2x 5		.
	x			2x 5
3.2.7.		(x 4)dx						.
3.2.7.		2						.
	2x			6x		8
3.2.10.				(x 1)dx					.
3.2.10.		x		2					.
		x		4x 10
3.2.13.			(5x 1)dx					.
3.2.13.		2						.
		x 4x 6
3.2.16.			(2x 1)dx				.
3.2.16.		2					.
		x		4x 3
3.2.19.			(2x 1)dx				.
3.2.19.		2					.
		3 x 2x
3.2.22.				(x	3)dx					.
3.2.22.				2	4x 5					.
			4x		4x 5

3.2.25.(x2 2)dx .

xx 2

3.2.28. (22x 1)dx . 5x 2x 1

3.2.5.	(x				5)dx				.
	2
	x				x 2
3.2.8.		(5x 2)dx												.
		2
	2x					5x 2
3.2.11.			(x 4)dx									.
		2
			x 2x 3
3.2.14.							xdx								.
		x		2		2x					4

3.2.17.				(2			x)dx						.
				2
			x 4x 5
3.2.20.		(x				4)dx					.
			3x				2	x

3.2.23.			(2x 3)dx												.
			2
			x 2x 7
3.2.26.			(3x 2)dx													.

			x		2	8x					17

3.2.29.		(x				4)dx				.
		2
		x				x 1

3.1.27.								dx							.
3.1.27.			x		2			6x	8						.
			x					6x	8
3.1.30.								dx							.
3.1.30.			x		2			5x	1						.
			x					5x	1
3.2.3.		(2x 1)dx								.
3.2.3.		2								.
	3x						2x 1
3.2.6.		(3x 2)dx										.
3.2.6.		2										.
	5x						3x 2
3.2.9.		(4x 1)dx									.
3.2.9.		2									.
	4x 4x 5
3.2.12.				(4			x 8)dx						.
3.2.12.				2									.
		x				6x 10
3.2.15.				(x 3)dx						.
3.2.15.				2						.
		x				5x 4
3.2.18.				(2x 1)dx										.
3.2.18.				2										.
		3x						6x 9
3.2.21.		(3x 1)dx							.
3.2.21.		2							.
		x 4x 2
3.2.24.				(x 5)dx								.
3.2.24.		2x				2		x 3				.
		2x						x 3
3.2.27.			(x				7)dx					.
3.2.27.			2				3x 1					.
	4x						3x 1
3.2.30.								xdx							.
3.2.30.		x		2		8x 20									.
		x				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 .

3.3.2.

dx .

3)(x 2)

4x 3)(x 5)

3.3.3.

43x 67

dx .

3.3.4.

2x2 8x 9

dx .

12)(x 1)

x 2)(x 3)

3.3.5.

12x

dx .

3.3.6.

2x 7

dx .

6x 5)(x 3)

6)(x 1)

113

3.3.7.

x2 8x 4

dx .

3.3.8.

5x 17

dx .

5x 6)(x 1)

4x 3)(x 5)

3.3.9.

6x2 6x 6

dx .

3.3.10.

37x 85

dx .

x 2)(x 1)

2x 3)(x

3.3.11.

3x2 3x 24

dx .

3.3.12.

6x2 4x 30

dx .

2x 3)(x

x 2)(x 3)

3.3.13.

3x2 15

dx .

3.3.14.

x2 19x 6

dx .

5x 6)(x

5x 6)(x 1)

3.3.15.

dx .

3.3.16.

4x2 32x 52

dx .

6x 5)(x

3.3.17.

2x2 41x 91

dx .

3.3.18.

dx .

2x 3)(x

2x 3)(x 2)

3.3.19.

dx .

3.3.20.

6x2

dx .

3x 2)(x

8x 15)(x 1)

3.3.21.

6x4

dx .

3.3.22.

2x2 26

dx .

1)(x 2)

4x 3)(x 5)

3.3.23.

2x2 12x 6

dx .

3.3.24.

20x2

dx .

8x 15)(x 1)

2x 3)(x 4)

3.3.25.

x 7

dx .

3.3.26.

6x 21

dx .

5x 6)(x

x 2)(x 1)

2.3.27.

2x4 3

dx .

3.3.28.

7x2 17x

dx .

5x 4)(x

2x 3)(x

3.3.29.

6x4 30x2 30

dx .

3.3.30.

3x2 17x 2

dx .

1)(x 2)

5x 6)(x

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

3.4.1.	x3 1
						dx .
	x	3		x	2	dx .
	x			x
3.4.3.				3x2 1				dx .
3.4.3.	(x		2	1)(x			1)	dx .
	(x			1)(x			1)

3.4.2.	x3		2x 2 2x 1				dx .
3.4.2.			x	3	x	2	dx .
			x		x
3.4.4.		x 2			dx .
3.4.4.	x	3	x	2	dx .
	x		x

114

x3 x2

3.4.5.4x 4 8x3 3x 3 dx .

x3 2x 2 x

3.4.7.2x 2 2x 1 dx .

x2 x3

3.4.9.

3 1

dx .

(x 1)

3.4.11.

2x2 5x 1

dx .

3.4.13.

3x 2 2

dx .

x(x 1)

3.4. 15.

x 2

dx .

3.4.17.

x3 3

dx .

1)(x

3.4.19.

dx .

3.4.21.

x3 4x 5

dx .

1)(x

3.4.23.

x 5

dx .

3.4.25.

dx .

1)(x

3.4.27.

2x2 1

dx .

3.4.29.

3x x2

2 dx .

x(x 1)

3.4.6.		x 2					dx .
3.4.6.	x	3	x		2		dx .
	x		x
3.4.8.						4x2
									dx .
	(x		2						dx .
	(x			2x 1)(x 1)
3.4.10.				1				dx .
3.4.10.		x	3	x		2		dx .
		x		x

3.4.12. x2 x 2 dx .

3.4.14.	4x			4 8x3 1			dx .
	(x		2	x)(x 1)

3.4.16.				1		dx .
	x	3		x	2

3.4.18.

6x 2x2

dx .

3.4.20.

4x 4

8x3

dx .

x(x 1)

3.4.22.

x 2 3x 2

dx .

3.4.24.

3x2 7x 2

dx .

x)(x

3.4.26.

2x3 4x 3

dx .

3.4.28.

2x3

5x2

dx .

3.4.30.

dx .

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 .	3.5.2.			12 6x	dx .
	(x	2	2x 5)(x 1)			(x	2	4x 13)(x 1)

115

3.5.3.	x2					6x 8								dx .								3.5.4.		4x 2													dx .
3.5.3.						x		3					8	dx .								3.5.4.	x	4							4x				2		dx .
						x							8										x								4x
3.5.5.						2x					2 2x 20						dx .					3.5.6.	x2				3x 2													dx .
3.5.5.	(x				2			2x 5)(x 1)									dx .					3.5.6.				x		3				1								dx .
	(x							2x 5)(x 1)																		x						1
3.5.7.	7x 10												dx .									3.5.8.								9(x 1)dx														.
3.5.7.		x		3							8		dx .									3.5.8.	(x			2			4x 13)(x 1)															.
		x									8												(x						4x 13)(x 1)
3.5.9.	4x x2 12														dx .							3.5.10.		3 9x											dx .
3.5.9.						x			3			8			dx .							3.5.10.		x			3			1					dx .
						x						8												x						1
3.5.11.							2x2 2x 20											dx .				3.5.12.					(4x 10)dx															.
3.5.11.			(x				2											dx .				3.5.12.		(x					2													.
			(x							2x 5)(x 1)														(x								2x 10)x
3.5.13.			6 9x										dx .									3.5.14.				(x					2 13x 40)dx														.
3.5.13.			x			3		8					dx .									3.5.14.		(x				2				4x 13)(x											1)		.
			x					8																(x								4x 13)(x											1)
3.5.15.				4x2 x 10												dx .						3.5.16.						6x								dx .
3.5.15.										x		3	8			dx .						3.5.16.				x		3				1				dx .
										x			8													x						1
3.5.17.						(x				2 4x 20)dx								.				3.5.18.				3x2 2x 1															dx .
3.5.17.			(x				2											.				3.5.18.											x	3		1					dx .
			(x							4x 13)(x 1)																							x			1
3.5.19.														8						dx .		3.5.20.								(4x 2							38)dx									.
3.5.19.				(x				2			6x									dx .		3.5.20.			(x					2																.
				(x							6x				13)(x 1)										(x								2x 2)(x										2)
3.5.21.									19x x						2 34					dx .		3.5.22.				2x 2 7x														dx .
3.5.21.				(x				2			4x									dx .		3.5.22.						x			3		8)							dx .
				(x							4x				13)(x 1)													x					8)
3.5.23.													36								dx .	3.5.24.														5x			13									dx .
3.5.23.				(x				2			2x										dx .	3.5.24.		(x					2			6x 13)(x												1)				dx .
				(x							2x				10)(x 2)									(x								6x 13)(x												1)
3.5.25.											x2 5x 40								dx .			3.5.26.								4x2 7x 5																dx .
3.5.25.				(x				2			2x				5)(x		2)		dx .			3.5.26.		(x 1)(x													2									dx .
				(x							2x				5)(x		2)							(x 1)(x																2x 5)
3.5.27.													x2 23						dx .			3.5.28.									4x2 3x 17																dx .
3.5.27.				(x				2			2x				5)(x		1)		dx .			3.5.28.				(x				2			2x														dx .
				(x							2x				5)(x		1)									(x							2x						5)(x 1)
3.5.29.							5x2 17x 36												dx .			3.5.30.													2x			22									dx .
3.5.29.			(x				2			6x 13)(x							1)		dx .			3.5.30.			(x				2				2x					5)(x					1)				dx .
			(x							6x 13)(x							1)								(x								2x					5)(x					1)

116

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 =
(csc x), sin mx, cos nx and so on. Since tg x =					cos x	, ctg x =	sin x	, sec x =
					cos x		sin x
1		and cosec x =	1	, any such product can be expressed in the form
	cos x	and cosec x =	sin x	, any such product can be expressed in the form
	cos x		sin x

sinnx cosmx for some integers n and m, positive, zero, or negative. The methods in this section thus show how to compute sinn xcosm 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 – x and let t = tg 2x .

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

T 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

117

4.2. Integration of sinnx cosmx.

Consider the integral sinn xcosm xdx , where n and m are nonnegative integers.

4.2.1. If m = 1 and n 1 then the integral becomes: sinn xcos 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
1 cos 2 x k	2
1 cos 2 x k

dx and is easier than is given.

4.2.6.If n = 0 and m 2 and m=2k is even positive integer then the given

integral becomes cos2k x dx .

Similarly replace cos2x by 1 cos2x . 2

4.2.7. More generally to find sinn xcosm 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 xcosm xdx , if m is odd, as is illustrated by example 9.

118

If both n and m are even in sinn xcosm 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 tg 2x t

The substitution

t tg

arctg t x 2arctg t dx

2dt

1 t2

thus leads to the formulas:

sin x

cos x

1 t 2

and dx

dt .

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.

119

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

Replace sec2 x by tg2x + 1.

and n even)

sin

n 1

sin

n 2

sinn x

xdx

cos xsin

xdx

cos

n 1

cos

n 2

cosn x

x dx n sin xcos

xdx.

secn x (n 2)

secn 2 x tg x

n 2

sec

n 2

x dx .

secn x dx =

n 1

120

T 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

cos2 xdx 1 cos2 2x dx 12 dx 12 cos 2xdx 12 x 14 sin 2x C.

3.I sin4 2xdx .

Solution. Replace sin2x by 1 cos2x , then
									2							2
				I sin			2		2				1 cos4x			2
				I sin					2x dx					2		dx
														2
		1	1	2 cos 4x cos				2	4x dx			1		2 cos 4x			1 cos8x
		4	1	2 cos 4x cos					4x dx			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 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

sin 8x C.

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

t 3

t 5

C cos x

cos3

x cos5

x C .

tg5 xdx .

121

Solution. Multiplying the integrand we get

tg5

xdx tg x (tg2

x)2 dx tg x (

1)2 dx

tg x

cos

tg x

dx tg xdx

sin x

dx 2 tg xd tg x

sin x dx

cos

cos x

d cos

tg2 x

d cos x

tg2

x 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 13 1 t2 3 dt
	1	1 3t	2	3t	4	t	6	dt	1		3	3	t	5	1	t	6
	3	1 3t		3t		t		dt	3	t t		5	t		6	t		3C
	3								3			5			6

13 cos 3x 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

122

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