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13. ∫32 sin 6 x cos4 xdx .

Solution. Multiplying the integrand we get

32 sin 6 x cos4 x = 2(2 sin x cos x)4 sin 2 x = sin 4 2x(1− cos 2x) = = sin 4 2x − sin 4 2x cos 2x = 14 (1− cos 4x)2 − sin 4 2x cos 2x =

=14 (1− 2 cos 4x + cos2 4x) − sin 4 2x cos 2x =

=14 − 12 cos 4x + 18 (1+ cos 8x) − sin 4 2x cos 2x .

Then

∫32 sin 6 x cos4 xdx =

∫ dx −

∫ cos 4xdx +

∫ dx +

∫cos 8xdx −

− ∫sin 4 2x cos 2xdx =

−

∫cos 4xd (4x) +

∫cos 8xd(8x) −

−

∫sin 4 2xd(sin 2x) =

−

sin 4x +

sin 8x −

sin 5

2x + C .

14. ∫

(sin x + cos x)

Solution. Integrand

R(sin x, cos x) =

is even for both cos x

(sin x + cos x)2

and sin x :

R(− sin x,− cos x) =

= R(sin x, cos x) .

(− sin x − cos x)2

(sin x + cos x)2

Multiplying the integrand we get

∫

= ∫

d(tg x + 1)

= −

+ C .

(sin x + cos x)

(tg x + 1)

cos

(tg x + 1)

tg x + 1

15. ∫

cos3 xdx

(1− sin x)

Solution. Integrand

R(sin x, cos x) =

cos3 x

is odd for cos x :

(1− sin x)3

R(sin x,− cos x) =

(− cos x)3

= −

cos3 x

= −R(sin x, cos x) .

− sin x)3

(1− sin x)3

121

Then

cos

xdx

sin x = t,

1+ t

∫

cos xdx = dt,

= ∫

1− t

dt =

∫

dt =

(1− t)

(1− sin x)

cos2 x = 1− t 2

= ∫

(t − 1) +

dt = ∫

+ 2∫

= ln

t − 1

−

+ C =

t −

(t − 1)

t − 1

(t − 1)

sin x − 1

+ C.

= ln

−

sin x − 1

16. ∫ tg3 x sec5 x dx.		and that tgx = sec2x − 1.
Solution. Recall d(secx) = tgx secx dx
Let t = secx and dt = secx tgx dx. Then ∫ tg3 x sec5 x dx =
∫ tg2 x sec4 x (tg x sec x)dx = ∫(sec2 x − 1)sec4 x (tg x sec x)dx =
= ∫ (t 2 − 1)t 4dt =∫ (t6 − t4 )dt =	t7	−	t5	+ C = sec7 x	− sec5 x	+ C .

7		5		7	5

17. I = ∫ tg2 x sec4 xdx.

Solution. Let us recall that dtgx = sec2x dx. So pair sec2x with dx to form

sec2x dx. This suggests the substitution

t = tgx; dt = sec2x dx. Recall also that

sec2x = tg2x + 1.

Then I = ∫ tg2 x sec2 x sec2 xdx = ∫t2 (t2 + 1)dt =

= ∫(t4 + t2 )dt =

+ C =

tg5 x

tg3 x

+ C.

18. ∫

sin

x cos x

Solution. Integrand R(sin x, cos x) =

is even for both sin x

sin 3

x cos x

and cos x :

R(− sin x,− cos x) =

= R(sin x, cos x) .

(− sin x)3

sin 3

(− cos x)

x cos x

Multiplying the integrand as function of tg x: ∫ R(tg x)d tg x . This time try

dt =	dx	and 1+ tg2 x =	1		. Consequently,
	cos2 x		cos2
				x
				122

∫

= ∫

(1+ tg2

x)d tg x

sin

x cos

x cos x

= ∫

d tg

+ ∫

d tg x

= −

ctg2 x + ln

tg x

+ C .

tg x

19. I = ∫

sin 2xdx

cos

+ sin

= t transform the integral to

Solution. The substitutions tg x

t = tg x;cos4 x =

;sin4

x =

;

I =

(1+ t

)

(1+ t

)

= ∫

1+ t

+ t

sin 2x =

;dx =

(1+ t2 )2

+ (1+ t2 )2

1+ t2

= ∫

2tdt

u = t2

= ∫

= arctg u + C = arctg(tg2 x) + C.

du = 2tdt

1+ u

1+ t

20. ∫

5 − 3cos x

Solution. The substitution

= t

transforms the integral into

∫

= ∫

2dt

= 2∫

5 − 3 cos x

− t2

(5 + 5t2 − 3 + 3t2 )

(1+ t

)

− 3

1+ t

= ∫

arctg(2t) + C =

arctg(2 tg

) + C .

+ 4t

2 − sin x

21. ∫

dx .

2 + cos x

Solution. Let

= t .

Then

x = 2 arctg t , dx =

, sin x =

;

cos x =

1− t 2

+ t 2

1+ t 2

Consequently,

1+ t 2

−

t 2 − t + 1

2 − sin x

2dt

∫

= ∫

1+ t

4∫

dt =

2 + cos x

1)(t

+ 3)

2 +

1− t

1+ t

1+ t 2

123

= 4∫

dt − 2∫

dt 2

arctg

−

+ 3

+ 1)(t

+ 3)

−∫

(t2

+ 3) − (t2 + 1)

arctg

− ∫

d(t 2 + 1)

∫

d(t 2 + 3)

+ 1)(t

+ 3)

+ 1

+ 3

arctg

− ln(t 2 + 1)

+ ln(t 2 + 3) + C .

Replacing t by

. Thus

∫

2 − sin x

dx =

arctg(

)

+ ln(2

+ cos x)

+ C .

2 + cos x

22. ∫

sin x

dx .

1+ sin x

Solution. We can use the substitution tg

= t , however there is another

way: by transforming the integrand into

sin x

sin x(1

− sin x)

sin x(1

− sin x)

sin x

−

sin 2

1+ sin x

+ sin x)(1− sin x)

1− sin 2

cos2 x

cos2

sin x

−

+ 1.

cos2

Then

cos2 x

sin x

∫

= ∫

dx −

∫

+ ∫ dx =

1+ sin x

cos

= −∫

d cos x

− tg x + x + C =

− tg x + x + C .

cos

cos x

23. ∫ sin 3 xdx .

3 cos4 x

Solution. Write the integral in the form:

∫	sin2	x sin xdx	= ∫	(1−cos2 x) sin xdx	.
	3 cos4 x			3 cos4 x

Since sin xdx =−d(cos x) , we could have used the substitution cos x = t . Thus

sin3 xdx

(1− t2 )dt

t2 dt

−

∫

= −∫

=∫

− ∫

=∫ t

dt − ∫ t

dt =

3 cos4 x

3 t4

124

+ 3t−

+ C =

cos

x + 3cos−

x + C.

t 3

24. ∫

cos x − sin x

dx .

sin x + cos x

Solution. Since cos x − sin x = (sin x + cos x)′ , namely

∫

cos x − sin x

dx = ∫

d(sin x + cos x)

= ln | sin x + cos x | +C .

sin x + cos x

Exercises for class and homework

Т.4

Find the integrals of trigonometric functions.

1. ∫sin 3 x sin 2xdx .

2. ∫ cos4 xdx .

3. ∫sin 3 xdx .

4. ∫ tg4 xdx .

5. ∫ ctg6 xdx .

6. ∫sin 3 x cos4 xdx .

7. ∫sin 6x sin 4xdx .

8. ∫cos 9x sin 5xdx .

9. ∫sin 6x cos2 xdx .

10. ∫

11. ∫

sin x + 2 cos 2x

12. ∫

sin x

dx .

1− ctg x

cos x − sin 2x

sin x + cos x

13. ∫

14. ∫

15. ∫

cos2

xdx

4 + 3 tg x

8 − 4 sin x + 7 cos x

sin

16. ∫

17. ∫ ctg3 xdx .

18. ∫

sin

x cos

sin

x cos x

19. ∫sin x sin 4x cos 7xdx .

20. ∫

cos2

xdx

21. ∫

sin

sin x + cos x

22. ∫

23. ∫

24. ∫

+ sin

4 sin

3 x cos5 x

+ tg x +

4 ctg x

Answers

sin5 x + C .

x +

sin 2x +

sin 4x + C . 3. − cos x +

cos3 x + C . 4.

− tg x +

tg3 x + C .

5. −

ctg5 x +

ctg3 x − ctg x − x + C .

6. −

cos5 x +

cos7 x + C .

125

7.		1		sin 2x −											1			sin10x + C .												8.	1		cos 4x −												1					cos14x + C .							9. −					1		cos 6x −						1		cos8x −
7.				sin 2x −										20				sin10x + C .												8.	8		cos 4x −												28					cos14x + C .							9. −					12		cos 6x −						32		cos8x −
	4													20																	8														28																	12								32
−	1			cos 4x + C .																								10.				x		+		1		ln				cos x − sin x													+ C .											11. C − ln								cos x − sin 2x									.
	1																															x				1
	16																														2					2
	16																														2					2																																							tg			x		− 5
			x					1																												4									3

12.							−				ln		cos x + sin x														+ C .			13.								x			+										ln	4cos x + 3sin x									+ C .									14. ln							2					+ C .
																																																																													2
			2						2																										25											25																													tg			x		− 3
			2						2																										25											25																																x
																																																																													2
15.			−				1	ctg3 x −									1				ctg5 x + C . 16.													sin x									−				2					−		1		+		5		ln			tg(		π		+	x
							1										1																	sin x													2							1				5							π			x	)	+ C .
																																2cos2 x															sin x										2											2	)	+ C .
					3										5																	2cos2 x															sin x							3sin3 x			2						4					2
17. −						1		ctg2 x − ln														sin x				+ C .								18. −									1				ctg4 x − ctg2 x − ln												ctg x					+ C .								19.					1				sin10x +
17. −						1		ctg2 x − ln														sin x				+ C .								18. −									1				ctg4 x − ctg2 x − ln												ctg x					+ C .								19.					1				sin10x +
					2																																				4																																				40
					2																																				4																																				40
+	1			sin10x +											1			sin 4x −									1		sin12x −						1		sin 2x + C .																							20. −						cos x					−	1	ln			tg			x		+ C .
+	1			sin10x +											1			sin 4x −									1		sin12x −						1		sin 2x + C .																							20. −						cos x					−	1	ln			tg			x		+ C .
40														16													48								8																															2sin2 x						2					2
21.							1									x			+				π	)	+ C .						22.										1							acrtg(						2 tg x) + C .														23. 4 4 tg x + C .
							1			ln			tg(			x							π																		1
										ln			tg(										8																		2
								2							2								8																		2
24.					4			x −				3			ln					tg x + 2							+			2						−				3				ln					cos x				+ C .
24.					4			x −				3			ln					tg x + 2							+			2						−				3				ln					cos x				+ C .
				25				25																				5(tg x + 2)											25
				25				25																				5(tg x + 2)											25

		Individual test problems
	Т.4	Individual test problems

4.1. Find the integrals of trigonometric functions.
4.1.1. а) ∫cos4 3x sin 2 3xdx ; б) ∫				tg xdx			;	в) ∫cos 3x cos xdx .
4.1.1. а) ∫cos4 3x sin 2 3xdx ; б) ∫			1	+ 4 tg	2		;	в) ∫cos 3x cos xdx .
			1	+ 4 tg		x

4.1.2. а) ∫cos3 x5 sin 4 xdx ;

4.1.3. а) ∫cos3 x sin8 xdx ;

4.1.4. а) ∫ cos3 xdx ;

3 sin 4 x

4.1.5. а) ∫ sin3 xdx ;

3 cos4 x

б) ∫		tg 2xdx			;
	4	+ tg	2	2x

б) ∫ sin3 x cos7 x

б) ∫ 1+16 sin2 x ;

б) ∫ cos2 xdx ; sin8 x

в) ∫sin x cos 4xdx .

; в) ∫cos 7x cos 5xdx .

в) ∫sin 5x cos 3xdx .

в) ∫sin 3x sin 2xdx .

126

4.1.6. а) ∫ cos3 xdx ;

3 sin 2 x

4.1.7. а) ∫ sin 3 xdx ;

3 cos4 x

4.1.8. а) ∫ 3sin 3 xdx ; cos4 x

4.1.9. а) ∫ sin 3 xdx ;

5 cos2 x

4.1.10. а) ∫cos4 x sin 3 xdx ;

4.1.11. а) ∫8 cos5 x sin3 xdx ;

4.1.12. а) ∫cos4 x sin 2 xdx ;

4.1.13. а) ∫3 cos2 x sin 3 xdx

4.1.14. а) ∫3 sin 2 x cos3 xdx ;

4.1.15. а) ∫cos4 x sin 5 xdx ;

4.1.16. а) ∫cos4 x sin 6 xdx ;

4.1.17. а) ∫cos2 2x sin 4 2xdx ;

4.1.18. а) ∫cos3 x sin3 xdx ;

4.1.19. а) ∫ cos3 xdx ;

5 sin3 x

4.1.20. а) ∫ sin3 xdx ;

3 cos2 x

б)

∫

sin 2xdx

;

в) ∫sin 3x cos xdx .

sin

x + cos

б) ∫

;

в) ∫sin 9x cos xdx .

7 + 8sin x + 6 cos x

б) ∫

tg xdx

;

в) ∫cos 2x cos 3xdx .

−

4 tg

б) ∫

(3 − sin x)dx

;

в) ∫sin 5x sin 7xdx .

3 + cos x

б) ∫

;

в) ∫sin 4x cos 2xdx .

+ cos

б) ∫

;

в) ∫sin x cos 9xdx .

+ 5 cos

б) ∫

cos2 xdx

;

в) ∫sin 3x cos 2xdx .

sin

;

б) ∫

;

в) ∫cos 5x cos xdx .

1+ 4 tg x

б) ∫

;

в) ∫ cos 4x cos 3xdx .

5 + 4 cos x

б) ∫ sin−5 x cos−5 xdx ;

в) ∫ sin 3x sin 8xdx .

б) ∫

(2 − 3sin x)dx

;

в) ∫ sin 6x sin 5xdx .

3 + 2 cos x

б) ∫

;

в) ∫ cos 3x cos 7xdx .

sin

x cos

б) ∫

;

в) ∫ cos 4x cos 9xdx .

− sin

б) ∫

; в) ∫ sin 7x cos 2xdx .

+ 4 sin x + 3cos x

б) ∫

;

в) ∫ sin 4x sin12xdx .

sin

x cos

127

4.1.21. а) ∫ cos3 2xdx ;

3 sin 2 2x

4.1.22. а) ∫ sin 3 2xdx ;

3 cos2 2x

4.1.23. а) ∫ sin 3 xdx ;

3 cos4 x

4.1.24. а) ∫ 3cos3 xdx ; sin 4 x

4.1.25.а) ∫cos2 x sin 4 xdx ;

4.1.26.а) ∫5 cos3 x sin5 xdx ;

4.1.27.а) ∫cos5 x sin 4 xdx ;

4.1.28.а) ∫cos2 3x sin 4 3xdx ;

б) ∫

(3 + sin x)dx

; в) ∫ sin 8x cos 2xdx .

4 + 3cos x

б) ∫

; в) ∫ sin x sin11xdx .

+ 8sin x + 5 cos x

б) ∫

;

в) ∫ sin12x cos 6xdx .

16 + tg x

б) ∫

; в) ∫ cos11x cos 2xdx .

sin

x cos

б) ∫

; в) ∫ sin 8x sin 3xdx .

sin

x cos

б) ∫

;

в) ∫ cos 5x cos 8xdx .

25 − cos

б) ∫

; в) ∫ sin 6x cos 4xdx .

sin

x cos

б) ∫

(2 + sin x)dx

;

в) ∫ cos 4x cos13xdx .

4 + cos x

4.1.29. а) ∫cos3 x sin 4 xdx ; б) ∫				dx				; в) ∫ sin13x cos 3xdx .
		2 + 4 sin x + cos x
4.1.30. а) ∫sin3 x cos8 xdx ;	б) ∫			dx			;	в) ∫ sin16x sin 2xdx .
			sin	9	5	x
				x cos
Topic 5. Integration of irrational functions
A rational function of x and	a − x2 , a + x2						,	x2 − a , n ax + b and so

on can be integrated by using the algebraic and trigonometric substitutions. Integration of binomial differentials. Euler substitutions. Method by Ostrogradskiy.

Literature: [1, section 6, ch. 6.6], [2, section 2, ch. 2.1], [3, гл. 7, § 1], [4, section 7, § 22], [6, section 8], [7, section 10, §§10—11], [9, 1 part, § 33].

128

	Main concepts
Т.5	Main concepts

5.1. Integration of the quadratic irrationality

Consider the integrals

∫	dx	, ∫	Ax + B	dx , ( a ≠ 0 , D ≠ 0 ).
	ax 2 + bx + c		ax 2 + bx
				+ c

These integrals can be found by the method illustrated in Examples 5, 6, 8 and 9 (see topic 3). First complete the square of the denominator and use the

substitution	x +	b	= t	. The first integral in this case transforms to the form:
substitution	x +	2a	= t	. The first integral in this case transforms to the form:
		2a
		∫	dz				= ln z + z 2 ± m2 + C , if a > 0 ,
		∫	z 2 ± m2
or
			∫		dz				z
			∫	m	2	− z		2	= arcsin m + C , if a < 0 ,
				m		− z

and the second integral is expressed as the sum of simpler integrals.

5.2. Integration of

	m		r
R x, x n		,..., x s ;

ax + b n

ax + b s

R x,(ax + b)n

,...,(ax

+ b)s

; R x,

,...,

cx + d

Any rational function of x and

can be transformed into rational

x n ,..., x s

function of t by the substitution

x = tk , dx = ktt−1dt.

Here

k is the smallest

multiple of the fractional denominators

,...,

A rational function of x and (ax + b)

,...,(ax + b)

can be transformed into

rational function of t by the substitution ax + b = t k, dx =

t k – 1dt. Here k is

the smallest multiple of the fractional denominators

,...,

ax +b

,...,

can be transformed into rational

The function R x,

cx +d

129

function of t by the substitution	ax + b	= tλ	. Here k is the smallest multiple of
	cx + d

the fractional denominators mn ,..., rs .

5.3. Integration of binomial differentials xm(a + bxn)pdx

Here m, n and p are rational numbers.

Case 1: p is an integer number. Let x = tk. Here k is a less general denominator of fractional powers m and n.

Case 2:

m +1

is an integer number, let a + bxn = tk. Here k is a denominator

of fraction p.

Case 3:

m +1

+ p

is an

integer number. Let

a + bxn

= tk . Here k is a

denominator of fraction p.

5.4. Integration of

R(x,

ax2 +bx +c )

We shall consider three methods:

The first Euler substitution

ax2 + bx + c = t ± x a, (a > 0).

The second Euler substitution

ax2 + bx + c = tx ±

c, (c > 0).

The third Euler substitution

ax2 + bx + c = t x − x

(if x1 is the root of

equation).

5.5. Integrals in the form ∫

(x + m)

+ bx + c

The substitution

x + m =

turns this integral into the easy integral.

5.6. Integrals in the form ∫

f (x)dx

F(x) ax

+ bx + c

Here

f (x)

is a rational function of a single variable. This integral can be

F(x)

computed by partial fractions of

f (x)

F(x)

130

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