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5.7. Three trigonometric substitutions
Integrals in the form	∫ R(x,	ax 2 + bx + c )dx can be found by using a
trigonometric substitution, if the integrand is a rational function of x and:
1.	a − x2	let	x =	a sin t.
2.	a + x2	let	x =	a tg t.
3.	x2 − a let		x =	a sect.

The motivation behind this general procedure is quite simple.

1. If you replace x in a − x2 by	x = a sin t you obtain:
a − x2 = a − ( a sin t)2	= a − a sin2 t = a 1− sin2 t =

= a cos2 t = a cost.

The important thing is that the square root sign disappears.


2.	a + x2 =	a + a tg2 t =			a	1+	sin2 t		=
2.	a + x2 =	a + a tg2 t =			a	1+	cos2 t		=
							cos2 t
		=	a	cos2 t + sin2 t				=	a	= a sec t.
		=	a		cos2 t			=	cost	= a sec t.
					cos2 t				cost
3.	x2 − a =	a	− a =		a	1− cos2 t			= a tg t.
		cos2 t				cos2 t

	Т.5			Typical problems
				Find the integrals
1. ∫				dx		.
1. ∫		1 − x − 2x2				.
Solution. Complete the square:
			2		2				x		1										1		2		1		1					9				1	2
			2		2				x		1										1				1		1					9				1
1−x −		2x		=−2 x		+				−			= −2				x +						−			−				=	2		−		x +			.

								2 2												4				16 2							16					4
								2 2												4				16 2							16					4
Then
Then
	∫			dx		= ∫								dx								= 1			∫				d(x + 1/ 4)						2	=
		1 − x − 2x2									9			− (x +	1			)	2	]			2				3			2				1	2
										2[																					− x		+
												16				4
												16				4												4						4
							arcsin x + 1/ 4																					4						4
				=	1		arcsin x + 1/ 4								=				1			arcsin 4x + 1								+ C .
						2								3 / 4						2						3
														131

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2. ∫

(x − 2)dx

2 + 4x + 17

derivative (4x 2 + 4x + 17)′ = 8x + 4 ,

Solution.

Given the

the

numerator

transforms to the form:

x − 2 =

(8x + 4) −

− 2 =

(8x +4)−

(4x2 +4x +17)′−

Then

(8x + 4) −

)dx

∫

(x − 2)dx

= ∫ (

∫ d(4x 2 + 4x + 17) −

4x 2 + 4x + 17

−

5 ∫

4x 2 + 4x + 17 − 5

∫

d(2x + 1)

+ 4x + 17

(2x + 1)

+ 16

= 1

4x 2 + 4x + 17 −

5 ln 2x + 1+ 4x 2

+ 4x + 17 + C.

∫ 2 u

used

the

given

table

integrals:

+ C

and

du =

∫

= ln

u + u2 ±a2

+C .

u2 ±a2

3. ∫

dx .

x + 3 x 2

Solution. The smallest multiple of 2 and 3 is 6. This suggests the substitution x = t 6 , dx = 6t 5 dt . Then

∫

3 x

dx = ∫

t 2

6∫

t 4

6∫

(t 4

− 1) + 1

dt =

+ t

t +

t + 1

x + 3 x 2

6∫

(t 2 + 1)(t − 1)(t + 1) + 1

= 6∫(t 2 + 1)(t −

6∫

1)dt +

t + 1

= 6∫ (t 3 − t 2 + t − 1)dt + 6∫

t 4

− 2t 3

+ 3t 2

− 6t + 6 ln

t + 1

+ C =

t + 1

3 3 x 2

− 2 x + 3 3

x − 6 6 x + 6 ln 6 x + 1 + C .

4. I = ∫

x34− 3 x dx.

132

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Solution.

I =

x = t12 ;

= ∫

(t18 − t4 )12t11dt

dx = 12t11dt;

6t3

= 2

∫

(t18

− t4 )t8dt = 2

∫

(t26

− t

12 )dt = 2

−

+ C =

= 2x2 12 x3

−

2x12 x

+ C =

2x2 4 x

−

2x12 x

+ C.

5. I = ∫

33 x − 1 + xdx.

3x − 1

3x − 1 = t6

t3 +

2t5dt

Solution.

I =

x =

∫

dx = 2t5dt

∫

(3t6

+ t9 + t3 )dt =

+ C =

(3x −1) 6 3x −1 +

(3x −1) 3

(3x −1)2

3 (3x −1)2 + C.

6. ∫

x 24+ 1+ x dx .

1+ x

Solution. Let1+ x = t 4 , then

x = t 4 − 1 ,

dx = 4t 3 dt .

Consequently,

∫ x 24+ 1+ x dx = ∫

(t 4 − 1)2 + t 2

4t 3dt = 4∫ ((t 4 − 1)2 + t 2 )t 2 dt =

1+ x

= 4∫ (t

+ t

)dt

+ C =

−

= 4 4

+ x)11

−

8 4

(1+ x)

7 +

4 4 (1+ x)3 + 4 4

(1+ x)5 + C .

7. ∫

4 (x −1)3 (x + 2)5

Solution.

this

case

4 (x − 1)3 (x + 2)5

= (x − 1)(x + 2) 4

x + 2

and an

x − 1

integrand is a rational function of

x and 4

x + 2 .

x − 1

133

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The substitution

x + 2

= t 4 is appropriate:

x − 1

x =

t 4

+ 2

x − 1 =

x + 2 =

3t 4

dx = −

12t 3dt

t 4

− 1

t 4

− 1

(t 4

− 1)2

Then

∫

4 (x −1)3 (x + 2)5

x + 2

(x − 1)(x + 2)

x − 1

= ∫

t 4

− 1

t 4

− 1

− 12t 3

dt =

−

∫

+ C =

x − 1

+ C .

− 1)

8.I = ∫ x − 2 − x + 1dx.

x x − 2

x − 2

−

x + 1

−

x +1

x − 2

Solution. I = ∫

dx = ∫

dx =

x −

x − 2

x + 1

= t2

x + 1 = xt2 − 2t2

x − 2

(1− t) t (t2 −1)dt

− x

= 2t

+ 1

x =

2t2 + 1

= −6∫

t2 −1

(2t2 + 1)(t2 −1)2

dx = −

6tdt

;

(t2 −1)2

= 6∫

(t −1) tdt

= 6∫

tdt

(2t2 + 1)(t −1)(t + 1)

(2t2 + 1)(t + 1)

∫

2t + 1

= 2

−

dt =

ln 2t

+ 1 +

2 arctg(

2t) − 2 ln t + 1 + C =

2t2 + 1

t + 1

	3x	x + 1
= ln

	x − 2	x − 2

2	+	2arctg 2(x + 1) + C.
+ 1
		x − 2

134

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Integration of the binomial differentials

9. ∫ x (3 − 3 x )2 dx .

Solution. Multiplying the integrand we get

1 1 2 1 5 7

∫ x (3 − 3 x )2 dx = ∫ x 2 (9 − 6x 3 + x 3 )dx = ∫ (9x 2 − 6x 6 + x 6 )dx =

																					3				36					11					6			13
																= 6x						2	−		36				x		6	+			6		x		6				+ C .
					dx											= 6x						2	−		11				x		6	+			13		x		6				+ C .
																									11										13
10. I = ∫														.
10. I = ∫			x (3 x + 1)2											.
Solution. This p is an integer number and is equal to																																																−2. Thus:
			I = ∫ x							−	1						1					−2												x = t			6	;					= 6∫						t	2	dt
										−	1						1					−2															6													2
											2						3
													x					+			1 dx =																																		=
													x					+			1 dx =																5	dt;									(1+ t					2		2	=
																															dx = 6t							dt;									(1+ t						)
				−		3t					+ 3arctgt + C = − 36 x																												+ 3arctg 6											x + C.
				−	1+ t2						+ 3arctgt + C = − 36 x																												+ 3arctg 6											x + C.
					1+ t2																									1+ 3							x
11. I = ∫			5		xdx								.
11. I = ∫			3 − 25 x3										.
			3 − 25 x3
Solution. Multiplying the integrand we get
																					5				xdx												1								3			1
													I = ∫								5				xdx								=		∫ x5 (3 − 2x5 )2 dx.
													I = ∫																				=		∫ x5 (3 − 2x5 )2 dx.
																							3 − 25 x3
Here m =		1		; n =					3		and								m + 1					= 2							– an integer number.
Here m =	5			; n =							and													= 2							– an integer number.
	5							5													n
																														3																3 − t2								5
																																																						5
																																				2																		3
																																				2																		3
			1												3			−		1							3 −			2x5 = t								x =																;
																		−		1							3 −			2x5 = t								x =																;
	∫																	2																														2
																		2																														2
Thus I =			x5					− 2			x										dx =																											2								=
Thus I =			x5		3			− 2			x				5						dx =																											2								=
																																				5						3 − t2
																																				5						3 − t2						3
																															dx = −							t											dt;
																															dx = −					3		t			2								dt;
																																				3					2
																	1																					2
								3 − t2													−1 5											3			− t2													5
								3 − t2									3				−1 5											3			− t2					3								5								2
																		t −																								dt = −								∫(3 − t )dt =
			= ∫							2								t −										t							2							dt = −						6		∫(3 − t )dt =
										2																3									2													6
			= −			5	t +					5				t3			+ C =							5					(3 − 25 x3 )3 −													5					3 − 25 x3 + C.
			= −				t +				18					t3			+ C =						18						(3 − 25 x3 )3 −																		3 − 25 x3 + C.
					2						18														18																2
																														135

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12. ∫

x11 1+ x

Solution. Transforming the integrand we get

I = ∫

= ∫ x −11 (1+ x 4 )−

x11

dx .

1+ x 4

m + 1

Here p = −

m = −11,

n = 4 . Thereby

+ p = −3

the integer

number, we have the third case. Transforming the integrand we get

= ∫

∫

x11 x 2

x −4 + 1

x13

x −4 + 1

This

suggests

the substitution

x −4 + 1 = t 2 ,

− 4x −5 dx = 2tdt .

Therefore,

x −4 = t 2 − 1,

x −5 dx = −

tdt . Multiplying both numerator and denominator by

x −5 , we get

x −5 dx

−

tdt

I = ∫

= ∫

x −4 + 1

(x −4 )−2

x −4 + 1

(t 2 − 1)−2

t 2

= −

∫ (t 2 − 1)

2 dt = −

∫ (t 4

− 2t 2 + 1)dt = −

t 5

t 3

−

t + C =

= −

+ x

4 5

+ x 4 3

−

1 1

+ x

+ C .

Euler substitutions

13. ∫

x2 + 2x + 2

Solution. Here a = 1

> 0 . We can use the substitution

x 2 + 2x + 2 = t + x .

Then

+ 2x +

2 = t

+ 2tx + x

, 2x

+ 2 = t

+ 2tx ,

t 2

−

2(1

− t)

dx =

− t 2 + 2t − 2

, 1+ x

+ 2x

+ 2

= 1+ t

t 2

−

= −

2(1− t)2

2(1− t)

Thereafter we get an integral

∫

= ∫

(−t 2

+ 2t − 2)

2(1− t)

∫

t 2 − 2t + 2

dt .

x2 + 2x + 2

2(1− t)

(−t

)

(1− t)t

136

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To express the integrand as the sum of partial fractions:

t 2 − 2t + 2

With equality

(1− t)t 2

t 2

1− t

t 2 − 2t + 2 = A(1− t) + Bt(1− t) + Ct 2

computing the unknown coefficients:

t 2 : − B + C = 1 , B = 0 .

t = 0 : A = 2 ;

t = 1: C = 1;

Consequently,

∫

t 2 − 2t + 2

dt = 2∫

+ ∫

= −

− ln

t − 1

+ C .

(1− t)t

1− t

Replacing t by

x2 + 2x + 2 − x . Thus

− ln1+ x − x 2 + 2x + 2 + C .

∫ 1+ x 2 + 2x + 2

x − x 2 + 2x + 2

14. I = ∫

9 − x2

dx.

x + 2

Solution. Here the integrand is polynomial in which a = −1 < 0 and c = 9.

We use the second Euler substitution: 9 − x2

= tx − 3

− 6(t2 −1)

9 − x

= t

− 6xt + 9 x =

6 t

dx =

dt.

t2 + 1

(t2 + 1)2

Hence

6t2

− 3 (t

− 1)dt

+ 1

−

I = −6

∫

= −9

∫

dt.

+ 1) (t

+ 3t)

+ 1)

+ 2

+ 1

Multiplying the integrand we get

I = −9

−

dt =

∫

9 t + 3t + 1 3 (t

+ 1)

9 t + 1

−

5 ln 2t + 3 −

−

− 4arctgt

+ C,

when t =

9 − x2 + 3 .

2t + 3 + 5

2 + 1

15. I = ∫

(x + 1) x

2 + 4x

+ 2

Solution. This suggests the substitution

x + 1 =

and dx = −

, hence

t 2

137

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I = −∫

dt2 = −

∫

− 1

+ 4

− 1

+ 2

= −∫

= − arcsin t − 1 + C

1+ 2t − t2

2 − (t − 1)2

− 1

= − arcsin

x + 1

+ C = arcsin

+ C .

2(x + 1)

Trigonometric substitution

16.

I = ∫

3 − 2x

− x 2

(x + 1)

dx .

Solution. Completing the square of the radicand:

3 − 2x − x 2 = 4 − (x + 1)2 .

Letting

x + 1 = 2 sin t , dx = 2 cos tdt , we obtain then

I = ∫

4 − (x + 1)2

= ∫

4 − 4 sin 2 t

2 cos tdt

= ∫

2 cos t

2 cos tdt =

(x + 1)

4 sin

= ∫

cos 2 t

= ∫

1− sin

2 t

= ∫

−∫ dt =−ctg t −t +C .

sin

Replacing t by arcsin

x + 1

. Thus

1− sin 2 t

1−

+ 1) 2

ctg t =

cos t

3 − 2x − x

sin t

x + 1

Consequently,

I = −

3 − 2x − x 2

− arcsin

x + 1

+ C .

x + 1

17.

I = ∫

xdx

−

9 + x

138

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x = 3tg t

3tg t

3dt

Solution.

I =

dx =

= ∫

cos2 t

cos

−

+ 9 tg2 t

t = arctg

= 3∫

sin tdt

z = cost

= −3∫

dz = − sin tdt

= I.

cos2 t(cost − 1)

z2 (z − 1)

This integral can be found by the partial fractions:

B C

(B + C)z2 + (A − B)z

− A

z2 (z − 1)

z − 1

z2 (z − 1)

B + C =

= 1;

∫

A − B =

= −1; then

I = −3

−

dz =

z − 1

= −1;

− A = 1

= −3

− ln

+ ln

z − 1

+ C = −

+ 3ln

cost

− 3ln

cost −1

+ C =

cost

t = arctg

cost =

= −

9 + x2 + 3ln

−

9 + x

1+ tg

9 + x

− 3ln 3 −

9 + x2

+ C = C −

9 + x2

+ 3ln 3 − 3ln

9 + x2

+ 3ln

9 + x2

−

9 + x2

− 3ln 3 −

9 + x2

= C −

9 + x2

− 3ln

3 −

+ 3ln 3.

Exercises for class and homework

Т.5

Find the integrals

(3x − 5)dx .

1. ∫

2. ∫

3. ∫

x 2 + 6x + 11

3 + 2x − x 2

2x 2 − 12x + 15

4. ∫

(6x +

1)dx

5. ∫

(2 sin x − 1) cos xdx

6. ∫

x − x 2

5 + 4 sin x − sin 2 x

x +

139

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7. ∫ 3 x + 1				dx	3 .
		x − 1	(x	− 1)
10.	∫		dx		.
		x( x + 5
				x 2 )
13.	∫	3	x	dx .
		1+ 3 x 2

16. ∫			dx		.
		x +
			x 2 − x + 4
19.	∫	x 2 dx			.
		x 2 + 2x + 5

22.	∫	x 2 dx			.
		x 2 − 2x + 2

25.	∫	x 4 dx			.
		x 2 + 4x + 5

28.∫ 2x + x 2 dx .

8.	∫	6 xdx3 .		9. ∫		x − 3		dx .
		1+	x			x		x
11.	∫	1−	x dx .	12. ∫	3 4 x + 1dx .
		1+	x			x
	∫	x7		15. ∫		dx
14.			dx .							.
14.			dx .		x(1+ x		3	)	1/ 4	.
		1+ x 2			x(1+ x			)
17.	∫	x (1+ 2 6 x )3dx .		18. ∫ 3		dx	3 .
						1+ x

20.∫

23.∫

26.∫

29.∫

	dx		.
(2x − 3) 4x − x 2			.
(2x − 3) 4x − x 2
	dx			.
(x − 1) 6x − x 2 − 5				.
(x − 1) 6x − x 2 − 5
dx		.
x 2 + x − x 2		.
x 2 + x − x 2
1+ x 2	dx .
2 + x 2	dx .
2 + x 2

21. ∫		dx		.
21. ∫	(6x − 8 − x2 )3			.
	(6x − 8 − x2 )3
24. ∫		dx	.
24. ∫	x	x2 + x + 1	.
	x	x2 + x + 1

27. ∫ (2x 2 − 3x)dx . x 2 − 2x + 5

∫xdx

30..

x− x 2 − 1

Answers

1. ln x + 3 +								x2 + 6x + 11 + C .																					2. arcsin								x − 1		+ C .								3.			3		2x2 − 12x + 15 +
1. ln x + 3 +								x2 + 6x + 11 + C .																					2. arcsin										+ C .								3.					2x2 − 12x + 15 +
																																						2										2
+2 2 ln \| x − 3 +										x2 − 6x + 15 / 2 \| +C . 4.																− 6					x − x2 + 4arcsin(2x + 1) + C . 5.																					−2		5 + 4t − t2 +
					t − 2																				6.										4							4						7.				3		3	x		+		1 4
+3arcsin(								) + C ,			where							t = sin t .										2			x − 4					x + 4ln1+								x	+ C .							16				x	−		1			−
+3arcsin(						3		) + C ,			where							t = sin t .										2			x − 4					x + 4ln1+								x	+ C .																	−
						3
3	3	x +			1 7							6		6		x	5	− 2			x + 6		6		x − 6arctg								6		x + C .				11.			(			x − 2)				1− x − arcsin										x + C .
28	3		x −		1		+ C . 8.					5				x		− 2			x + 6				x − 6arctg										x + C .				11.			(			x − 2)				1− x − arcsin										x + C .
28			x −		1							5
12.	12 3				4	x	+ 1)		7	−	3		4		x +				4	+ C . 13.						u		3	− 3u + C ,where u													= (1+ x					2 / 3 1/ 2					. 14.			t7	−			3	t	5	+
		7		(							3	(						1)																														)
				(							3	(						1)																														)							7			5
																										1																	15									5			7			5
+t3 − t + C ,							where				t =					1 + x2 . 17.								−			ln			t − 1			+ 8ln					2t + 1		−					ln	t + 1			+				+ C ,

																										2																2									t	+ 1
																										2																2									t	+ 1
where t =							x2 − x +				4 + 2					.	18.			1	ln	u2		+ u + 1							−	1		arctg				2u + 1														3	1+ x3					.
where t =										x						.	18.			6	ln	(u − 1)2									−	3		arctg				3			+ C , where u =													x				.
										x										6		(u − 1)2										3						3																x

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