knowrohit07/gita_dataset · Datasets at Hugging Face

Chapter stringclasses 18 values	sentence_range stringlengths 3 9	Text stringlengths 7 7.34k
1	3418-3421	INTEGRALS 299 10 The process of differentiation and integration are inverses of each other as discussed in Section 7 2 2 (i)
1	3419-3422	The process of differentiation and integration are inverses of each other as discussed in Section 7 2 2 (i) EXERCISE 7
1	3420-3423	2 2 (i) EXERCISE 7 1 Find an anti derivative (or integral) of the following functions by the method of inspection
1	3421-3424	2 (i) EXERCISE 7 1 Find an anti derivative (or integral) of the following functions by the method of inspection 1
1	3422-3425	EXERCISE 7 1 Find an anti derivative (or integral) of the following functions by the method of inspection 1 sin 2x 2
1	3423-3426	1 Find an anti derivative (or integral) of the following functions by the method of inspection 1 sin 2x 2 cos 3x 3
1	3424-3427	1 sin 2x 2 cos 3x 3 e 2x 4
1	3425-3428	sin 2x 2 cos 3x 3 e 2x 4 (ax + b)2 5
1	3426-3429	cos 3x 3 e 2x 4 (ax + b)2 5 sin 2x – 4 e3x Find the following integrals in Exercises 6 to 20: 6
1	3427-3430	e 2x 4 (ax + b)2 5 sin 2x – 4 e3x Find the following integrals in Exercises 6 to 20: 6 (4 3 ex+ 1) dx ∫ 7
1	3428-3431	(ax + b)2 5 sin 2x – 4 e3x Find the following integrals in Exercises 6 to 20: 6 (4 3 ex+ 1) dx ∫ 7 2 (1–12 ) x dx x ∫ 8
1	3429-3432	sin 2x – 4 e3x Find the following integrals in Exercises 6 to 20: 6 (4 3 ex+ 1) dx ∫ 7 2 (1–12 ) x dx x ∫ 8 2 ( ) ax bx c dx + + ∫ 9
1	3430-3433	(4 3 ex+ 1) dx ∫ 7 2 (1–12 ) x dx x ∫ 8 2 ( ) ax bx c dx + + ∫ 9 (22 x) x e dx + ∫ 10
1	3431-3434	2 (1–12 ) x dx x ∫ 8 2 ( ) ax bx c dx + + ∫ 9 (22 x) x e dx + ∫ 10 2 x –1 dx x       ∫ 11
1	3432-3435	2 ( ) ax bx c dx + + ∫ 9 (22 x) x e dx + ∫ 10 2 x –1 dx x       ∫ 11 3 2 2 5 4 x x – dx +x ∫ 12
1	3433-3436	(22 x) x e dx + ∫ 10 2 x –1 dx x       ∫ 11 3 2 2 5 4 x x – dx +x ∫ 12 3 3 4 x x dx x + + ∫ 13
1	3434-3437	2 x –1 dx x       ∫ 11 3 2 2 5 4 x x – dx +x ∫ 12 3 3 4 x x dx x + + ∫ 13 3 2 1 1 x x x – dx −x – + ∫ 14
1	3435-3438	3 2 2 5 4 x x – dx +x ∫ 12 3 3 4 x x dx x + + ∫ 13 3 2 1 1 x x x – dx −x – + ∫ 14 (1 – x) x dx ∫ 15
1	3436-3439	3 3 4 x x dx x + + ∫ 13 3 2 1 1 x x x – dx −x – + ∫ 14 (1 – x) x dx ∫ 15 ( 32 2 3) x x x dx + + ∫ 16
1	3437-3440	3 2 1 1 x x x – dx −x – + ∫ 14 (1 – x) x dx ∫ 15 ( 32 2 3) x x x dx + + ∫ 16 (2 3cos x) x – x e dx + ∫ 17
1	3438-3441	(1 – x) x dx ∫ 15 ( 32 2 3) x x x dx + + ∫ 16 (2 3cos x) x – x e dx + ∫ 17 (22 3sin 5 ) x – x x dx + ∫ 18
1	3439-3442	( 32 2 3) x x x dx + + ∫ 16 (2 3cos x) x – x e dx + ∫ 17 (22 3sin 5 ) x – x x dx + ∫ 18 sec (sec tan ) x x x dx + ∫ 19
1	3440-3443	(2 3cos x) x – x e dx + ∫ 17 (22 3sin 5 ) x – x x dx + ∫ 18 sec (sec tan ) x x x dx + ∫ 19 2 sec2 cosec x dx x ∫ 20
1	3441-3444	(22 3sin 5 ) x – x x dx + ∫ 18 sec (sec tan ) x x x dx + ∫ 19 2 sec2 cosec x dx x ∫ 20 2 2 – 3sin cos x x ∫ dx
1	3442-3445	sec (sec tan ) x x x dx + ∫ 19 2 sec2 cosec x dx x ∫ 20 2 2 – 3sin cos x x ∫ dx Choose the correct answer in Exercises 21 and 22
1	3443-3446	2 sec2 cosec x dx x ∫ 20 2 2 – 3sin cos x x ∫ dx Choose the correct answer in Exercises 21 and 22 21
1	3444-3447	2 2 – 3sin cos x x ∫ dx Choose the correct answer in Exercises 21 and 22 21 The anti derivative of 1 x x   +     equals (A) 1 1 3 2 1 2 C 3 x x + + (B) 2 2 23 1 C 3 2 x x + + (C) 3 1 2 2 2 2 C 3 x x + + (D) 3 1 2 2 3 1 C 2 2 x x + + 22
1	3445-3448	Choose the correct answer in Exercises 21 and 22 21 The anti derivative of 1 x x   +     equals (A) 1 1 3 2 1 2 C 3 x x + + (B) 2 2 23 1 C 3 2 x x + + (C) 3 1 2 2 2 2 C 3 x x + + (D) 3 1 2 2 3 1 C 2 2 x x + + 22 If 3 34 ( ) 4 d f x x dx x = − such that f (2) = 0
1	3446-3449	21 The anti derivative of 1 x x   +     equals (A) 1 1 3 2 1 2 C 3 x x + + (B) 2 2 23 1 C 3 2 x x + + (C) 3 1 2 2 2 2 C 3 x x + + (D) 3 1 2 2 3 1 C 2 2 x x + + 22 If 3 34 ( ) 4 d f x x dx x = − such that f (2) = 0 Then f (x) is (A) 4 13 129 8 x +x − (B) 3 14 129 8 x +x + (C) 4 13 129 8 x +x + (D) 3 14 129 8 x +x − 300 MATHEMATICS 7
1	3447-3450	The anti derivative of 1 x x   +     equals (A) 1 1 3 2 1 2 C 3 x x + + (B) 2 2 23 1 C 3 2 x x + + (C) 3 1 2 2 2 2 C 3 x x + + (D) 3 1 2 2 3 1 C 2 2 x x + + 22 If 3 34 ( ) 4 d f x x dx x = − such that f (2) = 0 Then f (x) is (A) 4 13 129 8 x +x − (B) 3 14 129 8 x +x + (C) 4 13 129 8 x +x + (D) 3 14 129 8 x +x − 300 MATHEMATICS 7 3 Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions
1	3448-3451	If 3 34 ( ) 4 d f x x dx x = − such that f (2) = 0 Then f (x) is (A) 4 13 129 8 x +x − (B) 3 14 129 8 x +x + (C) 4 13 129 8 x +x + (D) 3 14 129 8 x +x − 300 MATHEMATICS 7 3 Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions It was based on inspection, i
1	3449-3452	Then f (x) is (A) 4 13 129 8 x +x − (B) 3 14 129 8 x +x + (C) 4 13 129 8 x +x + (D) 3 14 129 8 x +x − 300 MATHEMATICS 7 3 Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions It was based on inspection, i e
1	3450-3453	3 Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions It was based on inspection, i e , on the search of a function F whose derivative is f which led us to the integral of f
1	3451-3454	It was based on inspection, i e , on the search of a function F whose derivative is f which led us to the integral of f However, this method, which depends on inspection, is not very suitable for many functions
1	3452-3455	e , on the search of a function F whose derivative is f which led us to the integral of f However, this method, which depends on inspection, is not very suitable for many functions Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms
1	3453-3456	, on the search of a function F whose derivative is f which led us to the integral of f However, this method, which depends on inspection, is not very suitable for many functions Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms Prominent among them are methods based on: 1
1	3454-3457	However, this method, which depends on inspection, is not very suitable for many functions Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms Prominent among them are methods based on: 1 Integration by Substitution 2
1	3455-3458	Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms Prominent among them are methods based on: 1 Integration by Substitution 2 Integration using Partial Fractions 3
1	3456-3459	Prominent among them are methods based on: 1 Integration by Substitution 2 Integration using Partial Fractions 3 Integration by Parts 7
1	3457-3460	Integration by Substitution 2 Integration using Partial Fractions 3 Integration by Parts 7 3
1	3458-3461	Integration using Partial Fractions 3 Integration by Parts 7 3 1 Integration by substitution In this section, we consider the method of integration by substitution
1	3459-3462	Integration by Parts 7 3 1 Integration by substitution In this section, we consider the method of integration by substitution The given integral ( ) ∫f x dx can be transformed into another form by changing the independent variable x to t by substituting x = g (t)
1	3460-3463	3 1 Integration by substitution In this section, we consider the method of integration by substitution The given integral ( ) ∫f x dx can be transformed into another form by changing the independent variable x to t by substituting x = g (t) Consider I = ( ) f x dx ∫ Put x = g(t) so that dx dt = g′(t)
1	3461-3464	1 Integration by substitution In this section, we consider the method of integration by substitution The given integral ( ) ∫f x dx can be transformed into another form by changing the independent variable x to t by substituting x = g (t) Consider I = ( ) f x dx ∫ Put x = g(t) so that dx dt = g′(t) We write dx = g′(t) dt Thus I = ( ) ( ( )) ( ) f x dx f g t g t dt = ′ ∫ ∫ This change of variable formula is one of the important tools available to us in the name of integration by substitution
1	3462-3465	The given integral ( ) ∫f x dx can be transformed into another form by changing the independent variable x to t by substituting x = g (t) Consider I = ( ) f x dx ∫ Put x = g(t) so that dx dt = g′(t) We write dx = g′(t) dt Thus I = ( ) ( ( )) ( ) f x dx f g t g t dt = ′ ∫ ∫ This change of variable formula is one of the important tools available to us in the name of integration by substitution It is often important to guess what will be the useful substitution
1	3463-3466	Consider I = ( ) f x dx ∫ Put x = g(t) so that dx dt = g′(t) We write dx = g′(t) dt Thus I = ( ) ( ( )) ( ) f x dx f g t g t dt = ′ ∫ ∫ This change of variable formula is one of the important tools available to us in the name of integration by substitution It is often important to guess what will be the useful substitution Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following examples
1	3464-3467	We write dx = g′(t) dt Thus I = ( ) ( ( )) ( ) f x dx f g t g t dt = ′ ∫ ∫ This change of variable formula is one of the important tools available to us in the name of integration by substitution It is often important to guess what will be the useful substitution Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following examples Example 5 Integrate the following functions w
1	3465-3468	It is often important to guess what will be the useful substitution Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following examples Example 5 Integrate the following functions w r
1	3466-3469	Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following examples Example 5 Integrate the following functions w r t
1	3467-3470	Example 5 Integrate the following functions w r t x: (i) sin mx (ii) 2x sin (x2 + 1) (iii) 4 2 tan xsec x x (iv) 1 2 sin (tan ) 1 x– x + Solution (i) We know that derivative of mx is m
1	3468-3471	r t x: (i) sin mx (ii) 2x sin (x2 + 1) (iii) 4 2 tan xsec x x (iv) 1 2 sin (tan ) 1 x– x + Solution (i) We know that derivative of mx is m Thus, we make the substitution mx = t so that mdx = dt
1	3469-3472	t x: (i) sin mx (ii) 2x sin (x2 + 1) (iii) 4 2 tan xsec x x (iv) 1 2 sin (tan ) 1 x– x + Solution (i) We know that derivative of mx is m Thus, we make the substitution mx = t so that mdx = dt Therefore, 1 sin sin mx dx t dt m = ∫ ∫ = – 1 m cos t + C = – 1 m cos mx + C INTEGRALS 301 (ii) Derivative of x2 + 1 is 2x
1	3470-3473	x: (i) sin mx (ii) 2x sin (x2 + 1) (iii) 4 2 tan xsec x x (iv) 1 2 sin (tan ) 1 x– x + Solution (i) We know that derivative of mx is m Thus, we make the substitution mx = t so that mdx = dt Therefore, 1 sin sin mx dx t dt m = ∫ ∫ = – 1 m cos t + C = – 1 m cos mx + C INTEGRALS 301 (ii) Derivative of x2 + 1 is 2x Thus, we use the substitution x2 + 1 = t so that 2x dx = dt
1	3471-3474	Thus, we make the substitution mx = t so that mdx = dt Therefore, 1 sin sin mx dx t dt m = ∫ ∫ = – 1 m cos t + C = – 1 m cos mx + C INTEGRALS 301 (ii) Derivative of x2 + 1 is 2x Thus, we use the substitution x2 + 1 = t so that 2x dx = dt Therefore, 2 sin (2 1) sin x x dx t dt + = ∫ ∫ = – cos t + C = – cos (x2 + 1) + C (iii) Derivative of x is 21 1 1 2 2 – x x =
1	3472-3475	Therefore, 1 sin sin mx dx t dt m = ∫ ∫ = – 1 m cos t + C = – 1 m cos mx + C INTEGRALS 301 (ii) Derivative of x2 + 1 is 2x Thus, we use the substitution x2 + 1 = t so that 2x dx = dt Therefore, 2 sin (2 1) sin x x dx t dt + = ∫ ∫ = – cos t + C = – cos (x2 + 1) + C (iii) Derivative of x is 21 1 1 2 2 – x x = Thus, we use the substitution 1 so that giving 2 x t dx dt x = = dx = 2t dt
1	3473-3476	Thus, we use the substitution x2 + 1 = t so that 2x dx = dt Therefore, 2 sin (2 1) sin x x dx t dt + = ∫ ∫ = – cos t + C = – cos (x2 + 1) + C (iii) Derivative of x is 21 1 1 2 2 – x x = Thus, we use the substitution 1 so that giving 2 x t dx dt x = = dx = 2t dt Thus, 4 2 4 2 tan sec 2 tan sec x x t t t dt dx t x = ∫ ∫ = 4 2 2 tan tsec t dt ∫ Again, we make another substitution tan t = u so that sec2 t dt = du Therefore, 4 2 4 2 tan sec 2 t t dt u du = ∫ ∫ = 5 2 C 5 u + = 2 tan5 C 5 t + (since u = tan t) = 2 tan5 C (since ) 5 x t x + = Hence, 4 2 tan xsec x dx x ∫ = 2 tan5 C 5 x + Alternatively, make the substitution tan x t = (iv) Derivative of 1 2 1 tan 1 – x x = +
1	3474-3477	Therefore, 2 sin (2 1) sin x x dx t dt + = ∫ ∫ = – cos t + C = – cos (x2 + 1) + C (iii) Derivative of x is 21 1 1 2 2 – x x = Thus, we use the substitution 1 so that giving 2 x t dx dt x = = dx = 2t dt Thus, 4 2 4 2 tan sec 2 tan sec x x t t t dt dx t x = ∫ ∫ = 4 2 2 tan tsec t dt ∫ Again, we make another substitution tan t = u so that sec2 t dt = du Therefore, 4 2 4 2 tan sec 2 t t dt u du = ∫ ∫ = 5 2 C 5 u + = 2 tan5 C 5 t + (since u = tan t) = 2 tan5 C (since ) 5 x t x + = Hence, 4 2 tan xsec x dx x ∫ = 2 tan5 C 5 x + Alternatively, make the substitution tan x t = (iv) Derivative of 1 2 1 tan 1 – x x = + Thus, we use the substitution tan–1 x = t so that 2 1 dx x + = dt
1	3475-3478	Thus, we use the substitution 1 so that giving 2 x t dx dt x = = dx = 2t dt Thus, 4 2 4 2 tan sec 2 tan sec x x t t t dt dx t x = ∫ ∫ = 4 2 2 tan tsec t dt ∫ Again, we make another substitution tan t = u so that sec2 t dt = du Therefore, 4 2 4 2 tan sec 2 t t dt u du = ∫ ∫ = 5 2 C 5 u + = 2 tan5 C 5 t + (since u = tan t) = 2 tan5 C (since ) 5 x t x + = Hence, 4 2 tan xsec x dx x ∫ = 2 tan5 C 5 x + Alternatively, make the substitution tan x t = (iv) Derivative of 1 2 1 tan 1 – x x = + Thus, we use the substitution tan–1 x = t so that 2 1 dx x + = dt Therefore , 1 sin (tan2 ) sin 1 – x dx t dt x = + ∫ ∫ = – cos t + C = – cos (tan –1x) + C Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique
1	3476-3479	Thus, 4 2 4 2 tan sec 2 tan sec x x t t t dt dx t x = ∫ ∫ = 4 2 2 tan tsec t dt ∫ Again, we make another substitution tan t = u so that sec2 t dt = du Therefore, 4 2 4 2 tan sec 2 t t dt u du = ∫ ∫ = 5 2 C 5 u + = 2 tan5 C 5 t + (since u = tan t) = 2 tan5 C (since ) 5 x t x + = Hence, 4 2 tan xsec x dx x ∫ = 2 tan5 C 5 x + Alternatively, make the substitution tan x t = (iv) Derivative of 1 2 1 tan 1 – x x = + Thus, we use the substitution tan–1 x = t so that 2 1 dx x + = dt Therefore , 1 sin (tan2 ) sin 1 – x dx t dt x = + ∫ ∫ = – cos t + C = – cos (tan –1x) + C Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique These will be used later without reference
1	3477-3480	Thus, we use the substitution tan–1 x = t so that 2 1 dx x + = dt Therefore , 1 sin (tan2 ) sin 1 – x dx t dt x = + ∫ ∫ = – cos t + C = – cos (tan –1x) + C Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique These will be used later without reference (i) ∫tan = log sec + C x dx x We have sin tan cos x x dx dx x = ∫ ∫ 302 MATHEMATICS Put cos x = t so that sin x dx = – dt Then tan log C log cos C dt x dx – – t – x t = = + = + ∫ ∫ or tan log sec C x dx x = + ∫ (ii) ∫cot = log sin + C x dx x We have cos cot sin x x dx dx x = ∫ ∫ Put sin x = t so that cos x dx = dt Then cot dt x dx t = ∫ ∫ = log t +C = log sin C x + (iii) ∫sec = log sec + tan + C x dx x x We have sec (sec tan ) sec sec + tan x x x x dx dx x +x = ∫ ∫ Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt Therefore, sec log + C = log sec tan C dt x dx t x x t = = + + ∫ ∫ (iv) ∫cosec = log cosec – cot + C x dx x x We have cosec (cosec cot ) cosec (cosec cot ) x x x x dx dx x +x = + ∫ ∫ Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt So cosec – –log\| \| – log\|cosec cot \| C dt x dx t x x t = = = + + ∫ ∫ = 2 2 cosec cot – log C cosec cot x x x x − + − = log cosec cot C x – x + Example 6 Find the following integrals: (i) 3 2 sin xcos x dx ∫ (ii) sin sin ( ) x dx x +a ∫ (iii) 1 1 tan dx x ∫+ INTEGRALS 303 Solution (i) We have 3 2 2 2 sin cos sin cos (sin ) x x dx x x x dx = ∫ ∫ = 2 2 (1– cos ) cos (sin ) x x x dx ∫ Put t = cos x so that dt = – sin x dx Therefore, 2 2 sin cos (sin ) x x x dx ∫ = 2 2 (1– t) t dt −∫ = 3 5 2 4 ( – ) C 3 5 t t – t t dt – –   = +     ∫ = 3 5 1 1 cos cos C 3 5 – x x + + (ii) Put x + a = t
1	3478-3481	Therefore , 1 sin (tan2 ) sin 1 – x dx t dt x = + ∫ ∫ = – cos t + C = – cos (tan –1x) + C Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique These will be used later without reference (i) ∫tan = log sec + C x dx x We have sin tan cos x x dx dx x = ∫ ∫ 302 MATHEMATICS Put cos x = t so that sin x dx = – dt Then tan log C log cos C dt x dx – – t – x t = = + = + ∫ ∫ or tan log sec C x dx x = + ∫ (ii) ∫cot = log sin + C x dx x We have cos cot sin x x dx dx x = ∫ ∫ Put sin x = t so that cos x dx = dt Then cot dt x dx t = ∫ ∫ = log t +C = log sin C x + (iii) ∫sec = log sec + tan + C x dx x x We have sec (sec tan ) sec sec + tan x x x x dx dx x +x = ∫ ∫ Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt Therefore, sec log + C = log sec tan C dt x dx t x x t = = + + ∫ ∫ (iv) ∫cosec = log cosec – cot + C x dx x x We have cosec (cosec cot ) cosec (cosec cot ) x x x x dx dx x +x = + ∫ ∫ Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt So cosec – –log\| \| – log\|cosec cot \| C dt x dx t x x t = = = + + ∫ ∫ = 2 2 cosec cot – log C cosec cot x x x x − + − = log cosec cot C x – x + Example 6 Find the following integrals: (i) 3 2 sin xcos x dx ∫ (ii) sin sin ( ) x dx x +a ∫ (iii) 1 1 tan dx x ∫+ INTEGRALS 303 Solution (i) We have 3 2 2 2 sin cos sin cos (sin ) x x dx x x x dx = ∫ ∫ = 2 2 (1– cos ) cos (sin ) x x x dx ∫ Put t = cos x so that dt = – sin x dx Therefore, 2 2 sin cos (sin ) x x x dx ∫ = 2 2 (1– t) t dt −∫ = 3 5 2 4 ( – ) C 3 5 t t – t t dt – –   = +     ∫ = 3 5 1 1 cos cos C 3 5 – x x + + (ii) Put x + a = t Then dx = dt
1	3479-3482	These will be used later without reference (i) ∫tan = log sec + C x dx x We have sin tan cos x x dx dx x = ∫ ∫ 302 MATHEMATICS Put cos x = t so that sin x dx = – dt Then tan log C log cos C dt x dx – – t – x t = = + = + ∫ ∫ or tan log sec C x dx x = + ∫ (ii) ∫cot = log sin + C x dx x We have cos cot sin x x dx dx x = ∫ ∫ Put sin x = t so that cos x dx = dt Then cot dt x dx t = ∫ ∫ = log t +C = log sin C x + (iii) ∫sec = log sec + tan + C x dx x x We have sec (sec tan ) sec sec + tan x x x x dx dx x +x = ∫ ∫ Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt Therefore, sec log + C = log sec tan C dt x dx t x x t = = + + ∫ ∫ (iv) ∫cosec = log cosec – cot + C x dx x x We have cosec (cosec cot ) cosec (cosec cot ) x x x x dx dx x +x = + ∫ ∫ Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt So cosec – –log\| \| – log\|cosec cot \| C dt x dx t x x t = = = + + ∫ ∫ = 2 2 cosec cot – log C cosec cot x x x x − + − = log cosec cot C x – x + Example 6 Find the following integrals: (i) 3 2 sin xcos x dx ∫ (ii) sin sin ( ) x dx x +a ∫ (iii) 1 1 tan dx x ∫+ INTEGRALS 303 Solution (i) We have 3 2 2 2 sin cos sin cos (sin ) x x dx x x x dx = ∫ ∫ = 2 2 (1– cos ) cos (sin ) x x x dx ∫ Put t = cos x so that dt = – sin x dx Therefore, 2 2 sin cos (sin ) x x x dx ∫ = 2 2 (1– t) t dt −∫ = 3 5 2 4 ( – ) C 3 5 t t – t t dt – –   = +     ∫ = 3 5 1 1 cos cos C 3 5 – x x + + (ii) Put x + a = t Then dx = dt Therefore sin sin ( ) sin ( ) sin x t – a dx dt x a t = + ∫ ∫ = sin cos cos sin sin t a – t a dt t ∫ = cos – sin cot a dt a t dt ∫ ∫ = 1 (cos ) (sin ) log sin C a t – a t   +   = 1 (cos ) ( ) (sin ) log sin ( ) C a x a – a x a   + + +   = 1 cos cos (sin ) log sin ( ) C sin x a a a – a x a – a + + Hence, sin sin ( ) x dx x +a ∫ = x cos a – sin a log \|sin (x + a)\| + C, where, C = – C1 sin a + a cos a, is another arbitrary constant
1	3480-3483	(i) ∫tan = log sec + C x dx x We have sin tan cos x x dx dx x = ∫ ∫ 302 MATHEMATICS Put cos x = t so that sin x dx = – dt Then tan log C log cos C dt x dx – – t – x t = = + = + ∫ ∫ or tan log sec C x dx x = + ∫ (ii) ∫cot = log sin + C x dx x We have cos cot sin x x dx dx x = ∫ ∫ Put sin x = t so that cos x dx = dt Then cot dt x dx t = ∫ ∫ = log t +C = log sin C x + (iii) ∫sec = log sec + tan + C x dx x x We have sec (sec tan ) sec sec + tan x x x x dx dx x +x = ∫ ∫ Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt Therefore, sec log + C = log sec tan C dt x dx t x x t = = + + ∫ ∫ (iv) ∫cosec = log cosec – cot + C x dx x x We have cosec (cosec cot ) cosec (cosec cot ) x x x x dx dx x +x = + ∫ ∫ Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt So cosec – –log\| \| – log\|cosec cot \| C dt x dx t x x t = = = + + ∫ ∫ = 2 2 cosec cot – log C cosec cot x x x x − + − = log cosec cot C x – x + Example 6 Find the following integrals: (i) 3 2 sin xcos x dx ∫ (ii) sin sin ( ) x dx x +a ∫ (iii) 1 1 tan dx x ∫+ INTEGRALS 303 Solution (i) We have 3 2 2 2 sin cos sin cos (sin ) x x dx x x x dx = ∫ ∫ = 2 2 (1– cos ) cos (sin ) x x x dx ∫ Put t = cos x so that dt = – sin x dx Therefore, 2 2 sin cos (sin ) x x x dx ∫ = 2 2 (1– t) t dt −∫ = 3 5 2 4 ( – ) C 3 5 t t – t t dt – –   = +     ∫ = 3 5 1 1 cos cos C 3 5 – x x + + (ii) Put x + a = t Then dx = dt Therefore sin sin ( ) sin ( ) sin x t – a dx dt x a t = + ∫ ∫ = sin cos cos sin sin t a – t a dt t ∫ = cos – sin cot a dt a t dt ∫ ∫ = 1 (cos ) (sin ) log sin C a t – a t   +   = 1 (cos ) ( ) (sin ) log sin ( ) C a x a – a x a   + + +   = 1 cos cos (sin ) log sin ( ) C sin x a a a – a x a – a + + Hence, sin sin ( ) x dx x +a ∫ = x cos a – sin a log \|sin (x + a)\| + C, where, C = – C1 sin a + a cos a, is another arbitrary constant (iii) cos 1 tan cos sin dx x dx x x x = + + ∫ ∫ = 1 (cos + sin + cos – sin ) 2 cos sin x x x x dx x x + ∫ 304 MATHEMATICS = 1 1 cos – sin 2 2 cos sin x x dx dx x x + + ∫ ∫ = C1 1 cos sin 2 2 2 cos sin x x – x dx x x + + + ∫
1	3481-3484	Then dx = dt Therefore sin sin ( ) sin ( ) sin x t – a dx dt x a t = + ∫ ∫ = sin cos cos sin sin t a – t a dt t ∫ = cos – sin cot a dt a t dt ∫ ∫ = 1 (cos ) (sin ) log sin C a t – a t   +   = 1 (cos ) ( ) (sin ) log sin ( ) C a x a – a x a   + + +   = 1 cos cos (sin ) log sin ( ) C sin x a a a – a x a – a + + Hence, sin sin ( ) x dx x +a ∫ = x cos a – sin a log \|sin (x + a)\| + C, where, C = – C1 sin a + a cos a, is another arbitrary constant (iii) cos 1 tan cos sin dx x dx x x x = + + ∫ ∫ = 1 (cos + sin + cos – sin ) 2 cos sin x x x x dx x x + ∫ 304 MATHEMATICS = 1 1 cos – sin 2 2 cos sin x x dx dx x x + + ∫ ∫ = C1 1 cos sin 2 2 2 cos sin x x – x dx x x + + + ∫ (1) Now, consider cos sin I cos sin x – x dx x x = + ∫ Put cos x + sin x = t so that (cos x – sin x) dx = dt Therefore 2 I log C dt t t = = + ∫ = 2 log cos sin C x x + + Putting it in (1), we get 1 2 C C 1 + + log cos sin 1 tan 2 2 2 2 dx x x x x = + + + ∫ = 1 2 C C +1 log cos sin 2 2 2 2 x x x + + + = 1 2 C C +1 log cos sin C C 2 2 2 2 x x x ,  + + = +     EXERCISE 7
1	3482-3485	Therefore sin sin ( ) sin ( ) sin x t – a dx dt x a t = + ∫ ∫ = sin cos cos sin sin t a – t a dt t ∫ = cos – sin cot a dt a t dt ∫ ∫ = 1 (cos ) (sin ) log sin C a t – a t   +   = 1 (cos ) ( ) (sin ) log sin ( ) C a x a – a x a   + + +   = 1 cos cos (sin ) log sin ( ) C sin x a a a – a x a – a + + Hence, sin sin ( ) x dx x +a ∫ = x cos a – sin a log \|sin (x + a)\| + C, where, C = – C1 sin a + a cos a, is another arbitrary constant (iii) cos 1 tan cos sin dx x dx x x x = + + ∫ ∫ = 1 (cos + sin + cos – sin ) 2 cos sin x x x x dx x x + ∫ 304 MATHEMATICS = 1 1 cos – sin 2 2 cos sin x x dx dx x x + + ∫ ∫ = C1 1 cos sin 2 2 2 cos sin x x – x dx x x + + + ∫ (1) Now, consider cos sin I cos sin x – x dx x x = + ∫ Put cos x + sin x = t so that (cos x – sin x) dx = dt Therefore 2 I log C dt t t = = + ∫ = 2 log cos sin C x x + + Putting it in (1), we get 1 2 C C 1 + + log cos sin 1 tan 2 2 2 2 dx x x x x = + + + ∫ = 1 2 C C +1 log cos sin 2 2 2 2 x x x + + + = 1 2 C C +1 log cos sin C C 2 2 2 2 x x x ,  + + = +     EXERCISE 7 2 Integrate the functions in Exercises 1 to 37: 1
1	3483-3486	(iii) cos 1 tan cos sin dx x dx x x x = + + ∫ ∫ = 1 (cos + sin + cos – sin ) 2 cos sin x x x x dx x x + ∫ 304 MATHEMATICS = 1 1 cos – sin 2 2 cos sin x x dx dx x x + + ∫ ∫ = C1 1 cos sin 2 2 2 cos sin x x – x dx x x + + + ∫ (1) Now, consider cos sin I cos sin x – x dx x x = + ∫ Put cos x + sin x = t so that (cos x – sin x) dx = dt Therefore 2 I log C dt t t = = + ∫ = 2 log cos sin C x x + + Putting it in (1), we get 1 2 C C 1 + + log cos sin 1 tan 2 2 2 2 dx x x x x = + + + ∫ = 1 2 C C +1 log cos sin 2 2 2 2 x x x + + + = 1 2 C C +1 log cos sin C C 2 2 2 2 x x x ,  + + = +     EXERCISE 7 2 Integrate the functions in Exercises 1 to 37: 1 2 2 1 +xx 2
1	3484-3487	(1) Now, consider cos sin I cos sin x – x dx x x = + ∫ Put cos x + sin x = t so that (cos x – sin x) dx = dt Therefore 2 I log C dt t t = = + ∫ = 2 log cos sin C x x + + Putting it in (1), we get 1 2 C C 1 + + log cos sin 1 tan 2 2 2 2 dx x x x x = + + + ∫ = 1 2 C C +1 log cos sin 2 2 2 2 x x x + + + = 1 2 C C +1 log cos sin C C 2 2 2 2 x x x ,  + + = +     EXERCISE 7 2 Integrate the functions in Exercises 1 to 37: 1 2 2 1 +xx 2 ( ) 2 log x x 3
1	3485-3488	2 Integrate the functions in Exercises 1 to 37: 1 2 2 1 +xx 2 ( ) 2 log x x 3 1 log x x x + 4
1	3486-3489	2 2 1 +xx 2 ( ) 2 log x x 3 1 log x x x + 4 sin xsin (cos ) x 5
1	3487-3490	( ) 2 log x x 3 1 log x x x + 4 sin xsin (cos ) x 5 sin ( ) cos ( ) ax b ax b + + 6
1	3488-3491	1 log x x x + 4 sin xsin (cos ) x 5 sin ( ) cos ( ) ax b ax b + + 6 ax +b 7
1	3489-3492	sin xsin (cos ) x 5 sin ( ) cos ( ) ax b ax b + + 6 ax +b 7 2 x x + 8
1	3490-3493	sin ( ) cos ( ) ax b ax b + + 6 ax +b 7 2 x x + 8 2 1 2 x x + 9
1	3491-3494	ax +b 7 2 x x + 8 2 1 2 x x + 9 2 (4 2) 1 x x x + + + 10
1	3492-3495	2 x x + 8 2 1 2 x x + 9 2 (4 2) 1 x x x + + + 10 1 x – x 11
1	3493-3496	2 1 2 x x + 9 2 (4 2) 1 x x x + + + 10 1 x – x 11 4 x x + , x > 0 12
1	3494-3497	2 (4 2) 1 x x x + + + 10 1 x – x 11 4 x x + , x > 0 12 1 3 5 3 ( x –1) x 13
1	3495-3498	1 x – x 11 4 x x + , x > 0 12 1 3 5 3 ( x –1) x 13 2 3 3 (2 3 ) x +x 14
1	3496-3499	4 x x + , x > 0 12 1 3 5 3 ( x –1) x 13 2 3 3 (2 3 ) x +x 14 1 x(log )m x , x > 0, ≠1 m 15
1	3497-3500	1 3 5 3 ( x –1) x 13 2 3 3 (2 3 ) x +x 14 1 x(log )m x , x > 0, ≠1 m 15 2 9 4 x – x 16
1	3498-3501	2 3 3 (2 3 ) x +x 14 1 x(log )m x , x > 0, ≠1 m 15 2 9 4 x – x 16 2 3 ex + 17
1	3499-3502	1 x(log )m x , x > 0, ≠1 m 15 2 9 4 x – x 16 2 3 ex + 17 x2 x e INTEGRALS 305 18
1	3500-3503	2 9 4 x – x 16 2 3 ex + 17 x2 x e INTEGRALS 305 18 1 2 1 tan– x e +x 19
1	3501-3504	2 3 ex + 17 x2 x e INTEGRALS 305 18 1 2 1 tan– x e +x 19 2 2 1 1 x x e – e + 20
1	3502-3505	x2 x e INTEGRALS 305 18 1 2 1 tan– x e +x 19 2 2 1 1 x x e – e + 20 2 2 2 2 x – x x – x e – e e e + 21
1	3503-3506	1 2 1 tan– x e +x 19 2 2 1 1 x x e – e + 20 2 2 2 2 x – x x – x e – e e e + 21 tan2 (2x – 3) 22
1	3504-3507	2 2 1 1 x x e – e + 20 2 2 2 2 x – x x – x e – e e e + 21 tan2 (2x – 3) 22 sec2 (7 – 4x) 23
1	3505-3508	2 2 2 2 x – x x – x e – e e e + 21 tan2 (2x – 3) 22 sec2 (7 – 4x) 23 1 2 sin 1 – x – x 24
1	3506-3509	tan2 (2x – 3) 22 sec2 (7 – 4x) 23 1 2 sin 1 – x – x 24 2cos 3sin 6cos 4sin x – x x x + 25
1	3507-3510	sec2 (7 – 4x) 23 1 2 sin 1 – x – x 24 2cos 3sin 6cos 4sin x – x x x + 25 2 2 1 cos (1 tan ) x – x 26
1	3508-3511	1 2 sin 1 – x – x 24 2cos 3sin 6cos 4sin x – x x x + 25 2 2 1 cos (1 tan ) x – x 26 cos x x 27
1	3509-3512	2cos 3sin 6cos 4sin x – x x x + 25 2 2 1 cos (1 tan ) x – x 26 cos x x 27 sin 2 cos 2 x x 28
1	3510-3513	2 2 1 cos (1 tan ) x – x 26 cos x x 27 sin 2 cos 2 x x 28 cos 1 sin x x + 29
1	3511-3514	cos x x 27 sin 2 cos 2 x x 28 cos 1 sin x x + 29 cot x log sin x 30
1	3512-3515	sin 2 cos 2 x x 28 cos 1 sin x x + 29 cot x log sin x 30 sin 1 cos x x + 31
1	3513-3516	cos 1 sin x x + 29 cot x log sin x 30 sin 1 cos x x + 31 ( ) 2 sin 1 cos x x + 32
1	3514-3517	cot x log sin x 30 sin 1 cos x x + 31 ( ) 2 sin 1 cos x x + 32 1 1 cot x + 33
1	3515-3518	sin 1 cos x x + 31 ( ) 2 sin 1 cos x x + 32 1 1 cot x + 33 1 1 –tan x 34
1	3516-3519	( ) 2 sin 1 cos x x + 32 1 1 cot x + 33 1 1 –tan x 34 tan sin cos x x x 35
1	3517-3520	1 1 cot x + 33 1 1 –tan x 34 tan sin cos x x x 35 ( ) 2 1 log x x + 36

1

3418-3421

INTEGRALS 299 10 The process of differentiation and integration are inverses of each other as discussed in Section 7 2 2 (i)

1

3419-3422

The process of differentiation and integration are inverses of each other as discussed in Section 7 2 2 (i) EXERCISE 7

1

3420-3423

2 2 (i) EXERCISE 7 1 Find an anti derivative (or integral) of the following functions by the method of inspection

1

3421-3424

2 (i) EXERCISE 7 1 Find an anti derivative (or integral) of the following functions by the method of inspection 1

1

3422-3425

EXERCISE 7 1 Find an anti derivative (or integral) of the following functions by the method of inspection 1 sin 2x 2

1

3423-3426

1 Find an anti derivative (or integral) of the following functions by the method of inspection 1 sin 2x 2 cos 3x 3

1

3424-3427

1 sin 2x 2 cos 3x 3 e 2x 4

1

3425-3428

sin 2x 2 cos 3x 3 e 2x 4 (ax + b)2 5

1

3426-3429

cos 3x 3 e 2x 4 (ax + b)2 5 sin 2x – 4 e3x Find the following integrals in Exercises 6 to 20: 6

1

3427-3430

e 2x 4 (ax + b)2 5 sin 2x – 4 e3x Find the following integrals in Exercises 6 to 20: 6 (4 3 ex+ 1) dx ∫ 7

1

3428-3431

(ax + b)2 5 sin 2x – 4 e3x Find the following integrals in Exercises 6 to 20: 6 (4 3 ex+ 1) dx ∫ 7 2 (1–12 ) x dx x ∫ 8

1

3429-3432

sin 2x – 4 e3x Find the following integrals in Exercises 6 to 20: 6 (4 3 ex+ 1) dx ∫ 7 2 (1–12 ) x dx x ∫ 8 2 ( ) ax bx c dx + + ∫ 9

1

3430-3433

(4 3 ex+ 1) dx ∫ 7 2 (1–12 ) x dx x ∫ 8 2 ( ) ax bx c dx + + ∫ 9 (22 x) x e dx + ∫ 10

1

3431-3434

2 (1–12 ) x dx x ∫ 8 2 ( ) ax bx c dx + + ∫ 9 (22 x) x e dx + ∫ 10 2 x –1 dx x       ∫ 11

1

3432-3435

2 ( ) ax bx c dx + + ∫ 9 (22 x) x e dx + ∫ 10 2 x –1 dx x       ∫ 11 3 2 2 5 4 x x – dx +x ∫ 12

1

3433-3436

(22 x) x e dx + ∫ 10 2 x –1 dx x       ∫ 11 3 2 2 5 4 x x – dx +x ∫ 12 3 3 4 x x dx x + + ∫ 13

1

3434-3437

2 x –1 dx x       ∫ 11 3 2 2 5 4 x x – dx +x ∫ 12 3 3 4 x x dx x + + ∫ 13 3 2 1 1 x x x – dx −x – + ∫ 14

1

3435-3438

3 2 2 5 4 x x – dx +x ∫ 12 3 3 4 x x dx x + + ∫ 13 3 2 1 1 x x x – dx −x – + ∫ 14 (1 – x) x dx ∫ 15

1

3436-3439

3 3 4 x x dx x + + ∫ 13 3 2 1 1 x x x – dx −x – + ∫ 14 (1 – x) x dx ∫ 15 ( 32 2 3) x x x dx + + ∫ 16

1

3437-3440

3 2 1 1 x x x – dx −x – + ∫ 14 (1 – x) x dx ∫ 15 ( 32 2 3) x x x dx + + ∫ 16 (2 3cos x) x – x e dx + ∫ 17

1

3438-3441

(1 – x) x dx ∫ 15 ( 32 2 3) x x x dx + + ∫ 16 (2 3cos x) x – x e dx + ∫ 17 (22 3sin 5 ) x – x x dx + ∫ 18

1

3439-3442

( 32 2 3) x x x dx + + ∫ 16 (2 3cos x) x – x e dx + ∫ 17 (22 3sin 5 ) x – x x dx + ∫ 18 sec (sec tan ) x x x dx + ∫ 19

1

3440-3443

(2 3cos x) x – x e dx + ∫ 17 (22 3sin 5 ) x – x x dx + ∫ 18 sec (sec tan ) x x x dx + ∫ 19 2 sec2 cosec x dx x ∫ 20

1

3441-3444

(22 3sin 5 ) x – x x dx + ∫ 18 sec (sec tan ) x x x dx + ∫ 19 2 sec2 cosec x dx x ∫ 20 2 2 – 3sin cos x x ∫ dx

1

3442-3445

sec (sec tan ) x x x dx + ∫ 19 2 sec2 cosec x dx x ∫ 20 2 2 – 3sin cos x x ∫ dx Choose the correct answer in Exercises 21 and 22

1

3443-3446

2 sec2 cosec x dx x ∫ 20 2 2 – 3sin cos x x ∫ dx Choose the correct answer in Exercises 21 and 22 21

1

3444-3447

2 2 – 3sin cos x x ∫ dx Choose the correct answer in Exercises 21 and 22 21 The anti derivative of 1 x x   +     equals (A) 1 1 3 2 1 2 C 3 x x + + (B) 2 2 23 1 C 3 2 x x + + (C) 3 1 2 2 2 2 C 3 x x + + (D) 3 1 2 2 3 1 C 2 2 x x + + 22

1

3445-3448

Choose the correct answer in Exercises 21 and 22 21 The anti derivative of 1 x x   +     equals (A) 1 1 3 2 1 2 C 3 x x + + (B) 2 2 23 1 C 3 2 x x + + (C) 3 1 2 2 2 2 C 3 x x + + (D) 3 1 2 2 3 1 C 2 2 x x + + 22 If 3 34 ( ) 4 d f x x dx x = − such that f (2) = 0

1

3446-3449

21 The anti derivative of 1 x x   +     equals (A) 1 1 3 2 1 2 C 3 x x + + (B) 2 2 23 1 C 3 2 x x + + (C) 3 1 2 2 2 2 C 3 x x + + (D) 3 1 2 2 3 1 C 2 2 x x + + 22 If 3 34 ( ) 4 d f x x dx x = − such that f (2) = 0 Then f (x) is (A) 4 13 129 8 x +x − (B) 3 14 129 8 x +x + (C) 4 13 129 8 x +x + (D) 3 14 129 8 x +x − 300 MATHEMATICS 7

1

3447-3450

The anti derivative of 1 x x   +     equals (A) 1 1 3 2 1 2 C 3 x x + + (B) 2 2 23 1 C 3 2 x x + + (C) 3 1 2 2 2 2 C 3 x x + + (D) 3 1 2 2 3 1 C 2 2 x x + + 22 If 3 34 ( ) 4 d f x x dx x = − such that f (2) = 0 Then f (x) is (A) 4 13 129 8 x +x − (B) 3 14 129 8 x +x + (C) 4 13 129 8 x +x + (D) 3 14 129 8 x +x − 300 MATHEMATICS 7 3 Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions

1

3448-3451

If 3 34 ( ) 4 d f x x dx x = − such that f (2) = 0 Then f (x) is (A) 4 13 129 8 x +x − (B) 3 14 129 8 x +x + (C) 4 13 129 8 x +x + (D) 3 14 129 8 x +x − 300 MATHEMATICS 7 3 Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions It was based on inspection, i

1

3449-3452

Then f (x) is (A) 4 13 129 8 x +x − (B) 3 14 129 8 x +x + (C) 4 13 129 8 x +x + (D) 3 14 129 8 x +x − 300 MATHEMATICS 7 3 Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions It was based on inspection, i e

1

3450-3453

3 Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions It was based on inspection, i e , on the search of a function F whose derivative is f which led us to the integral of f

1

3451-3454

It was based on inspection, i e , on the search of a function F whose derivative is f which led us to the integral of f However, this method, which depends on inspection, is not very suitable for many functions

1

3452-3455

e , on the search of a function F whose derivative is f which led us to the integral of f However, this method, which depends on inspection, is not very suitable for many functions Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms

1

3453-3456

, on the search of a function F whose derivative is f which led us to the integral of f However, this method, which depends on inspection, is not very suitable for many functions Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms Prominent among them are methods based on: 1

1

3454-3457

However, this method, which depends on inspection, is not very suitable for many functions Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms Prominent among them are methods based on: 1 Integration by Substitution 2

1

3455-3458

Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms Prominent among them are methods based on: 1 Integration by Substitution 2 Integration using Partial Fractions 3

1

3456-3459

Prominent among them are methods based on: 1 Integration by Substitution 2 Integration using Partial Fractions 3 Integration by Parts 7

1

3457-3460

Integration by Substitution 2 Integration using Partial Fractions 3 Integration by Parts 7 3

1

3458-3461

Integration using Partial Fractions 3 Integration by Parts 7 3 1 Integration by substitution In this section, we consider the method of integration by substitution

1

3459-3462

Integration by Parts 7 3 1 Integration by substitution In this section, we consider the method of integration by substitution The given integral ( ) ∫f x dx can be transformed into another form by changing the independent variable x to t by substituting x = g (t)

1

3460-3463

3 1 Integration by substitution In this section, we consider the method of integration by substitution The given integral ( ) ∫f x dx can be transformed into another form by changing the independent variable x to t by substituting x = g (t) Consider I = ( ) f x dx ∫ Put x = g(t) so that dx dt = g′(t)

1

3461-3464

1 Integration by substitution In this section, we consider the method of integration by substitution The given integral ( ) ∫f x dx can be transformed into another form by changing the independent variable x to t by substituting x = g (t) Consider I = ( ) f x dx ∫ Put x = g(t) so that dx dt = g′(t) We write dx = g′(t) dt Thus I = ( ) ( ( )) ( ) f x dx f g t g t dt = ′ ∫ ∫ This change of variable formula is one of the important tools available to us in the name of integration by substitution

1

3462-3465

The given integral ( ) ∫f x dx can be transformed into another form by changing the independent variable x to t by substituting x = g (t) Consider I = ( ) f x dx ∫ Put x = g(t) so that dx dt = g′(t) We write dx = g′(t) dt Thus I = ( ) ( ( )) ( ) f x dx f g t g t dt = ′ ∫ ∫ This change of variable formula is one of the important tools available to us in the name of integration by substitution It is often important to guess what will be the useful substitution

1

3463-3466

Consider I = ( ) f x dx ∫ Put x = g(t) so that dx dt = g′(t) We write dx = g′(t) dt Thus I = ( ) ( ( )) ( ) f x dx f g t g t dt = ′ ∫ ∫ This change of variable formula is one of the important tools available to us in the name of integration by substitution It is often important to guess what will be the useful substitution Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following examples

1

3464-3467

We write dx = g′(t) dt Thus I = ( ) ( ( )) ( ) f x dx f g t g t dt = ′ ∫ ∫ This change of variable formula is one of the important tools available to us in the name of integration by substitution It is often important to guess what will be the useful substitution Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following examples Example 5 Integrate the following functions w

1

3465-3468

It is often important to guess what will be the useful substitution Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following examples Example 5 Integrate the following functions w r

1

3466-3469

Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following examples Example 5 Integrate the following functions w r t

1

3467-3470

Example 5 Integrate the following functions w r t x: (i) sin mx (ii) 2x sin (x2 + 1) (iii) 4 2 tan xsec x x (iv) 1 2 sin (tan ) 1 x– x + Solution (i) We know that derivative of mx is m

1

3468-3471

r t x: (i) sin mx (ii) 2x sin (x2 + 1) (iii) 4 2 tan xsec x x (iv) 1 2 sin (tan ) 1 x– x + Solution (i) We know that derivative of mx is m Thus, we make the substitution mx = t so that mdx = dt

1

3469-3472

t x: (i) sin mx (ii) 2x sin (x2 + 1) (iii) 4 2 tan xsec x x (iv) 1 2 sin (tan ) 1 x– x + Solution (i) We know that derivative of mx is m Thus, we make the substitution mx = t so that mdx = dt Therefore, 1 sin sin mx dx t dt m = ∫ ∫ = – 1 m cos t + C = – 1 m cos mx + C INTEGRALS 301 (ii) Derivative of x2 + 1 is 2x

1

3470-3473

x: (i) sin mx (ii) 2x sin (x2 + 1) (iii) 4 2 tan xsec x x (iv) 1 2 sin (tan ) 1 x– x + Solution (i) We know that derivative of mx is m Thus, we make the substitution mx = t so that mdx = dt Therefore, 1 sin sin mx dx t dt m = ∫ ∫ = – 1 m cos t + C = – 1 m cos mx + C INTEGRALS 301 (ii) Derivative of x2 + 1 is 2x Thus, we use the substitution x2 + 1 = t so that 2x dx = dt

1

3471-3474

Thus, we make the substitution mx = t so that mdx = dt Therefore, 1 sin sin mx dx t dt m = ∫ ∫ = – 1 m cos t + C = – 1 m cos mx + C INTEGRALS 301 (ii) Derivative of x2 + 1 is 2x Thus, we use the substitution x2 + 1 = t so that 2x dx = dt Therefore, 2 sin (2 1) sin x x dx t dt + = ∫ ∫ = – cos t + C = – cos (x2 + 1) + C (iii) Derivative of x is 21 1 1 2 2 – x x =

1

3472-3475

Therefore, 1 sin sin mx dx t dt m = ∫ ∫ = – 1 m cos t + C = – 1 m cos mx + C INTEGRALS 301 (ii) Derivative of x2 + 1 is 2x Thus, we use the substitution x2 + 1 = t so that 2x dx = dt Therefore, 2 sin (2 1) sin x x dx t dt + = ∫ ∫ = – cos t + C = – cos (x2 + 1) + C (iii) Derivative of x is 21 1 1 2 2 – x x = Thus, we use the substitution 1 so that giving 2 x t dx dt x = = dx = 2t dt

1

3473-3476

Thus, we use the substitution x2 + 1 = t so that 2x dx = dt Therefore, 2 sin (2 1) sin x x dx t dt + = ∫ ∫ = – cos t + C = – cos (x2 + 1) + C (iii) Derivative of x is 21 1 1 2 2 – x x = Thus, we use the substitution 1 so that giving 2 x t dx dt x = = dx = 2t dt Thus, 4 2 4 2 tan sec 2 tan sec x x t t t dt dx t x = ∫ ∫ = 4 2 2 tan tsec t dt ∫ Again, we make another substitution tan t = u so that sec2 t dt = du Therefore, 4 2 4 2 tan sec 2 t t dt u du = ∫ ∫ = 5 2 C 5 u + = 2 tan5 C 5 t + (since u = tan t) = 2 tan5 C (since ) 5 x t x + = Hence, 4 2 tan xsec x dx x ∫ = 2 tan5 C 5 x + Alternatively, make the substitution tan x t = (iv) Derivative of 1 2 1 tan 1 – x x = +

1

3474-3477

Therefore, 2 sin (2 1) sin x x dx t dt + = ∫ ∫ = – cos t + C = – cos (x2 + 1) + C (iii) Derivative of x is 21 1 1 2 2 – x x = Thus, we use the substitution 1 so that giving 2 x t dx dt x = = dx = 2t dt Thus, 4 2 4 2 tan sec 2 tan sec x x t t t dt dx t x = ∫ ∫ = 4 2 2 tan tsec t dt ∫ Again, we make another substitution tan t = u so that sec2 t dt = du Therefore, 4 2 4 2 tan sec 2 t t dt u du = ∫ ∫ = 5 2 C 5 u + = 2 tan5 C 5 t + (since u = tan t) = 2 tan5 C (since ) 5 x t x + = Hence, 4 2 tan xsec x dx x ∫ = 2 tan5 C 5 x + Alternatively, make the substitution tan x t = (iv) Derivative of 1 2 1 tan 1 – x x = + Thus, we use the substitution tan–1 x = t so that 2 1 dx x + = dt

1

3475-3478

Thus, we use the substitution 1 so that giving 2 x t dx dt x = = dx = 2t dt Thus, 4 2 4 2 tan sec 2 tan sec x x t t t dt dx t x = ∫ ∫ = 4 2 2 tan tsec t dt ∫ Again, we make another substitution tan t = u so that sec2 t dt = du Therefore, 4 2 4 2 tan sec 2 t t dt u du = ∫ ∫ = 5 2 C 5 u + = 2 tan5 C 5 t + (since u = tan t) = 2 tan5 C (since ) 5 x t x + = Hence, 4 2 tan xsec x dx x ∫ = 2 tan5 C 5 x + Alternatively, make the substitution tan x t = (iv) Derivative of 1 2 1 tan 1 – x x = + Thus, we use the substitution tan–1 x = t so that 2 1 dx x + = dt Therefore , 1 sin (tan2 ) sin 1 – x dx t dt x = + ∫ ∫ = – cos t + C = – cos (tan –1x) + C Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique

1

3476-3479

Thus, 4 2 4 2 tan sec 2 tan sec x x t t t dt dx t x = ∫ ∫ = 4 2 2 tan tsec t dt ∫ Again, we make another substitution tan t = u so that sec2 t dt = du Therefore, 4 2 4 2 tan sec 2 t t dt u du = ∫ ∫ = 5 2 C 5 u + = 2 tan5 C 5 t + (since u = tan t) = 2 tan5 C (since ) 5 x t x + = Hence, 4 2 tan xsec x dx x ∫ = 2 tan5 C 5 x + Alternatively, make the substitution tan x t = (iv) Derivative of 1 2 1 tan 1 – x x = + Thus, we use the substitution tan–1 x = t so that 2 1 dx x + = dt Therefore , 1 sin (tan2 ) sin 1 – x dx t dt x = + ∫ ∫ = – cos t + C = – cos (tan –1x) + C Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique These will be used later without reference

1

3477-3480

Thus, we use the substitution tan–1 x = t so that 2 1 dx x + = dt Therefore , 1 sin (tan2 ) sin 1 – x dx t dt x = + ∫ ∫ = – cos t + C = – cos (tan –1x) + C Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique These will be used later without reference (i) ∫tan = log sec + C x dx x We have sin tan cos x x dx dx x = ∫ ∫ 302 MATHEMATICS Put cos x = t so that sin x dx = – dt Then tan log C log cos C dt x dx – – t – x t = = + = + ∫ ∫ or tan log sec C x dx x = + ∫ (ii) ∫cot = log sin + C x dx x We have cos cot sin x x dx dx x = ∫ ∫ Put sin x = t so that cos x dx = dt Then cot dt x dx t = ∫ ∫ = log t +C = log sin C x + (iii) ∫sec = log sec + tan + C x dx x x We have sec (sec tan ) sec sec + tan x x x x dx dx x +x = ∫ ∫ Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt Therefore, sec log + C = log sec tan C dt x dx t x x t = = + + ∫ ∫ (iv) ∫cosec = log cosec – cot + C x dx x x We have cosec (cosec cot ) cosec (cosec cot ) x x x x dx dx x +x = + ∫ ∫ Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt So cosec – –log| | – log|cosec cot | C dt x dx t x x t = = = + + ∫ ∫ = 2 2 cosec cot – log C cosec cot x x x x − + − = log cosec cot C x – x + Example 6 Find the following integrals: (i) 3 2 sin xcos x dx ∫ (ii) sin sin ( ) x dx x +a ∫ (iii) 1 1 tan dx x ∫+ INTEGRALS 303 Solution (i) We have 3 2 2 2 sin cos sin cos (sin ) x x dx x x x dx = ∫ ∫ = 2 2 (1– cos ) cos (sin ) x x x dx ∫ Put t = cos x so that dt = – sin x dx Therefore, 2 2 sin cos (sin ) x x x dx ∫ = 2 2 (1– t) t dt −∫ = 3 5 2 4 ( – ) C 3 5 t t – t t dt – –   = +     ∫ = 3 5 1 1 cos cos C 3 5 – x x + + (ii) Put x + a = t

1

3478-3481

Therefore , 1 sin (tan2 ) sin 1 – x dx t dt x = + ∫ ∫ = – cos t + C = – cos (tan –1x) + C Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique These will be used later without reference (i) ∫tan = log sec + C x dx x We have sin tan cos x x dx dx x = ∫ ∫ 302 MATHEMATICS Put cos x = t so that sin x dx = – dt Then tan log C log cos C dt x dx – – t – x t = = + = + ∫ ∫ or tan log sec C x dx x = + ∫ (ii) ∫cot = log sin + C x dx x We have cos cot sin x x dx dx x = ∫ ∫ Put sin x = t so that cos x dx = dt Then cot dt x dx t = ∫ ∫ = log t +C = log sin C x + (iii) ∫sec = log sec + tan + C x dx x x We have sec (sec tan ) sec sec + tan x x x x dx dx x +x = ∫ ∫ Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt Therefore, sec log + C = log sec tan C dt x dx t x x t = = + + ∫ ∫ (iv) ∫cosec = log cosec – cot + C x dx x x We have cosec (cosec cot ) cosec (cosec cot ) x x x x dx dx x +x = + ∫ ∫ Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt So cosec – –log| | – log|cosec cot | C dt x dx t x x t = = = + + ∫ ∫ = 2 2 cosec cot – log C cosec cot x x x x − + − = log cosec cot C x – x + Example 6 Find the following integrals: (i) 3 2 sin xcos x dx ∫ (ii) sin sin ( ) x dx x +a ∫ (iii) 1 1 tan dx x ∫+ INTEGRALS 303 Solution (i) We have 3 2 2 2 sin cos sin cos (sin ) x x dx x x x dx = ∫ ∫ = 2 2 (1– cos ) cos (sin ) x x x dx ∫ Put t = cos x so that dt = – sin x dx Therefore, 2 2 sin cos (sin ) x x x dx ∫ = 2 2 (1– t) t dt −∫ = 3 5 2 4 ( – ) C 3 5 t t – t t dt – –   = +     ∫ = 3 5 1 1 cos cos C 3 5 – x x + + (ii) Put x + a = t Then dx = dt

1

3479-3482

These will be used later without reference (i) ∫tan = log sec + C x dx x We have sin tan cos x x dx dx x = ∫ ∫ 302 MATHEMATICS Put cos x = t so that sin x dx = – dt Then tan log C log cos C dt x dx – – t – x t = = + = + ∫ ∫ or tan log sec C x dx x = + ∫ (ii) ∫cot = log sin + C x dx x We have cos cot sin x x dx dx x = ∫ ∫ Put sin x = t so that cos x dx = dt Then cot dt x dx t = ∫ ∫ = log t +C = log sin C x + (iii) ∫sec = log sec + tan + C x dx x x We have sec (sec tan ) sec sec + tan x x x x dx dx x +x = ∫ ∫ Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt Therefore, sec log + C = log sec tan C dt x dx t x x t = = + + ∫ ∫ (iv) ∫cosec = log cosec – cot + C x dx x x We have cosec (cosec cot ) cosec (cosec cot ) x x x x dx dx x +x = + ∫ ∫ Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt So cosec – –log| | – log|cosec cot | C dt x dx t x x t = = = + + ∫ ∫ = 2 2 cosec cot – log C cosec cot x x x x − + − = log cosec cot C x – x + Example 6 Find the following integrals: (i) 3 2 sin xcos x dx ∫ (ii) sin sin ( ) x dx x +a ∫ (iii) 1 1 tan dx x ∫+ INTEGRALS 303 Solution (i) We have 3 2 2 2 sin cos sin cos (sin ) x x dx x x x dx = ∫ ∫ = 2 2 (1– cos ) cos (sin ) x x x dx ∫ Put t = cos x so that dt = – sin x dx Therefore, 2 2 sin cos (sin ) x x x dx ∫ = 2 2 (1– t) t dt −∫ = 3 5 2 4 ( – ) C 3 5 t t – t t dt – –   = +     ∫ = 3 5 1 1 cos cos C 3 5 – x x + + (ii) Put x + a = t Then dx = dt Therefore sin sin ( ) sin ( ) sin x t – a dx dt x a t = + ∫ ∫ = sin cos cos sin sin t a – t a dt t ∫ = cos – sin cot a dt a t dt ∫ ∫ = 1 (cos ) (sin ) log sin C a t – a t   +   = 1 (cos ) ( ) (sin ) log sin ( ) C a x a – a x a   + + +   = 1 cos cos (sin ) log sin ( ) C sin x a a a – a x a – a + + Hence, sin sin ( ) x dx x +a ∫ = x cos a – sin a log |sin (x + a)| + C, where, C = – C1 sin a + a cos a, is another arbitrary constant

1

3480-3483

(i) ∫tan = log sec + C x dx x We have sin tan cos x x dx dx x = ∫ ∫ 302 MATHEMATICS Put cos x = t so that sin x dx = – dt Then tan log C log cos C dt x dx – – t – x t = = + = + ∫ ∫ or tan log sec C x dx x = + ∫ (ii) ∫cot = log sin + C x dx x We have cos cot sin x x dx dx x = ∫ ∫ Put sin x = t so that cos x dx = dt Then cot dt x dx t = ∫ ∫ = log t +C = log sin C x + (iii) ∫sec = log sec + tan + C x dx x x We have sec (sec tan ) sec sec + tan x x x x dx dx x +x = ∫ ∫ Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt Therefore, sec log + C = log sec tan C dt x dx t x x t = = + + ∫ ∫ (iv) ∫cosec = log cosec – cot + C x dx x x We have cosec (cosec cot ) cosec (cosec cot ) x x x x dx dx x +x = + ∫ ∫ Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt So cosec – –log| | – log|cosec cot | C dt x dx t x x t = = = + + ∫ ∫ = 2 2 cosec cot – log C cosec cot x x x x − + − = log cosec cot C x – x + Example 6 Find the following integrals: (i) 3 2 sin xcos x dx ∫ (ii) sin sin ( ) x dx x +a ∫ (iii) 1 1 tan dx x ∫+ INTEGRALS 303 Solution (i) We have 3 2 2 2 sin cos sin cos (sin ) x x dx x x x dx = ∫ ∫ = 2 2 (1– cos ) cos (sin ) x x x dx ∫ Put t = cos x so that dt = – sin x dx Therefore, 2 2 sin cos (sin ) x x x dx ∫ = 2 2 (1– t) t dt −∫ = 3 5 2 4 ( – ) C 3 5 t t – t t dt – –   = +     ∫ = 3 5 1 1 cos cos C 3 5 – x x + + (ii) Put x + a = t Then dx = dt Therefore sin sin ( ) sin ( ) sin x t – a dx dt x a t = + ∫ ∫ = sin cos cos sin sin t a – t a dt t ∫ = cos – sin cot a dt a t dt ∫ ∫ = 1 (cos ) (sin ) log sin C a t – a t   +   = 1 (cos ) ( ) (sin ) log sin ( ) C a x a – a x a   + + +   = 1 cos cos (sin ) log sin ( ) C sin x a a a – a x a – a + + Hence, sin sin ( ) x dx x +a ∫ = x cos a – sin a log |sin (x + a)| + C, where, C = – C1 sin a + a cos a, is another arbitrary constant (iii) cos 1 tan cos sin dx x dx x x x = + + ∫ ∫ = 1 (cos + sin + cos – sin ) 2 cos sin x x x x dx x x + ∫ 304 MATHEMATICS = 1 1 cos – sin 2 2 cos sin x x dx dx x x + + ∫ ∫ = C1 1 cos sin 2 2 2 cos sin x x – x dx x x + + + ∫

1

3481-3484

Then dx = dt Therefore sin sin ( ) sin ( ) sin x t – a dx dt x a t = + ∫ ∫ = sin cos cos sin sin t a – t a dt t ∫ = cos – sin cot a dt a t dt ∫ ∫ = 1 (cos ) (sin ) log sin C a t – a t   +   = 1 (cos ) ( ) (sin ) log sin ( ) C a x a – a x a   + + +   = 1 cos cos (sin ) log sin ( ) C sin x a a a – a x a – a + + Hence, sin sin ( ) x dx x +a ∫ = x cos a – sin a log |sin (x + a)| + C, where, C = – C1 sin a + a cos a, is another arbitrary constant (iii) cos 1 tan cos sin dx x dx x x x = + + ∫ ∫ = 1 (cos + sin + cos – sin ) 2 cos sin x x x x dx x x + ∫ 304 MATHEMATICS = 1 1 cos – sin 2 2 cos sin x x dx dx x x + + ∫ ∫ = C1 1 cos sin 2 2 2 cos sin x x – x dx x x + + + ∫ (1) Now, consider cos sin I cos sin x – x dx x x = + ∫ Put cos x + sin x = t so that (cos x – sin x) dx = dt Therefore 2 I log C dt t t = = + ∫ = 2 log cos sin C x x + + Putting it in (1), we get 1 2 C C 1 + + log cos sin 1 tan 2 2 2 2 dx x x x x = + + + ∫ = 1 2 C C +1 log cos sin 2 2 2 2 x x x + + + = 1 2 C C +1 log cos sin C C 2 2 2 2 x x x ,  + + = +     EXERCISE 7

1

3482-3485

Therefore sin sin ( ) sin ( ) sin x t – a dx dt x a t = + ∫ ∫ = sin cos cos sin sin t a – t a dt t ∫ = cos – sin cot a dt a t dt ∫ ∫ = 1 (cos ) (sin ) log sin C a t – a t   +   = 1 (cos ) ( ) (sin ) log sin ( ) C a x a – a x a   + + +   = 1 cos cos (sin ) log sin ( ) C sin x a a a – a x a – a + + Hence, sin sin ( ) x dx x +a ∫ = x cos a – sin a log |sin (x + a)| + C, where, C = – C1 sin a + a cos a, is another arbitrary constant (iii) cos 1 tan cos sin dx x dx x x x = + + ∫ ∫ = 1 (cos + sin + cos – sin ) 2 cos sin x x x x dx x x + ∫ 304 MATHEMATICS = 1 1 cos – sin 2 2 cos sin x x dx dx x x + + ∫ ∫ = C1 1 cos sin 2 2 2 cos sin x x – x dx x x + + + ∫ (1) Now, consider cos sin I cos sin x – x dx x x = + ∫ Put cos x + sin x = t so that (cos x – sin x) dx = dt Therefore 2 I log C dt t t = = + ∫ = 2 log cos sin C x x + + Putting it in (1), we get 1 2 C C 1 + + log cos sin 1 tan 2 2 2 2 dx x x x x = + + + ∫ = 1 2 C C +1 log cos sin 2 2 2 2 x x x + + + = 1 2 C C +1 log cos sin C C 2 2 2 2 x x x ,  + + = +     EXERCISE 7 2 Integrate the functions in Exercises 1 to 37: 1

1

3483-3486

(iii) cos 1 tan cos sin dx x dx x x x = + + ∫ ∫ = 1 (cos + sin + cos – sin ) 2 cos sin x x x x dx x x + ∫ 304 MATHEMATICS = 1 1 cos – sin 2 2 cos sin x x dx dx x x + + ∫ ∫ = C1 1 cos sin 2 2 2 cos sin x x – x dx x x + + + ∫ (1) Now, consider cos sin I cos sin x – x dx x x = + ∫ Put cos x + sin x = t so that (cos x – sin x) dx = dt Therefore 2 I log C dt t t = = + ∫ = 2 log cos sin C x x + + Putting it in (1), we get 1 2 C C 1 + + log cos sin 1 tan 2 2 2 2 dx x x x x = + + + ∫ = 1 2 C C +1 log cos sin 2 2 2 2 x x x + + + = 1 2 C C +1 log cos sin C C 2 2 2 2 x x x ,  + + = +     EXERCISE 7 2 Integrate the functions in Exercises 1 to 37: 1 2 2 1 +xx 2

1

3484-3487

(1) Now, consider cos sin I cos sin x – x dx x x = + ∫ Put cos x + sin x = t so that (cos x – sin x) dx = dt Therefore 2 I log C dt t t = = + ∫ = 2 log cos sin C x x + + Putting it in (1), we get 1 2 C C 1 + + log cos sin 1 tan 2 2 2 2 dx x x x x = + + + ∫ = 1 2 C C +1 log cos sin 2 2 2 2 x x x + + + = 1 2 C C +1 log cos sin C C 2 2 2 2 x x x ,  + + = +     EXERCISE 7 2 Integrate the functions in Exercises 1 to 37: 1 2 2 1 +xx 2 ( ) 2 log x x 3

1

3485-3488

2 Integrate the functions in Exercises 1 to 37: 1 2 2 1 +xx 2 ( ) 2 log x x 3 1 log x x x + 4

1

3486-3489

2 2 1 +xx 2 ( ) 2 log x x 3 1 log x x x + 4 sin xsin (cos ) x 5

1

3487-3490

( ) 2 log x x 3 1 log x x x + 4 sin xsin (cos ) x 5 sin ( ) cos ( ) ax b ax b + + 6

1

3488-3491

1 log x x x + 4 sin xsin (cos ) x 5 sin ( ) cos ( ) ax b ax b + + 6 ax +b 7

1

3489-3492

sin xsin (cos ) x 5 sin ( ) cos ( ) ax b ax b + + 6 ax +b 7 2 x x + 8

1

3490-3493

sin ( ) cos ( ) ax b ax b + + 6 ax +b 7 2 x x + 8 2 1 2 x x + 9

1

3491-3494

ax +b 7 2 x x + 8 2 1 2 x x + 9 2 (4 2) 1 x x x + + + 10

1

3492-3495

2 x x + 8 2 1 2 x x + 9 2 (4 2) 1 x x x + + + 10 1 x – x 11

1

3493-3496

2 1 2 x x + 9 2 (4 2) 1 x x x + + + 10 1 x – x 11 4 x x + , x > 0 12

1

3494-3497

2 (4 2) 1 x x x + + + 10 1 x – x 11 4 x x + , x > 0 12 1 3 5 3 ( x –1) x 13

1

3495-3498

1 x – x 11 4 x x + , x > 0 12 1 3 5 3 ( x –1) x 13 2 3 3 (2 3 ) x +x 14

1

3496-3499

4 x x + , x > 0 12 1 3 5 3 ( x –1) x 13 2 3 3 (2 3 ) x +x 14 1 x(log )m x , x > 0, ≠1 m 15

1

3497-3500

1 3 5 3 ( x –1) x 13 2 3 3 (2 3 ) x +x 14 1 x(log )m x , x > 0, ≠1 m 15 2 9 4 x – x 16

1

3498-3501

2 3 3 (2 3 ) x +x 14 1 x(log )m x , x > 0, ≠1 m 15 2 9 4 x – x 16 2 3 ex + 17

1

3499-3502

1 x(log )m x , x > 0, ≠1 m 15 2 9 4 x – x 16 2 3 ex + 17 x2 x e INTEGRALS 305 18

1

3500-3503

2 9 4 x – x 16 2 3 ex + 17 x2 x e INTEGRALS 305 18 1 2 1 tan– x e +x 19

1

3501-3504

2 3 ex + 17 x2 x e INTEGRALS 305 18 1 2 1 tan– x e +x 19 2 2 1 1 x x e – e + 20

1

3502-3505

x2 x e INTEGRALS 305 18 1 2 1 tan– x e +x 19 2 2 1 1 x x e – e + 20 2 2 2 2 x – x x – x e – e e e + 21

1

3503-3506

1 2 1 tan– x e +x 19 2 2 1 1 x x e – e + 20 2 2 2 2 x – x x – x e – e e e + 21 tan2 (2x – 3) 22

1

3504-3507

2 2 1 1 x x e – e + 20 2 2 2 2 x – x x – x e – e e e + 21 tan2 (2x – 3) 22 sec2 (7 – 4x) 23

1

3505-3508

2 2 2 2 x – x x – x e – e e e + 21 tan2 (2x – 3) 22 sec2 (7 – 4x) 23 1 2 sin 1 – x – x 24

1

3506-3509

tan2 (2x – 3) 22 sec2 (7 – 4x) 23 1 2 sin 1 – x – x 24 2cos 3sin 6cos 4sin x – x x x + 25

1

3507-3510

sec2 (7 – 4x) 23 1 2 sin 1 – x – x 24 2cos 3sin 6cos 4sin x – x x x + 25 2 2 1 cos (1 tan ) x – x 26

1

3508-3511

1 2 sin 1 – x – x 24 2cos 3sin 6cos 4sin x – x x x + 25 2 2 1 cos (1 tan ) x – x 26 cos x x 27

1

3509-3512

2cos 3sin 6cos 4sin x – x x x + 25 2 2 1 cos (1 tan ) x – x 26 cos x x 27 sin 2 cos 2 x x 28

1

3510-3513

2 2 1 cos (1 tan ) x – x 26 cos x x 27 sin 2 cos 2 x x 28 cos 1 sin x x + 29

1

3511-3514

cos x x 27 sin 2 cos 2 x x 28 cos 1 sin x x + 29 cot x log sin x 30

1

3512-3515

sin 2 cos 2 x x 28 cos 1 sin x x + 29 cot x log sin x 30 sin 1 cos x x + 31

1

3513-3516

cos 1 sin x x + 29 cot x log sin x 30 sin 1 cos x x + 31 ( ) 2 sin 1 cos x x + 32

1

3514-3517

cot x log sin x 30 sin 1 cos x x + 31 ( ) 2 sin 1 cos x x + 32 1 1 cot x + 33

1

3515-3518

sin 1 cos x x + 31 ( ) 2 sin 1 cos x x + 32 1 1 cot x + 33 1 1 –tan x 34

1

3516-3519

( ) 2 sin 1 cos x x + 32 1 1 cot x + 33 1 1 –tan x 34 tan sin cos x x x 35

1

3517-3520

1 1 cot x + 33 1 1 –tan x 34 tan sin cos x x x 35 ( ) 2 1 log x x + 36