Chapter 5: Indefinite Integral PDF
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This document provides an introduction to indefinite integrals, outlining definitions, comparisons to differentiation, properties, and key integration formulae. It includes examples and integration methods. Specifically focusing on fundamental integration techniques, substitution methods.
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60 244 Indefinite Integral Introduction. E3 A function (x ) is called a primitive or an antiderivative of a function f (x ) if ' (x ) f (x ). x5 d x5 is a primitive of x 4 because 5 dx 5 For example, x4 Then d [ (x ) c] ' (x ) f (x ) dx ID Let (x ) be a primitive of...
60 244 Indefinite Integral Introduction. E3 A function (x ) is called a primitive or an antiderivative of a function f (x ) if ' (x ) f (x ). x5 d x5 is a primitive of x 4 because 5 dx 5 For example, x4 Then d [ (x ) c] ' (x ) f (x ) dx ID Let (x ) be a primitive of a function f (x ) and let c be any constant. [ ' (x ) f (x )] U (x ) c is also a primitive of f (x ). Thus, if a function f (x ) possesses a primitive, then it possesses infinitely many primitives D YG which are contained in the expression ( x ) c, where c is a constant. For example Note : If x5 x5 x5 , 2, 1 etc. are primitives of x 4. 5 5 5 F1 ( x ) and F2 (x ) are two antiderivatives of a function f (x) on an interval [a, b], then the difference between them is a constant. 5.1 Definition. U Let f (x ) be a function. Then the collection of all its primitives is called the indefinite integral f (x ) dx. ST of f (x ) and is denoted by Thus, d ( (x ) c) f (x ) f (x ) dx (x ) c dx where (x ) is primitive of f (x ) and c is an arbitrary constant known as the constant of integration. Here is the integral sign, f (x ) is the integrand, x is the variable of integration and dx is the element of integration. The process of finding an indefinite integral of a given function is called integration of the function. Indefinite Integral 245 It follows from the above discussion that integrating a function f (x ) means finding a function (x ) such that d ( (x )) f (x ). dx 60 5.2 Comparison between Differentiation and Integration. (1) Differentiation and integration both are operations on functions and each gives rise to a function. (2) Each function is not differentiable or integrable. E3 (3) The derivative of a function, if it exists, is unique. The integral of a function, if it exists, is not unique. ID (4) The derivative of a polynomial function decreases its degree by 1, but the integral of a polynomial function increases its degree by 1. (5) The derivative has a geometrical meaning, namely, the slope of the tangent to a curve at a point on it. The integral has also a geometrical meaning, namely, the area of some region. (6) The derivative is used in obtaining some physical quantities like velocity, acceleration U etc. of a particle. The integral is used in obtaining some physical quantities like centre of mass, momentum etc. D YG (7) Differentiation and integration are inverse of each other. 5.3 Properties of Integrals. (1) The differentiation of an integral is the integrand itself or the process of differentiation and integration neutralize each other, i.e., d dx f(x)dx f(x). (2) The integral of the product of a constant and a function is equal to the product of the U constant and the integral of the function, i.e., c f (x )dx c f (x )dx. (3) Integral of the sum or difference of two functions is equal to the sum or difference of f (x ) f (x )dx f (x )dx f (x )dx ST their integrals, i.e., 1 In k 1 2 1 2 the general form,. f1 (x ) k 2. f2 (x ) k 3. f3 (x ) .....dx k 1 f1 (x )dx k 2 f2 (x )dx k 3 f3 (x )dx ....... 5.4 Fundamental Integration Formulae. (1) (i) (iii) x n dx 1 x x n 1 d x n 1 xn c, n 1 n 1 dx n 1 dx 2 x c, (ii) dx x c (iv) (ax b )n dx 1 (ax b )n 1. c a n 1 246 Indefinite Integral e (4) (5) sin x dx cos x c (6) cos x dx sin x c (7) sec (8) cosec (9) sec x tan x dx sec x c dx e x c a x dx d ax ax dx log e a d (sin x ) cos x dx x dx tan x c 2 1 d ( cos x ) sin x dx x dx cot x c d (tan x ) sec 2 x dx d ( cot x ) cosec 2 x dx d (sec x ) sec x tan x dx d (cosec x ) cosec x cot x dx tan x dx log | cos x | c log | sec x | c d (log cos x ) tan x dx (12) cot x dx log | sin x | c log | cosec x | c d (log sin x ) cot x dx (13) sec x dx log | sec x tan x | c log tan 4 2 c (14) cosec x dx log | cosec x cot x | c log tan 2 c cosec x cot x dx cosec x c (11) D YG (10) U dx sin 1 x c cos 1 x c ST (15) (16) (17) , d x (e ) e x dx 1 ax b dx a (log | ax b | c ID 2 ax c log e a (ii) U x d 1 (log | x |) dx x 60 (3) E3 1 x dx log | x | c (2) (i) 1x 2 dx a x 2 dx 1 x 2 2 sin 1 x x c cos 1 c a a x x tan 1 x c cot 1 x c d log(sec x tan x ) sec x dx d (log | cosec x cot x |) cosec x dx d 1 , d (cos 1 x ) 1 (sin 1 x ) 2 dx dx 1x 1 x2 d 1 x sin dx a , d (cos 1 x ) 1 a x 2 2 dx a 1 a x2 2 d 1 (tan 1 x ) dx 1 x2 d 1 (cot 1 x ) dx 1 x2 (18) a 2 dx 1 x 1 x tan 1 c cot 1 c 2 a a a a x d a 1 x tan 2 dx a a x2 d 1 x a cot 2 dx a a x2 Indefinite Integral 247 dx (19) x (20) x x 1 2 sec 1 x c cosec 1 x c dx x a 2 2 x 1 x 1 sec 1 c cosec 1 c a a a a d 1 d 1 (sec 1 x ) (cosec 1 x ) 2 dx x x 1 dx x x 2 1 d x a d a 1 x (cosec 1 ) sec 2 2 dx a dx a x x a x x 2 a2 60 Note : In any of the fundamental integration formulae, if x is replaced by ax + b, then the same formulae is applicable but we must divide by coefficient of x or derivative of 1 (ax + b) i.e., a. In general, if f (x )dx (x ) c , then f (ax b) dx (ax b) c a sin(ax b) dx 1 cos(ax b) c, a E3 1 sec(ax b) dx a log | sec(ax b) tan( ax b)| c Some more results: a (iii) (iv) x a dx 2 x log {| x x 2 a 2 |} c sinh 1 c a x 2 a2 a 2 x 2 dx 1 1 x x a 2 x 2 a 2 sin 1 c 2 2 a x 2 a 2 dx when x a x log{| x x 2 a 2 |} c cos h 1 c a dx 2 1 1 1 1 x x x 2 a 2 a 2 log{ x x 2 a 2 } c x x 2 a 2 a 2 cosh 1 c 2 2 2 2 a x 2 a 2 dx 1 1 1 1 x x x 2 a 2 a 2 log{ x x 2 a 2 } c x x 2 a 2 a 2 sinh 1 c 2 2 2 2 a Important Tips ST (vii) 1 1 x 1 ax dx tanh 1 c log c, 2 a a 2a ax x 2 U (vi) when x a ID (ii) (v) 1 1 x 1 x a coth 1 c log c, 2 a a 2a x a a 2 U x D YG (i) etc. The signum function has an antiderivative on any interval which doesn't contain the point x = 0, and does not possess an anti-derivative on any interval which contains the point. The antiderivative of every odd function is an even function and vice-versa Example: 1 (x 1) 2 x (x 2 1) dx is equal to (b) log e x 2 tan 1 x (a) log e x Solution: (b) (x 1) 2 x (x 2 1) [MP PET 2003] dx x 2 1 2x x ((x 1) 2 dx x2 1 x (x 2 1) dx 2 (c) log e x (x x 2 1) 1 x 1 (d) log e { x (x 2 1)} 2 dx dx dx 2 log e x 2 tan 1 x 2 x x 1 248 Indefinite Integral (a) a log x Solution: (c) Example: 3 1 x x x2 x 1 x (b) a log x a x x 1 x x x2 Solution: (b) Example: 4 sin x 1 x 4 b 2 b c k x 3x3 (c) a log x b c c dx a log x 3 k 4 x 3x x (b) dx 2 (1 x )3 / 2 c 3 1 x 1 x x ( x 1 x) dx 1 x c (c) 1 x dx x cos 4 x dx [Rajasthan PET 2003] cos 2 x c 2 (sin x cos 4 x )dx (sin 2 x cos 2 x )(sin 2 x cos 2 x )dx (b) sin 2 x c 2 (c) D YG sin 2 x c 2 x 1 sin dx 4 U (d) x [Karnataka CET 2003] (c) x x (d) 8 cos sin c 8 8 1 x x sin cos c 8 8 8 2 x x x x cos 2 2 sin. cos dx sin 8 8 8 8 sin 8 cos 8 dx x x x (b) sin cos c 8 8 cos 2 x c 2 sin 2 x c 2 x 1 sin dx 4 x x (a) 8 sin cos c 8 8 Solution: (a) (d) 2(1 x )3 / 2 c 2 (1 x ) 3 / 2 c 3 (sin 2 x cos 2 x )dx (cos 2 x sin 2 x )dx cos 2 x dx Example: 5 (c) None of these [EAMCET 2003] (a) 4 b c k x 3x3 dx 1 1 x c 2 (a) Solution: (b) b c k 2 x 3x3 ax 3 bx 2 c dx x4 I dx equals to 60 x4 E3 ax 3 bx 2 c ID U Example: 2 x x sin cos 8 8 2 cos x / 8 sin x / 8 1/8 1/8 ST x x 8 sin cos c 8 8 5.5 Integration by Substitution (1) When integrand is a function i.e., f [( x )] '( x ) dx : Here, we put ( x ) t, so that ' (x )dx dt and in that case the integrand is reduced to f (t)dt. In this method, the integrand is broken into two factors so that one factor can be expressed in terms of the function whose differential coefficient is the second factor. Indefinite Integral 249 2 dx x (1 x ) I [Rajasthan PET 2002] 1 tan 1 ( x ) c 2 (a) Solution: (b) 2 I dx x (1 x ) 1t dt 2 (b) tan 1 ( x ) c 1 x t , put (c) 2 tan 1 ( x ) c dx dt 2 x tan 1 t c tan 1 (d) None of these 60 Example: 6 x c E3 (2) When integrand is the product of two factors such that one is the derivative of the f'( x ). f ( x ). dx others i.e., I (2 x 3 m 3 x 2m 6 x m ) (a) 6(m 1) 3m x 2m x m ) (2 x 2m 3 x m 6)1 / m dx , m 1 m ( x 0) = [IIT 2002] m (2 x 3 m 3 x 2m 6 x m ) m 1 (b) 6(m 1) m 1 m m (2 x 3 m 3 x 2m 6 x m ) m 1 (d) 6(m 1) D YG (2 x 3 m 3 x 2m 6 x m ) (c) 6(m 1) Solution: (a) (x For any natural number m U Example: 7 ID In this case we put f (x ) t and convert it into a standard integral. I (x 3 m 1 x 2m 1 x m 1 )(2 x 3m 3 x 2m 6 x m )1 / m dx 3m 3 x 2m 6 x m t 6 m (x 3 m 1 x 2m 1 x m 1 )dx dt Put 2 x 1 I 1 1 1/m 1 tm 1 t dt c t (m 1) / m c 6m 6 m 1 1 6(m 1) m U Again put the value of t, then we get option (a). x 2 tan 1 x 3 1 x6 dx is equal to ST Example: 8 (a) tan 1 (x 3 ) c Solution: (b). 3 x 2 dx dt x2 1x dx dt 1 3 3 1 x 6 x 5 1 / x dx k.5 1 / x c, then k is 3 (a) Solution: (a) 1 (tan 1 x 3 ) 2 c 6 (c) 1 (tan 1 x 3 ) 2 c 2 (d) 1 (tan 1 x 3 ) 2 c 2 (d) 2 log 5 Put tan 1 x 3 t 1 Example: 9 (b) 2 1 2 log 5 6 tdt 1 t 2 (tan 1 x 3 ) 2. c 3 2 6 2 (b) 2 log 5 Put x 2 t 2 x 3 dx dt x 3 dx (c) dt 2 2 log 5 250 Indefinite Integral 2 x 3 5 1 / x dx 2 1 1 1 1 5t. 5 1 / x c. On comparing, k 5 t dt c 2 log e 5 2 log e 5 2 2 log e 5 (3) Integral of a function of the form f (ax b ) : Here we put ax b t and convert it into tan(3 x 5) sec(3 x 5)dx Solution: (b) (b) sec x tan x dx sec x c sec(3 x 5) tan(3 x 5)dx 1 sec( 3 x 5) c 3 sec( 3 x 5 ) c 3 (4) If integral of a function of the form a sin 2 x 2 sin x b 2 cos 2 x 2 dx is equal to 1991] 1 log( a 2 sin 2 x b 2 cos 2 x ) c b a2 D YG (a) 2 (c) log( a 2 sin 2 x b 2 cos 2 x ) c Solution: (b) 2 sin 2 x 1 dx 2 sin 2 x b 2 cos 2 x (a b 2 ) 1e dx t dt 1 log( a 2 sin 2 x b 2 cos 2 x ) c a b2 2 (d) None of these 1 1 log( a 2 sin 2 x b 2 cos 2 x ) c log t c 2 a2 b 2 a b2 [Rajasthan PET 1990, 94; MP PET 1991; Roorkee 1977; SCRA 1999] U x (b) log(1 e x ) (a) log(1 e x ) 1e dx The value of the integral 1 x tan x dx (a) log | x cos x sin x | c (b) log | cos x x | + c ST Example: 13 I x x is equal to (c) log | cos x x sin x | c 1 x tan x dx cos x x sin x dx x x cos x Put (cos x x sin x ) t ( sin x x cos x sin x )dx dt x cos x dx dt I t dt (d) log( e x e 2 x ) I= x (c) log(1 e x ) dt e x log t log(1 e x ) dx , Put e x 1 t e x dx dt I t e 1 Solution: (b) Solution: (c) (b) [MNR 1987; Rajasthan PET Put a 2 sin 2 x b 2 cos 2 x t (a 2. 2 sin x cos x b 2. 2 cos x sin x ) dx dt sin 2 x (a 2 b 2 )dx dt a Example: 12 (d) None of these f'( x ) dx log[ f ( x )] c f(x) U Example: 11 (c) tan( 3 x 5) c E3 (a) sec( 3 x 5) c [MP PET 1988] ID Example: 10 1 f (x )dx (x ), then, f (ax b)dx a (ax b) +c 60 standard integral. Obviously if log | t | c log | cos x x sin x | c (d) None of these Indefinite Integral 251 f ( x )n 1 n f ( x ) f' ( x ) dx (5) If integral of a function of the form, x 1 x 2 dx (b) Solution: (d) Put 1 x 2 t 2 xdx dt , Example: 15 x 1 x 2 dx 1 2 2 9 2 sin 3 x c (b) 2 (2 sin 3 x )2 / 3 c 3 2 sin 3 x. cos 3 x dx 1 3 t dt tan x D YG tan x tan x. sin x cos x dx 2 (1 x 3 )3 / 2 c 9 ST (a) Solution: (c) x5 1 x 1 3 2 (c) 1 t c (d) tan x tan x sin x cos x 3 (t 1) dt t x 3.x 2 dx 1 x 1 3 dx sec 2 x dx tan x dt 2 t1 / 2 c 2 tan x c 2 2 (1 x 3 )3 / 2 (1 x 3 )1 / 2 c 9 3 (d) None of these dt 3 1 t3 / 2 2 2 t 1 dt 2 t c (1 x 3 )3 / 2 (1 x 3 )1 / 2 c 3 3 / 2 9 3 t (7) Standard substitutions Integrand form (i) a2 x 2 , 1 a x 2 2 , a2 x 2 +c sec x dx. Put 1 x 3 t 3 x 2 dx dt x 2 dx 3 2 [IIT 1975, 85] (b) 2 2 (1 x 3 )3 / 2 (1 x 3 )1 / 2 c 9 3 (c) 2 (2 sin 3 x )3 / 2 c 9 dx 2 f(x) c dx U 1 x3 f ( x) sin x. sec x Put t tan x dt sec 2 x dx , then it reduces to, x5 f'(x) (b) 2 tan x c tan x dx sin x. cos x (d) [MP PET 2002; Rajasthan PET 1985, 88; Bihar CEE 1974] (a) 2 sec x c Example: 17 2 (2 sin 3 x )3 / 2 c 3 [IIT 1976] U sin x. cos x dx 1 (1 x 2 ) 3 / 2 c 3 2 1 t3 / 2. c . (2 sin 3 x )3 / 2 c. 9 3 3/2 (6) If the integral of a function of the form, Solution: (b) (d) 1 1 t3/2. c (1 x 2 ) 3 / 2 c 3 2 3/2 (c) Put (2 sin 3 x ) t 3 cos 3 x dx dt Example: 16 t 1 / 2 dt 2 sin 3 x. cos 3 x dx (a) Solution: (d) (c) 3(1 x 2 )3 / 2 c 1 x2 c 60 1 x c 2 E3 1 2x 2 (a) [MP PET 1989] ID Example: 14 n 1 c n 1 Substitution x a sin , x a cos 252 Indefinite Integral , x 2 a2 x a x , x a x a , x (vii) ax , ax a x ax (viii ) x , x (a (a) Solution: (b) I x (a x ) dx x a sin 2 1 x (a x ) , x (a x ) x a sec 2 1 x (x a), x (x a) x a cos 2 x cos 2 sin 2 (x )( x ), ( ) x 2 )3 / 2 2 x a tan 2 1 is equal to x +c (a 2 x 2 )1 / 2 (b) x +c a 2 (a 2 x 2 )1 / 2 D YG Example: 18 x a sec or x a cosh x (a x ), ax , x x , ax (vi) 2 ax , x x , ax (v) 1 2 E3 x 2 a2 , (iv) x a ID (iii) x a tan or x a sinh , x 2 a2 2 2 (a 60 1 x 2 a2 , U (ii) (c) 1 +c a 2 (a 2 x 2 )1 / 2 [Rajasthan PET 2000] (d) None of these dx 2 x 2 )3 / 2 Put x a tan dx a sec 2 d I (a 1 a2 a sec 2 d a 2 tan 2 )3 / 2 d a sec 2 d a (sec 3 2 ) 3/2 sec a cos d I a 1 1 2 2 1 a2 sec 2 d sec 3 sin c x +c a 2 (x 2 a 2 )1 / 2 U I 2 Example: 19 1 x dx 1 x ST (a) sin 1 x Solution: (d) 1 1 1 x 2 C (b) sin 1 x 1 x 2 C (c) sin 1 x 1 x 2 C 2 2 Put x cos 2 , then I 2 [IIT 1971] 1 cos 1 (d) sin 1 x 1 x 2 C x dx 2 sin 2 d 2 1 cos 2. sin 2 d I 2 1 cos 2 2 sin 2 / 2 cos 2 . sin 2.d sin 2 C1 I 2 tan .2 sin cos d I 2.2 sin 2 d I 2 (1 cos 2 )d I 2 2 I 2 sin 2 C1 I cos 1 x 1 x 2 C1 I I sin 1 x 1 x 2 C 2 sin 1 x 1 x 2 C 1 [where C C1 2 ] Indefinite Integral 253 Trick : Rationalization of denominator and put 1 x 2 t 2. 5.6 Integration by Parts. (1) When integrand involves more than one type of functions : We may solve such integrals by a rule which is known as integration by parts. We know that, d (uv) udv vdu du u v dx u v dx {dx. vdx }dx I II i.e., the integral of the E3 If u and v are two functions of x, then 60 d dv du d (uv ) udv vdu (uv) u v dx dx dx product of two functions = (First function) × (Integral of second function) – Integral of {(Differentiation of first function) × (Integral of second function)} Integration with the help of above rule is called integration by parts. Before applying this rule ID proper choice of first and second function is necessary. Normally we use the following methods : (i) In the product of two functions, one of the function is not directly integrable (i.e., log | x |, sin 1 x , cos 1 x , tan 1 x......etc), then we take it as the first function and the remaining U function is taken as the second function. (ii) If there is no other function, then unity is taken as the second function e.g. In the sin 1 x dx , log x dx , 1 is taken as the second function. D YG integration of (iii) If both of the function are directly integrable then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable. Usually, we use the following preference order for the first function. (Inverse, Logarithmic, Algebraic, Trigonometric, exponential). This rule is simply called as “I LATE”. Important Tips x n. e ax dx , then In x n e ax n In 1 a a If In If In (log x ) dx , then In x log x x U ST (log x )2 (log x )3 1 ....................... dx , then In log(log x ) log x log x 2.(2 !) 3(3 !) If In If In (log x )n dx ; then In x (log x )n n. In 1 Successive integration by parts can be performed when one of the functions is x n (n is positive integer) which will be successively differentiated and the other is either of the following sin ax , cos ax , e ax , e ax , (x a)m which will be successively integrated. Chain rule : u. v dx u v 1 u' v 2 u"v 3 u v 4 ................. (1)n 1 u n 1 v n (1)n differential coefficient of u and v n stands for nth integral of v. u n. v n dx Where u n stands for nth 254 Indefinite Integral If In (log x )n dx , then In nI n1 (a) x (log x )n Solution: (a) [Karnataka CET 2003] (c) (log x )n 1 (b) (x log x )n (d) n(log x )n In (log x )n dx , In1 (log x )n1 dx Now In (log x )n.1 dx (log x )n. x n (log x )n 1. 1. x dx (log x )n. x n (log x )n1 dx x In x (log x )n n In 1 In n In 1 x (log x )n x sin x dx x cos x A, then A If (b) cos x constant (a) sin x constant Solution: (a) Since [MP PET 1992, 2000; Rajasthan PET 1997] E3 Example: 21 (c) Constant x dx [Rajasthan PET 1996, 2003; MP PET 1987, 97] (a) tan x log cos x c x sec Example: 23 [ f(x)g"(x) f "(x) g(x)]dx Solution: (c) f (x ) g' (x ) is equal to (b) f ' (x ) g(x ) f (x ) g' (x ) [MP PET 2001] (c) f (x ) g' (x ) f ' (x ) g(x ) (d) f (x ) g' (x ) f ' (x ) g(x ) f '(x) g'(x)dx ] [g(x) f '(x) g'(x)f '(x)dx ] f(x) g'(x) f '(x) g(x) sin 1 x cos 1 1 1 x cos x 2 [(2 x 1) sin 1 x x (1 x )] x c (2 x 1) sin x x (1 x ) x c ST (c) 2 1 sin 1 x cos 1 x 1 x cos 1 x sin dx is equal to [IIT 1983; WBJEE 1992] x U sin (a) Solution: (a) (d) [ f(x) g"(x) f "(x) g(x)]dx f(x) g"(x)dx f "(x) g(x)dx [ f (x ) g' (x ) Example: 24 x tan x log sec x c x dx x tan x tan x dx x tan x log(cos x ) c D YG (a) x2 sec 2 x log cos x c (c) 2 (b) Solution: (d) 2 ID 2 U x sec (d) None of these x sin x dx x cos x A x cos x sin x constant x cos x A Equating it, we get A sin x constant Example: 22 60 Example: 20 (b) dx 2 [(2 x 1) sin 1 x x (1 x )] x c (d) None of these 2 sin 1 x dx cos 1 x dx sin 1 x cos 1 x 2 Now we solve first and second expressions separately. For first expression, Put x sin 2 cos 2 1 2 x . sin 2 d cos 1 1 x dx dx sin 2 d cos 2 1 cos 2 sin 2 (2 x 1) 1 cos 2 d c1 sin 2 2 2 4 2 For second expression, sin x dx x 1 2 x (1 x ) c1 Indefinite Integral 255 Put x cos 2 dx 2 cos sin d sin 2 d 1 Therefore, I I 2 2 2 (2 x 1) {sin 1 2 sin 2 c2 4 x cos 1 (2 x 1) sin 1 x } x (1 x ) c x 2 dx (x sin x cos x )2 sin x cos x x sin x cos x (a) x sin x cos x x sin x cos x (b) (c) t 1 2 1 dt t 1 x . (x sin x cos x ) cos x (x sin x cos x ). 1 x 2 dx (x sin x cos x )2 cos x. 1 x ( sin x ) cos 2 x dx I U Integrate by parts, sin x x cos x x sin x cos x ID Differentiation of x sin x cos x is x cos x. Then, I I x x (1 x ) x c [MNR 1989; Rajasthan PET 2000] Solution: (c) (2 x 1) 2. x (1 x ) c I (2 x 1) sin 1 2 2 x E3 Example: 25 cos 2 x dx . sin 2 d 60 cos (d) None of these (x sin x cos x ) x cos x 2. x dx cos x 1 x. sec 2 x dx x sin x cos x cos x x x sin 2 x sin x cos x sin x cos x x (1 sin 2 x ) sin x x cos x 1 x sin x I I. (x sin x cos x ) cos x (x sin x cos x ) cos x x sin x cos x x sin x cos x cos x cos x D YG I (2) Integral is of the form e x f ( x ) f '( x )dx : If the integral is of the form e f (x ) f ' (x )dx, x then by breaking this integral into two integrals integrate one integral by parts and keeping other integral as it is, by doing so, we get e f (x ) f ' (x )dx e f (x ) c (ii) e mf (x ) f ' (x )dx e f (x ) c x (i) x mx e mx f (x ) f ' (x ) dx e mx f (x ) c m m U (iii) mx e x (1 cot x cot 2 x )dx ST Example: 26 [MP PET 2002] (a) e x cot x c Solution: (c) e Example: 27 e (a) Solution: (b) x x (b) e x cosec x c (cosec 2 x cot x )dx e x (c) e x cot x c (d) e x cosec x c [ cot x cosec 2 x ] e x cot x c 1 sin x dx 1 cos x e x tan 1 sin x 1 cos x [Rajasthan PET 2000] x +c 2 (b) e x cot 1 2 sin 2 x 2 x +c 2 (c) x x cos 2 2 1 cosec 2 x cot x x 2 2 2 2 sin 2 2 2 sin 1 x x e tan + c 2 2 (d) 1 x x e cot + c 2 2 256 Indefinite Integral x x x 1 sin x 1 ex dx e x cosec 2 cot dx e x cot c 2 2 2 1 cos x 2 Solution: (b) 1 x c (b) x e 2 tan Put tan 1 x t I I 1 c (d) None of these e 2t (1 tan t)2 dt = e 2 t (sec 2 t 2 tan t) dt 2 1 x c dx is equal to (b) (x 3)e x dx (x 4 )2 ex c x 3 (x 4 1)e x dx (x 4 ) 2 (c) of the 1 c (x 4 )2 (x 4 )e x ex dx dx 2 (x 4 ) (x 4 ) 2 ex c (x 4 )2 (d) ex dx (x 4 ) ex (x 4 ) 2 dx ex ex ex dx dx c c 2 2 x 4 (x 4 ) (x 4 ) D YG is [Karnataka CET 2000] ex c x 4 Integral x e 2t e 2t e 2 t sec 2 t dt 2 tan t sec 2 t. dt I e 2 t sec 2 t dt e 2 t tan t e 2 t sec 2 t dt c 2 2 ex (x 4 ) (3) 1 (x 3)e x (c) 2 x e tan U Solution: (a) c e 2 tan x (1 x )2 dx 1 x2 (x 4 ) (a) x 1 dx dt 1 x2 I e 2 t tan t c I x e 2 tan Example: 29 1 60 (a) x e tan [WB JEE 1998] E3 1 e 2 tan x (1 x )2 dx 1 x2 ID Example: 28 form [ x f ( x ) f(x )] dx : If the integral is of the form [x f ' (x) f (x)] dx then by breaking this integral into two integrals, integrate one integral by parts and keeping other integral as it is, by doing so, we get, [x f ' (x ) f (x )] dx x f (x ) c Example: 30 If d f (x ) x cos x sin x and f (0) 2, then f (x ) dx (b) x cos x sin x 2 U (a) x sin x Solution: (c) [MP PET 1989] (d) x cos x 2 (c) x sin x 2 d f (x ) x cos x sin x f (x ) (x cos x sin x ) dx x sin x c dx ST Since f (0 ) 2 c 2 , f (x ) x sin x 2 Example: 31 x sin x 1 cos x dx [Roorkee 1980; AMU 1997; UPSEAT 1999] (a) x tan x / 2 c Solution: (b) x sin x 1 cos x dx 2 x sec 1 (b) x tan x / 2 c 2 Trick : By inspection, (c) x tan x c 1 x tan x / 2 x x dx tan dx 2 2 2 1/2 1 x tan x c 2 tan 2 dx tan 2 dx x tan 2 c x x x d x x 1 x 2 sin x / 2 x sin x 2 x tan x tan c x sec 2 dx 2 2 2 2 cos x / 2 cos x / 2 1 cos x Hence the result. (4) Integrals of the form (d) e ax sin bxdx , e ax cos bx dx : Indefinite Integral 257 Working rule : To evaluate e ax sin bx dx or e ax cos bx dx , proceed as follows (i) Put the given integral equal to I. (ii) Integrate by parts, taking e ax as the first function. (iii) Again, integrate by parts taking e ax as the first function. This will involve I. Let I e ax sin bx dx. Then I e ax. sin bx dx e ax. I II 60 (iv) Transpose and collect terms involving I and then obtain the value of I. cos bx cos bx ae ax. dx b b 1 ax a ax 1 ax a e ax. sin bx sin bx e. cos bx ae ax. dx e. cos bx e. cos bx dx I II b b b b b b 1 ax a a2 1 ax a a2 e. cos bx 2 e ax. sin bx 2 I e. cos bx 2 e ax. sin bx 2 e ax. sin bx dx b b b b b b E3 e ax (a sin bx b cos bx ) c a2 b 2 e ax sin bx Similarly e ax. cos bx dx e ax a2 b 2 sin( bx tan 1 U Thus, ID a 2 e ax e ax I (a sin bx b cos bx ) c ( a sin bx b cos bx ) b2 b2 a2 b 2 e ax (a cos bx b sin bx ) c a2 b 2 D YG I I. b cos bx tan 1 c a a b e ax 2 ax 2 Note : e ax. sin(bx c)dx 2e 2 a sin(bx c) b cos(bx c) k a b e ax. cos(bx c) dx b ) c a e ax a cos(bx c) b sin( bx c) k a2 b 2 b sin (bx c) tan 1 k a a2 b 2 e ax b cos (bx c) tan 1 k a a2 b 2 e ax Important Tips x. e ax cos(bx c) dx U x e ax sin(bx c) dx 2 ST ax a. sin(bx c)dx (log a) a x. cos(bx c) dx Example: 32 If e x 2 b2 (log a) sin(bx c) b cos(bx c) k ax (log a) cos(bx c) b sin(bx c) k (log a)2 b 2 sin x dx 1 x e. a c, then a 2 e x sin x dx e x 1 1 2 [MP PET 1989,99; Rajasthan PET 1985; IIT 1978] (c) tan x c (b) cos x sin x (a) sin x cos x Solution: (a) x. e ax e ax a cos( bx c ) b sin( bx c ) [(a 2 b 2 ) cos(bx c) 2ab sin(bx c)] k a2 b 2 (a 2 b 2 )2 x ax x e ax a sin(bx c) b cos(bx c) 2 e 2 2 (a2 b 2 ) sin bx (bx c) 2ab cos bx (bx c) k 2 a b (a b ) 2 (1. sin x 1. cos x ) c Clearly a (sin x cos x ) x e (sin x cos x ) c 2 (d) None of these 258 Indefinite Integral If u e (a) 2 e Solution: (c) u ax cos bx dx and v e ax sin bx dx , then (a 2 b 2 ) (u 2 v 2 ) (b) (a 2 b 2 )e 2 ax ax sin bx a b b e ax cos bx dx e ax e ax. bx dx (d) (a 2 b 2 )e 2 ax (c) e 2 ax e ax sin bx a v b b bu av e ax sin bx......(i) Similarly, bv au e ax cos bx.....(ii) Squaring (i) and (ii) and adding. We get, (a 2 b 2 ) (u 2 v 2 ) e 2 ax. 60 Example: 33 ax (2) Integral of the form ax (3) Integral of the form (4) Integral of the form dx ax bx c 2 U (7) Integral of the form (px q). U px q ax 2 bx c ax dx. 2 f (x ) dx , where f (x ) is a polynomial of degree 2 or greater bx c x2 1 dx , x4 k x2 1 (6) Integral of the form (i) constant px q dx. bx c 2 D YG (5) Integral of the form than 2. dx , where ax 2 bx c can not be resolved into factors. bx c 2 ID (1) Integral of the form E3 5.7 Evaluation of the Various forms of Integrals by use of Standard Results. ax 2 bx c dx (ii) (8) x2 1 dx , x4 k x2 1 Integral where of k is any the form ax 2 bx c dx ST (9) Integral of the form dx P (1) Integrals of the form Q ax dx 2 bx c , where ax 2 bx c can not be resolved into factors. 2 2 b2 b b 2 4 ac b c b c We have, ax 2 bx c a x 2 . x a x 2 a x 2a 4 a 2 2a a a a 4a Case (i) : When b 2 4 ac 0 Indefinite Integral 259 ax dx 2 1 a bx c dx 2 b 2 4 ac b x 2 a 2 a 2 form , b b 2 4 ac 1 1 2a 2a . log c 2 2 a b 4 ac b b 4 ac 2. x 2a 2a 2a x 1 b 2 4 ac log x 2ax b b 2 4 ac dx 2 4 ac b 2 b x 2a 2a b x 1 2a tan 4 ac b 2 2a 1 1. a 4 ac b 2 2a c form , follows : x a dx 2 2 2ax b tan 1 c 4 ac b 2 4 ac b 2 2 ax dx 2 : To evaluate this form of integrals proceed as bx c D YG Working rule for evaluating 2 E3 bx c 1 a ID c U dx 2 2 2ax b b 2 4 ac Case (ii) : When b 2 4 ac 0 ax a dx 2 60 (i) Make the coefficient of x 2 unity by taking ‘a’ common from ax 2 bx c. (ii) Express the terms containing x 2 and x in the form of a perfect square by adding and subtracting the square of half of the coefficient of x. (iii) Put the linear expression in x equal to t and express the integrals in terms of t. dx dx or or 2 2 x a2 a form. After using the standard formulae, express the results in terms of x. U (iv) The resultant integrand will be either in 2x 2 1 (a) Solution: (d) I 7 1 2 2 dx x 1 x 2 dx x 1 2 2 a dx 2 x2 standard [Rajasthan PET 1997] 4x 1 c tan 1 7 ST Example: 34 x I (b) 1 2 1 2 7 4x 1 c (c) tan 1 7 4x 1 1 c tan 1 2 7 dx 7 1 x 4 4 2 2 1 x4 4x 1 2 1 4 1 I.. tan tan 1 c I c 2 7 7 ( 7 / 4) 7 Example: 35 sin 2 cos x dx equals x 4 sin x 5 (a) tan 1 (sin x ) c (b) tan 1 (sin x 2) c (c) tan 1 (sin x 1) c (d) None of these (d) 260 Indefinite Integral Solution: (b) I= sin 2 cos x dx x 4 sin x 5 (sin x 2) cos x 2 1 dx Put sin x 2 t cos x dx dt I= t dt 2 tan 1 t c tan 1 (sin x 2) c. 1 px q 60 px q is ax bx c bx c effected by breaking px q into two parts such that one part is the differential coefficient of the ax (2) Integral of the form dx : The integration of the function 2 If M and px q M N are two constants, E3 denominator and the other part is a constant. then we express 2 px q d (ax 2 bx c) N M.(2ax b) N (2aM)x Mb N. dx ID Comparing the coefficients of x and constant terms on both sides, we have, p 2aM M p b. 2a Thus, M and N are known. Hence, p p (2 ax b ) q b px q 2a 2a dx dx ax 2 bx c ax 2 bx c p 2ax b p dx p log | ax 2 bx c | dx q b 2 2 2a ax bx c 2a ax bx c 2a U and q Mb N N q Mb q D YG as the given integral p 2a is p dx q b C 2 2a ax bx c The integral on R.H.S. can be evaluated by the method discussed in previous section. (i) If b 2 4 ac 0, then (ii) b 2 4 ac 0, then If p (2aq bp) 2ax b b 2 4 ac log | ax 2 bx c | log k 2a 2a b 2 4 ac 2ax b b 2 4 ac U px q 2 px q p (2aq bp) 2ax b dx log | ax 2 bx c | tan 1 k 2 2a bx c a 4 ac b 4 ac b 2 2 bx c dx 2 sin 2 cos d 2 4 sin ST ax ax Example: 36 Solution: (a) 6 cos (a) 2 log | sin 2 4 sin 5 | 7 tan 1 (sin 2) c (b) 2 log | sin 2 4 sin 5 | 7 tan 1 (sin 2) c (c) 2 log | sin 2 4 sin 5 | 7 tan 1 (sin 2) c (d) 2 log | sin 2 4 sin 5 | 7 tan 1 (sin 2) c I 2(2 sin cos ) cos d I 2 ) 4 sin 6 (1 sin Put sin t cos d dt , I Let 4 t 1 M (4 sin 1) cos sin t 2 2 4 sin 5 d 4t 1 dt 4t 5 d 2 (t 4 t 5) N 4 t 1 M(2t 4) N dt Indefinite Integral 261 Comparing the coefficient of t and constant terms on both side, then M 2, N 7 I I2 t 2 2t 4 dt 4t 5 t 2(2 t 4 ) 7 t2 4t 5 dt dt 7 dt I 2 log | t 2 4 t 5 | 7 4t 5 (t 2)2 1 2 (3) Integral of the form 60 I 2 log | t 2 4 t 5 | 7 tan 1 (t 2) c 2 log | sin 2 4 sin 5 | 7 tan 1 (sin 2) c dx : To evaluate this form of integrals proceed as ax 2 bx c E3 follows : a common from ax 2 bx c. (i) Make the coefficient of x 2 unity by taking ax 2 bx c (ii) Put x 2 X 2 A 2 or 1 a dx. b c 2 x x a a ID dx Then, b c x , by the method of completing the square in the form, a a A 2 X 2 or U X 2 A 2 where, X is a linear function of x and A is a constant. consideration dx a x 2 2 D YG (iii) After this, use any of the following standard formulae according to the case under x sin 1 c a dx x a 2 2 a 0, b 2 4 ac 0, then If a 0, b 2 4 ac 0, then U Note : If ST If a 0, b 2 4 ac 0 Example: 37 dx 2 3x x 2 I= dx ax 2 bx c dx ax bx c 2 dx ax 2 bx c 1 a dx x a 2 dx 2 3x x 2 Put x 3 / 2 t dx dt log | x x 2 a 2 | c. 2ax b k. sin 1 2 a b 4 ac 2ax b 1 sinh 1 a 4 ac b 2 1 cosh 1 2ax b b 2 4 ac k k [EAMCET 1996] 2x 3 c (b) sec 1 17 2 equals 2x 3 c (a) tan 1 17 Solution: (c) log | x x 2 a 2 | c and dx 2 2 17 x 3 2 2 2x 3 c (c) sin 1 17 2x 3 c (d) cos 1 17 262 Indefinite Integral I 17 t2 2 dx x 4x 2 2 60 Solution: (a) 2 (a) log | x 2 x 2 2 4 x | c (b) log | x 2 x 2 2 4 x | c (c) log | x 2 x 2 2 4 x | c (d) log | x 2 x 2 2 4 x | c I dx (x 2) 2 I 2 dt t 2 ( 2 )2 E3 Example: 38 2x 3 t c sin 1 sin 1 17 c 17 / 2 dx Put x 2 t dx dt I log | t t 2 2 | c I log | x 2 x 2 4 x 2 | c px q ax 2 bx c dx : To evaluate this form of integrals, first we write, ID (4) Integral of the form d (ax 2 bx c) N px q M (2ax b) N dx Where M and N are constants. U px q M By equating the coefficients of x and constant terms on both sides, we get p bp and q bM N N q 2a 2a In this way, the integral breaks up into two parts given by px q ax bx c 2 Now, I1 p 2a dx p 2a 2ax b bp dx q 2a ax bx c 2 2ax b ax 2 bx c dx ax bx c 2 I1 I 2 , (say) dx ax 2 bx c (2ax b)dx dt, we U Putting 1/2 p p t p t 1 / 2 dt . C1 ax 2 bx c C 1 1 2a 2a a 2 ST I1 D YG p 2aM M and I 2 is calculated as in the previous section. Note : Example: 39 px q ax 2 bx c 5 2x 6 x x2 dx = p 2aq bp ax 2 bx c a 2a dx ax 2 bx c. dx = 2x 1 (a) 2 6 x x 2 4 sin 1 c 5 2x 1 (b) 2 6 x x 2 4 sin 1 c 5 2x 1 (c) 2 6 x x 2 4 sin 1 c 5 2x 1 (d) 2 6 x x 2 4 sin 1 c 5 have, Indefinite Integral 263 Solution: (b) 5 2x I 6 x x2 Let 5 2 x M dx d (6 x x 2 ) N 5 2 x M(1 2 x ) N dx Equating the coefficients of x and constant terms on both sides, we get Now, I1 (1 2 x ) 4 6xx 1 2x 6 x x2 dx 2 1 2x 6xx dx 4 2 dx dx Putting 6 x x 2 t (1 2 x )dx dt , we have, I dt t 2 t C1 2 6 x x 2 C1 and I2 dx 1 1 6 x2 x 4 4 dx dx 6xx 2 dx 6 (x 2 x ) du ID I1 I1 4 I 2 , (say) 6 x x2 E3 Hence, I 60 2 2 M M 1 and 5 M N N 5 1 4 5 2 x (1 2 x ) 4 25 1 x 4 2 2 2 5 u2 2 1 where, u x 2 U 2 1 u 2x 1 sin 1 C 2 sin 1 x C 2 sin 1 C2 5 / 2 5 2 5 2x 1 C (where, C C1 4 C2 ) 5 Example: 40 (a) (c) x2 x 2x 4 2 equals (x 1) x 2 2 x 4 3 sinh 1 c 3 (b) (x 1) x 2 2 x 4 3 sinh 1 c 3 (x 1) x 2 2 x 4 3 cosh 1 c 3 (d) (x 1) x 2 2 x 4 3 cosh 1 c 3 x2 dx U Solution: (a) D YG I I1 4 I 2 2 6 x x 2 4 sin 1 x 2x 4 2 x2 (x 1)2 3 dx 2 x 1 3 (x 1)2 3 2 dx x 1 x 2x 4 2 dx 3 dx (x 1)2 ( 3 ) 2 ST Put x 2 2 x 4 t 2 in the first expression 2(x 1)dx 2tdt (x 1) dx tdt x2 x 2x 4 x 2x 4 dx dx x 1 x 2 2 x 4 3 sinh 1 c 3 2 x2 2 (5) Integrals of the form than 2: tdt 3 t ax 2 dx (x 1)2 ( 3 )2 f(x) dx, where bx c f(x) is a polynomial of degree 2 or greater 264 Indefinite Integral To evaluate the integrals of the above form, divide the numerator by the denominator. Then, the integrals take the form given by R( x ) f (x ) dx Q(x ) 2 ax bx c ax bx c 2 where, Q(x ) is a polynomial and R(x ) is a linear polynomial in x. ax 2 f (x ) R( x ) dx Q(x )dx dx 2 bx c ax bx c 60 Then, we have x 2 2x 4 x3 (a) 3 dx equals. x2 2 (x 3 8 ) (x 1) Solution: (a) Example: 42 The value of x 2x 4 2 2x c dx [Rajasthan PET 1989] (b) x 3 x 2 2 x c 2x 3 3x 2 5 x 6 dx is x 2 3x 2 | x 1 (x 2)2 2x 3 3x 2 5 x 6 dx x 2 3x 2 = (2 x 3) x x 1 c x2 (b) x 3 x 2 4 ln | x 1 | 12 ln( x 2) c 8x dx = 3x 2 U 2 c (x 2 3 x ) 4 ln | x 2 3 x 2 | 12 ST (d) None of these (2 x 3) dx dx 12 2 (x 1) (x 2) 3x 2 (2 x 3) 4 x x 1 12 x 2 dx = (x 2 3 x ) 4 ln( x 1) 4 ln( x 2) 12 ln dx x 1 c x2 = x 2 3 x 8 ln(x 1) 16 ln(x 2) c = (x 2 3 x ) 8 ln (6) Integrals of the form I= (d) None of these D YG (c) (x 2 3 x ) 8 ln of the form (x 3 x 2 x ) c 3 (x 2)(x 2 2 x 4 )(x 1) x3 x2 2x c (x 2)(x 1) dx (x 2 x 2) dx = 2 3 2 x 2x 4 (a) x 2 3 x 4 ln | x 2 3 x 2 | 12 ln Solution: (c) (c) ID (x 3 8 ) (x 1) U Example: 41 E3 The integrals on R.H.S. can be obtained by the methods discussed earlier. x2 1 dx and x 4 kx 2 1 | x 1| c. (x 2) 2 x2 1 dx : x 4 kx 2 1 x2 1 dx , proceed as follows x 4 kx 2 1 To evaluate the integral Indefinite Integral 265 (i) Divide the numerator and denominator by x 2 to get I 1 x k 2 x 2 dt , which can be integrand as usual. 2k x2 1 dx , we divide the numerator and denominator by x 2 and get 4 2 x kx 1 dx 2 ID t 1 1 1 1 t 1 2 dx dt and x 2 2 2 t 2 x 2 2 t 2 2. x x x x Then, we put x t Thus, we have t 2 dt , which can be evaluated as usual. 2k U I dx. E3 (iii) To evaluate I 1 x2 1 x k 2 x 2 1 1 1 1 t 1 2 dx dt and x 2 2 2 t 2 x 2 2 t 2 2. x x x x Then, the given integral reduces to the form I 1 1 x2 60 (ii) Put x 1 D YG Important Tips Algebraic twins: x 2 4 1 dx 2x 2 x 1 4 x2 1 dx 4 1 x2 1 x dx 1 4 x2 1 dx , 4 1 x x x2 1 x 4 U We know the result of I1 1 dx x x2 1 x 4 2x 2 4 1 k x2 dx and I2 dx 1 dx , (x 4 2 dx 1 kx 2) x2 1 dx , so for 4 1 x x x2 4 1 dx and for x dx 4 1 , we can use the result of I1 I 2 I I2 and 1. 2 2 ST Trigonometric twins: Example: 43 x x2 1 4 x 1 2 1 x2 1 1 x 2 cot x dx , (sin dx 4 x cos x ) 4 , sin dx 6 x cos x 6 , sin x cos x a b sin x cos dx dx x2 x 1 1 log 2 c 2 x x 1 (a) Solution: (d) x2 1 tan x dx , dx (b) x2 x 1 1 log 2 c 2 x x 1 1 x 1 x2 dx log 2 2 1 x x 1 x 1 x2 x 1 (c) log 2 c x x 1 (d) x2 x 1 1 log 2 c 2 x x 1 1 1 1 x2 1 x 1 x2 x 1 x c log 2 c c log 2 1 2 2 x x 1 x 1 x 1 x Indefinite Integral 265 tan x dx tan x 1 1 tan x 2 tan x 1 tan 1 2 2 log tan x 2 tan x 1 c 2 tan x 1 (a) 2 tan x 1 1 tan x 2 tan x 1 tan 1 log c 2 tan x 2 tan x 1 2 tan x 2 2 1 (b) 60 Example: 44 tan x 1 c (c) log 2 tan x Let I tan x dx tan x t tan x t 2 , we have sec 2 xdx 2tdt (1 tan 2 x )dx 2tdt Putting 2t i.e., (1 t 4 )dx 2 tdt dx tan x dx 1 t 2 2t 1t 1 t 2 dt 1 t2 1 t 2 4 dt 2t 2 1t 1 4 dt 1 (t 2 1) (t 2 1) dt t4 1 t 2 dt 1 1 t2 t2 2 1 t 2 t dt D YG t. 1 U I dt 1 t4 ID Solution: (a) E3 (d) None of these Again putting t u 1 1 t 1 dt du t2 t2 1 t 4 1 dt t2 1 t 4 1 dt 1 t2 2 1 t 2 t dt. in the first integral and t v 1 1 t 1 dt dv t2 in the second integral, we have, I du 2 ( 2) 2 v dv 2 ( 2) 2 1 t t t 1 log tan 2 2 2 2 t 1 1 U u 1 2 Example: 45 1 2 2 t2 1 1 t 1 log t t 2 1 c c tan 1 1 2 t2 t 2 1 t 2 2 2 2 t tan x 1 1 tan x 2 tan x 1 tan 1 log c 2 tan x 2 2 tan x 2 tan x 1 ST u 1 log v 2 c tan 1 2 2 2 2 v 2 1 tan x cot x dx [IIT 1989; Roorkee 1984; Karnataka CET 1991; Pb. CET 1997] (a) (c) tan x cot x c 2 tan 1 2 tan x cot x c tan 1 2 (b) tan x cot x c 2 tan 1 2 (d) None of these 266 Indefinite Integral tan x cot x dx Put tan x t 2 sec 2 x dx 2 t dt dx I 2t dt 1 t4 1 1 1 2 dt 1 2 1 2t t t dt 2 dt 2 t . 2 1 t 1 t4 2 1 t 2 22 t 2 t t Put t 1 1 p 1 2 dt dp t t p 2 dp 2 2 2 tan 1 2 tan x cot x c 2 tan 1 2 c 2 tan 1 1 t t c 2 ax 2 bx c dx : To evaluate this form of integrals, express ax 2 bx c in the form a (x )2 2 ID (7) Integrals of the forms p 2 2 60 I= E3 Solution: (a) by the method of completing the square and apply the standard result discussed in the above section according to the case as may be. (a) (b) (2ax b) ax 2 bx c 4 ac b 2 4a 8a x 2 8 x 12 dx dx ax bx c 2 [CBSE 1999] D YG Example: 46 ax 2 bx c dx U Note : 1 (x 4 ) x 2 8 x 12 2 log| x 4 x 2 8 x 12 | c 2 1 (x 4 ) x 2 8 x 12 2 log| x 4 x 2 8 x 12 | c 2 (c) (x 4 ) x 2 8 x 12 log| x 4 x 2 8 x 12 | c (d) (x 4 ) x 2 8 x 12 log| x 4 x 2 8 x 12 | c x 8 x 12 dx (x 8 x 16) 4 dx (x 4) 2 dx t 2 dx t 2 Let I 2 U Solution: (b) 2 2 2 2 2 (putting x 4 t dx dt ) dt. ST 2 2 Example: 47 Solution: (a) 1 1 1 t t 2 2 2 .2 2 log | t t 2 2 2 | c (x 4 ) (x 4 )2 4 2 log | x 4 (x 4 )2 4 | c 2 2 2 1 (x 4 ) x 2 8 x 12 2 log | x 4 x 2 8 x 12 | c. 2 2ax x 2 dx (a) 1 1 x a (x a) 2 ax x 2 a 2 sin 1 c 2 2 a (b) 1 1 x a (x a) 2 ax x 2 a 2 sin 1 c 2 2 a (c) 1 1 x a (x a) 2ax x 2 a 2 cos 1 c 2 2 a (d) 1 1 x a (x a) 2ax x 2 a 2 cos 1 c 2 2 a 2ax x 2 dx a2 a2 2ax x 2 dx a2 (x 2 2ax a2 ) dx a 2 (x a)2 dx Indefinite Integral 267 1 1 (x a) (x a) 2ax x 2 a 2 sin 1 c 2 2 a (8) Integrals of the form (px q) ax 2 bx c dx : To evaluate this form of integral, proceed as follows: d (ax 2 bx c) N px q M (2ax b) N dx 60 (i) First express ( px q ) as px q M E3 Where, M and N are constant. (ii) Compare the coefficients of x and constant terms on both sides, will get p p and q Mb N N q Mb q p 2a M M b. 2a 2a (iii) Now, write the given integral p p (px q) ax 2 bx c dx (2ax b) ax 2 bx c dx q b ax 2 bx c dx 2a 2a p p I1 q b I 2 , (say). 2a 2a ID as Example: 48 (2 x 3) x 2 4 x 3 dx U (iv) To evaluate I1 , put ax 2 bx c t and to evaluate I 2 , follows the method discussed in (7) (b) log | (x 2) ( x 2 4 x 3) | c D YG (a) log | (x 2) ( x 2 4 x 3) | c (c) log | (x 2) ( x 2 4 x 3) | c Solution: (d) Let 2 x 3 M d ( x 2 4 x 3) N dx (d) None of these 2 x 3 M (2 x 4 ) N Equating the coefficients of x and constant terms on both sides, we get 2 2 M M 1 and 3 4 M N N 3 4 1 1 2 x 3 (2 x 4) 1 (2 x 4 ) U Hence, I [(2 x 4 ) 1] x 2 4 x 3dx (2 x 4 ) x 2 4 x 3 dx Now, I1 x 2 4 x 3 dx I1 I 2 , (say) x 2 4 x 3 dx ST Putting x 2 4 x 3 t (2 x 4 )dx dt , we have I1 t 1 / 2 dt 1 1 (x 2) (x 2)2 12 .12 log x 2 (x 2)2 12 2 2 I I1 I2 Example: 49 Solution: (d) 2 t3/2 c 1 ( x 2 4 x 3) 3 / 2 c 1 I 2 3 3/2 (2 x 5) x 2 4 x 3 dx c2 (x 2) 2 1 2 dx 1 1 (x 2) x 2 4 x 3 log x 2 x 2 4 x 3 c 2 2 2 2 2 1 1 (x 4 x 3) 3 / 2 (x 2) x 2 4 x 3 log x 2 x 2 4 x 3 c, 3 2 2 x 2 4 x 3 dx [CBSE 1995] (a) log | x 2 x 2 4 x 3 | c (b) log | x 2 x 2 4 x 3 | c (c) log | x 2 x 2 4 x 3 | c (d) None of these Let 2 x 5 M (where, c c1 c 2 ), d (x 2 4 x 3) N 2 x 5 M (2 x 4 ) N dx 268 Indefinite Integral Equating the coefficients of x and constant terms on both sides, we get 2 2M M 1 5 4 M N N 4 M 5 1 x 2 4 x 3 dx I (2 x 4 ) x 2 4 x 3 dx I1 (2 x 4 ) x 2 4 x 3 dx I (x 2)2 12 I2 1 (x 2) x 2 4 x 3 2 2 2 (x 4 x 3)3 / 2 c1 I2 3 x 2 4 x 3 dx x 2 4 x 3 dx I I1 I2 (say) 1 1 2 2 (x 2) x 2 4 x 3. 1 log (x 2) x 4 x 3 2 2 1 log (x 2) x 2 4 x 3 c 2 , Therefore I I1 I2 2 2 2 1 1 (x 4 x 3)3 / 2 (2) x 2 4 x 3 log (x 2) x 2 4 x 3 c 3 2 2 dx P (Where c c1 c2 ) , (where P and Q and linear or quadratic expressions in Q ID (9) Integrals of the form x 2 4 x 4 4 3 dx 60 (2 x 4) 1 E3 Hence, I x): To evaluate such types of integrals, we have following substitutions according to the nature of expressions of P and Q in x : (i) When Q is linear and P is linear or quadratic, we put Q t 2. Example: 50 ( x 3) dx (a) Solution: (c) D YG U 1 (ii) When P is linear and Q is quadratic, we put P . t 1 (iii) When both P and Q are quadratic, we put x . t x 1 x 1 2 1 c log x 1 2 2 (b) [Pb. CET 1991] x 1 2 1 c (c) 1 log x 1 2 c log x 1 2 x 1 2 2 2 (d) None of these Put x 1 t 2 dx 2t dt ( x 3) dx (t U x 1 t 2 dt 22 ST 2 2. x 1 t 2 t dt 2 4 )t x 1 2 1 1 t2 c. log c log x 1 2 2.2 2 t2 5.8 Integrals of the form dx a b cos x and dx a b sin x. To evaluate such form of integrals, proceed as follows: x x 1 tan 2 2 tan 2 and sin x 2. (1) Put cos x 2 x 2 x 1 tan 1 tan 2 2 (2) Replace 1 tan 2 x x in the numerator by sec 2. 2 2 2 x t2 1 x 3 t2 4 Indefinite Integral 269 (3) Put tan x 1 x t so that sec 2 dx dt. 2 2 2 (4) Now, evaluate the integral obtained which will be of the form at dt 2 bt c by the method discussed earlier. dx Case II: When a b , then Case III: When a b , then a2 b 2 dx a b cos x dx b 2 a2 1 x ba 2 c x b a tan b a 2 b a tan 1 log x a b cos x a tan 2 c. dx a b sin x Case I: When a b or a 0 and a b , then 2 dx a b sin x D YG 2 ab x tan 1 tan c ab 2 2 U (ii) dx a b cos x E3 Case I: When a b , then 60 a b cos x ID (i) Case II: When a 2 b 2 , then dx a b sin x 2 a2 b 2 tan 1 x a tan 2 b c a2 b 2 x b) ( b 2 a 2 ) 2 log c 2 2 x 2 2 b a (a tan b) b a 2 (a tan 1 Case III: When a 2 b 2 In this case, either b a or b a dx 1 x 1 (a) When b a , then cot c [tan x sec x ] c a b sin x a a 4 2 U ST (b) When b a , then Example: 51 1 x 3 4 cos x dx (a) (c) Solution: (b) dx a b sin x a tan 4 2 c. [Rajasthan PET 1997] tan( x / 2) 7 c log tan( x / 2) 7 7 (b) 7 tan( x / 2) c log 7 tan( x / 2) 7 (d) 1 1 If b a then, dx a b cos x b a tan log b a tan b 2 a2 1 tan( x / 2) 7 c log tan( x / 2) 7 7 1 7 tan( x / 2) c log 7 tan( x / 2) 7 1 x x ba tan 7 1 2 2 c log c I x x 7 ba tan 7 2 2 270 Indefinite Integral dx (c) a 4 then [Roorkee 1979] , b3 (b) a , b3 4 , b arbitrary constant (d) a , b arbitrary constant 4 4 If a b , then x a b sin x a cot 4 2 c dx 1 x x dx Hence a and b arbitrary constant. 4 5 4 cos x dx If a b, then [Roorkee1983; Rajasthan PET 1990, 97] 2 1 tan 1 tan x c 3 3 (a) dx 5 4 cos x 3 tan dx 2 1 1 tan 1 tan x c 3 3 (b) a b cos x 2 1 x tan c. 3 2 2 x 1 tan 1 tan c 3 3 2 dx dx a b cos x c sin x , a sin x b cos x D YG (1) Integral of the form (c) (d) 1 x 1 tan 1 tan c 3 3 2 ab x. tan 1 tan c ab 2 a b 2 2 1 5.9 Integrals of the form and c r sin . x = tan c 2 4 ID Example: 53 x 1 sin x cot 4 2 = tan 2 4 2 c = tan 4 2 c Solution: (c) 60 (a) a Solution: (d) x 1 sin x tan 2 a b, E3 If U Example: 52. dx a b cos x c sin x : To evaluate such integrals, we put So that, r 2 b 2 c 2 and tan 1 b r cos dx c dx . I a r cos( x ) a r(cos cos x sin sin x ) b Again, Put x t dx dt , we have I dt a r cos t ST U Which can be evaluated by the method discussed earlier. dx (2) Integral of the form : To evaluate this type of integrals we substitute a sin x b cos x a r cos , b r sin and so r a 2 b 2 , tan 1 So, dx a sin x b cos x Note b a 1 dx 1 cos ec(x )dx r sin( x ) r 1 x log tan r 2 2 b x 1 log tan tan 1 c a 2 2 a b 1 2 2 : The integral of the above form can be evaluated by using cos x sin x 2 tan x / 2. 1 tan 2 x / 2 1 tan 2 x / 2 and 1 tan 2 x / 2 Indefinite Integral 271 Important Tips If a b , a 2 b 2 c 2 , then If a b, dx 2 2 2 2 (a b) tan x / 2 c b 2 c 2 a 2 dx 1 k log a b cos x c sin x (a b) tan x / 2 c b 2 c 2 a 2 b 2 c 2 a2 (b a) tan x / 2 c b 2 c 2 a 2 k. log (b a) tan x / 2 c b 2 c 2 a 2 b c a 1 a b cos x c sin x Example: 54 (a b) tan x / 2 c tan 1 k a b c a 2 b 2 c 2 a b cos x c sin x dx 2 2 2 sin x cos x dx [BIT Ranhi 1990; Rajasthan PET 1990, 97, 99; Karnataka CET 1999] (b) log tan ( / 8 x / 2) c (a) log tan ( / 8 x / 2) c (c) 1 log tan ( / 8 x / 2) c (d) None of these 2 2 tan x / 2 1 tan x / 2 2 1 tan x / 2 1 tan 2 x / 2 Put tan x / 2 t I2 2t 1 t 1 sec 2 x / 2 dx dt 2 dt ( I2 2 2 2 (t I 1 dt 2 ) (t 1) 2 dt 2 2 t 1) where c1 2 2 1 1 c I 1 log 2 1 log ( 2 1) c I 2 2 [( 2 1) tan x / 2] [ 2 1] [( 2 1) tan x / 2] [ 2 1] c log tan( / 8 x / 2) c1 log ( 2 1) c 2 I : 1 2 dx 1 sin x 2 U 1 cos x 1 2 sin( / 4 x ) dx 1 2 cosec ( / 4 x) dx 2 log tan ( / 8 x / 2) c. 1 cos x sin x dx ST 2 2 t 1 2 2 1 ( 2 1) tan x / 2 Trick 1 2 t 1 log tan / 8 tan x / 2 log 2 sec 2 x / 2 dx 2 tan x / 2 1 tan 2 x / 2 D YG 2 ID dx U Solution: (c) 60 If a b, a 2 b 2 c 2 , then E3 Example: 55 (a) log 1 cot Solution: (c) Given I [EAMCET 2002] x c 2 (b) log 1 tan x c 2 (c) log 1 cot x c 2 1 cos x sin x I 1 (1 tan I 2 tan dx dx 2 2 tan x / 2 x / 2) (1 tan 2 x / 2) 1 tan 2 x / 2 sec 2 x / 2. dx 2 x / 2 2 tan x / 2 I 1 tan 2 1 / 2. sec 2 x / 2. dx tan 2 x / 2 tan x / 2 sec 2 x / 2. dx x / 2 1 tan 2 x / 2 2 tan x / 2 (d) log 1 tan x c 2 272 Indefinite Integral tan x / 2 t Put 1 1 1 sec 2 x / 2.dx dt 2 I therefore t dt t 2 I t (t 1) dt t t 1 dt I t 1 t dt dt Example: 56 The antiderivative of f (x ) 1 5 log 1 tan x / 2 5 3 (a) Solution: (b) y t 1 c t I log 1 whose graph passes through the point (0, 0) is 3 5 sin x 3 cos x (b) 1 5 log 1 tan x / 2 5 3 sec 2 x / 2 dx (3 5 sin x 3 cos x) 10 tan x / 2 6 dx tan x / 2 1 c I log 1 cot x / 2 c tan x / 2 60 I log( t 1) log t c I log 1 5 (c) log 1 cot x / 2 5 3 a b cos x , a b sin x , asin x b cos x , (a sin x b cos x) , a b sin x c cos dx dx dx 2 2 2 dx 2 dx 2 2 2 x U form ID 1 5 log 1 tan x / 2. 5 3 5.10 Integrals of the (d) None of these 1 5 1 log( 5 tan x / 2 3) c log tan x / 2 1 c 5 3 5 Passes through (0, 0) c 0 then y E3 To evaluate the above forms of integrals proceed as follows: D YG (1) Divide both the numerator and denominator by cos 2 x. (2) Replace sec 2 x in the denominator, if any by (1 tan 2 x ). (3) Put tan x t sec 2 xdx dt. (4) Now, evaluate the integral thus obtained, by the method discussed earlier. Example: 57 1 3 sin dx x 1 tan 1 (3 tan 2 x ) c 3 U (a) 2 Solution: (b) sec 2 x dx sec x 3 tan 2 x 2 (b) 1 tan 1 (2 tan x ) c 2 [Roorkee 1989; CBSE PMT 1992; DCE 2001] (c) tan 1 (tan x ) c (d) None of these sec 2 x dx 1 4 tan 2 x ST Put 2 tan x t sec 2 x dx I Example: 58 (a) Solution: (c) 1t 1 2 4 sin dt 2 dt 2 2 1 1 tan 1 t c I tan 1 (2 tan x ) c. 2 2 dx x 5 cos 2 x 2 tan x c tan 1 5 5 1 sec 2 x dx 4 tan 2 x 5 [AISSE 1986; CBSE PMT 1992] (b) tan x c tan 1 5 5 1 (c) 2 tan x c (d) None of these tan 1 2 5 5 1 Put 2 tan x t 2 sec 2 x dx dt Indefinite Integral 273 dt 2 ( 5) 2 (2 sin x cos x) 2 tan x c tan 1 2 5 5 1 dx dx (2 sin x cos x )2 1 log( 2 tan x 1) c 2 dx 2 sin x (2 cot x )2 Put (2 cot x ) t cosec 2 x dx dt Example: 60 1 log 2 3 1 (c) 1 3 tan x (2 cot x ) dt t 1 1 c c 2 cot x t log 2 3 1 3 tan x tan 1 ( 3 tan x ) c (d) None of these cos 3 x dx 4 cos cos x 3 cos x dx x 3 cos x Dividing the numerator and denominator by cos 2 x , we have I dt 1 1 3t 2 3 1 t dt 2 1 3 3 1 I log 2 3 Example: 61 4 sin 2 1 3 tan x 1 3 tan x t 1 1 1 1 3t. log 3 c log c 1 3 2. 1 2 3 1 3t t 3 3 (b) U 2 2 3 t dx equals x 4 sin x. cos x 5 cos 2 x 1 1 tan 1 tan x c 4 2 dx 2 4 sin x 4 sin x. cos x 5 cos 2 x ST I= dx 2 x 3 c. 1 (a) tan 1 tan x c 2 Solution: (b) 1 dt 4 cos 1 D YG I 1 1 c 2 2 tan x 1 c 1 3 tan x U Let I 1 (b) 6 Solution: (a) 2 2 c 1 3 tan x (d) cosec 2 x dx cos x dx cos 3 x (a) 1 c 2 cot x (c) 60 (b) E3 [MP PET 1994] 2 1 1 2 2 tan x 1 (a) Solution: (c) t ID Example: 59 1 2 Put tan x 1 1 t sec 2 x dx dt 4 2 5.11 Integrals of the form (1) Integrals of the form 4 tan t 1 (c) 4 tan 1 tan x c 2 sec 2 x dx 2 x 5 4 tan x 1 4 1 1 dt tan 1 t c I 4 4 1 2 a sin x b cos x and c sin x d cos x (d) None of these sec 2 x dx 2 1 tan x 1 2 tan 1 1 tan x c. 2 a sin x b cos x q c sin x d cos x r a sin x b cos x dx : Such rational functions of sin x and cos x may x c sin x d cos be integrated by expressing the numerator of the integrand as follows: Numerator = M (Diff. of denominator)+N (Denominator) 274 Indefinite Integral d (c sin x d cos x ) N (c sin x d cos x ) dx The arbitary constants M and N are determined by comparing the coefficients of sin x and cos x from two sides of the above identity. Then, the given integral is a sin x b cos x M (c cos x d sin x ) N (c sin x d cos x ) c cos x d sin x I dx dx M dx N 1dx c sin x d cos x c sin x d cos x c sin x d cos x M log | c sin x d cos x | Nx c. i.e., a sin x b cos x M (2) Integrals of the form a sin x b cos x q c sin x d cos x r 60 dx : To evaluate this type of integrals, we E3 express the numerator as follows: Numerator M (Denominat or) N (Different iation of denominato r) P i.e., (c sin x b cos x q) M (c sin x d cos x r) N (c cos x d sin x ) P. where M, N, P are constants to be determined by comparing the coefficients of sin x , cos x and P D YG U dx c sin x d cos x r. U Mx N log | Denominato r | ST dx c sin x d cos x r ID constant term on both sides. Diff.of denominato r a sin x b cos x q dx M dx N dx c sin x d cos x r Denominato r 274 Indefinite Integral Important Tips ac bd ad bc x 2 log | c cos x d sin x | c. 2 d2 c d2 a cos x b sin x c cos x d sin x dx c Solution: (a) 3 cos x 3 sin x 4 sin x 5 cos x dx [EAMCET 1991] 60 Example: 62 27 3 x log( 4 sin x 5 cos x ) c 41 41 (a) 27 3 x log( 4 sin x 5 cos x ) c 41 41 (b) (c) 27 3 x log( 4 sin x 5 cos x ) c 41 41 (d) None of these E3 d (4 sin x 5 cos x ) N (4 sin x 5 cos x ) dx 3 cos x 3 sin x M (4 cos x 5 sin x ) N (4 sin x 5 cos x ) 3 cos x 3 sin x M I 1 cot x dx 1 1 x log | sin x cos x | c 2 2 1 1 x log | sin x cos x | c (c) 2 2 (a) Solution: (b) 3 4 cos x 5 sin x 41 dx 41 4 sin x 5 cos x dx 27 27 3 x log( 4 sin x 5 cos x ) c. 41 41 D YG Example: 63 3 27 (4 cos x 5 sin x ) (4 sin x 5 cos x ) 41 41 dx I 4 sin x 5 cos x U I ID Comparing the coefficient of sin x and cos x on both sides. 3 27 5 M 4 N 3 and 4 M 5 N 3 M , N 41 41 Here, I (b) [CBSE 1992; PB 1993; HB 1994] 1 1 x log | sin x cos x | c 2 2 (d) None of these 1 cot x 1 cos x sin x cos x dx dx dx sin x sin x d (sin x cos x ) N (sin x cos x ) sin x M (cos x sin x ) N (sin x cos x ) dx Comparing the coefficients of sin x and cos x of both the sides, we have 1 M N and 0 M N U Let sin x M ST Solving these equations, we have M sin x Hence, I 1 1 and N 2 2 1 1 (cos x sin x ) (sin x cos x ) 2 2 1 1 (cos x sin x ) (sin x cos x ) 1 2 2 dx 2 sin x cos x cos x sin x sin x cos x dx 2 1 dx 1 1 1 log | sin x cos x | x c. 2 2 5.12 Integration of Rational Functions by using Partial Fractions. (1) Proper rational functions: Functions of the form f (x ) , where f (x ) and g(x ) are g( x ) polynomial and g(x ) 0, are called rational functions of x. If degree of f (x ) is less than degree of g(x ) , then f (x ) is called a proper rational function. g( x ) Indefinite Integral 275 (2) Improper rational function : If degree of f (x ) is greater than or equal to degree of g(x ) , f (x ) , is called an improper rational function and every improper rational function can be g( x ) transformed to a proper rational function by dividing the numerator by the denominator. then (x 5) x3 x 2 5x 6 is an improper rational function and can be expressed as 60 For example, 19 x 30 19 x 30 , which is the sum of a polynomial (x 5) and a proper function 2. 2 x 5x 6 x 5x 6 E3 (3) Partial fractions: Any proper rational function can be broken up into a group of different rational fractions, each having a simple factor of the denominator of the original rational function. Each such fraction is called a partial fraction. If by some process, we can break a given rational function f (x ) into different fractions, g( x ) resolution or decomposition of ID whose denominators are the factors of g(x ) , then the process of obtaining them is called the f (x ) into its partial fractions. g( x ) U Depending on the nature of the factors of the denominator, the following cases arise. D YG Case I: When the denominator consists of non-repeated linear factors: To each factor (x a) occurring once in the denominator of a proper fraction, there corresponds a A partial fraction of the form , where A is a constant to be determined. x a Case II: When the denominator consists of linear factors, some repeated: To each factor (x a) occurring r times in the denominator of a proper rational function, corresponds a sum of r partial fractions of the form. A1 A2 Ar ........... 2 x a (x a) (x a)r linear single linear there U Where A' s are constants to be determined. Of course, Ar is not equal to zero. Case III: When the denominator consists of quadratic factors: To each irreducible non ST repeated quadratic factor ax 2 bx c, there corresponds a partial fraction of the form Ax B , where A and B are constants to be determined. 2 ax bx c To each irreducible quadratic factor ax 2 bx c occurring r times in the denominator of a proper rational fraction there corresponds a sum of r partial fractions of the form A1 x B 1 A2 x B2 Ar x Br ................. 2 2 2 ax bx c (ax bx c) (ax 2 bx c)r Where, A's and B's are constants to be determined. (4) General methods of finding the constants (i) In the given proper fraction, first of all factorize the denominator. 276 Indefinite Integral (ii) Express the given proper fraction into its partial fractions according to rules given above and multiply both the sides by the denominator of the given fraction. 60 (iii) Equate the coefficients of like powers of x in the resulting identity and solve the equations so obtained simultaneously to find the various constant is short method. Sometimes, we substitute particular values of the variable x in the identity obtained after clearing of fractions to find some or all the constants. For non-repeated linear factors, the values of x used as those for which the denominator of the corresponding partial fractions become zero. Note : If the given fraction is improper, then before finding partial fractions, the given fraction must be expressed as sum of a polynomial and a proper fraction by division. (1 sin x )(2 sin x ) dx cos x 2 sin x (b) log c 1 sin x (a) log [(1 sin x )(2 sin x )] c 1 sin x (c) log c 2 sin x (d) None of these Put sin x t cos x dx dt U Solution: (c) [Roorkee 1979] ID Example: 64 E3 (5) Special cases: Some times a suitable substitution transform the given function to a rational fraction which can be integrated by breaking it into partial fractions. 1 1 (1 sin x )(2 sin x ) (1 t)(2 t) t 1 t 2 dt cos x dx D YG dt t 1 log( t 1) log( t 2) c = log c t 2 1 sin x log c. 2 sin x Example: 65 3x 1 (x 2) (x 2) (a) 5 x2 7 log c 16 x 2 4 (x 2) (b) 16 x 2 7 log c 5 x 2 4 (x 2) (d) None of these 3x 1 U (c) Solution: (b) [Pb. CET 1996] 2 We have, (x 2) (x 2) 2 5 x 2 7 log c 16 x 2 4 (x 2) A B C 2 (x 2) (x 2) (x 2) ST 3 x 1 A(x 2)(x 2) B(x 2) C(x 2) 2.......(i) 5 7 ,C 16 4 Putting x 2 and - 2 successively in equation (i), we get B Now, we put x 0 and get A 3x 1 (x 2) (x 2) dx 2 5 16 Example: 66 5 16 x 2 4 (x 2) dx 7 dx 2 5 16 x2 dx 5 7 1 5 log( x 2) log( x 2) c 16 4 (x 2) 16 5 x 2 7 log c 16 x 2 4 ( x 2) x2 1 x (x 2 dx 1) (a) log x2 1 c x [MP PET 1999] (b) log x2 1 c x (c) log x x 1 2 c (d) log x x 1 2 c Indefinite Integral 277 Solution: (a) I Example: 67 If 1 2x dx I log( x 2 1) log x c 2 1 x x (x x2 1 c. x 2x 2 3 2 x 1 x dx a log b tan 1 c then values of a and b are 1) (x 2 4 ) x 1 2 (b) 1, 1 (a) 1, 1 (c) 1 / 2, 1 / 2 (d) 1 / 2, 1 / 2 Put x y 2 2x 2 3 2y 3 2y 3 A B (y 1) (y 4 ) (y 1) (y 4 ) (x 1) (x 2 4 ) (y 1) (y 4 ) 2 2 y 3 A(y 4 ) B(y 1) A B 2 and 4 A B 3 A 1 and B 1 1 1 1 a and b . 2 2 I 1 2 1 x4 dx (x 1) (x 2 1) dx x 1 4 2 I dx 1 x 1 1 x log tan 1 c 2 x 1 2 2 (b) (c) x (x 2) log (x 1) log (x 2 1) tan 1 x c 2 2 4 2 (d) None of these U x (x 2) log (x 1) log (x 2 1) tan 1 x c 2 2 4 2 x4 dx (x 1)(x 2 1) x4 1 dx (x 1) (x 2 1) (x 1) (x dx 2 1) (x 1) (x 1) (x 2 1) dx (x 1) (x 2 1) (x 1)(x x 1 1 1 x log( x 1) log( x 2 1) tan 1 2 2 4 2 2 x (x 2) log( x 1) log( x 2 1) tan 1 x c. 2 2 4 2 tan tan 3 d 1 tan 3 U 1 dx (x 1) (x 2 1) [Roorkee 1986] x (x 2) log (x 1) log (x 2 1) tan 1 x c 2 2 4 2 (a) (x 1) dx Example: 69 x ID 1 I D YG Solution: (a) y 1 dx y 4 dx E3 Comparing the coefficient of y and constant terms Example: 68 [Rajasthan PET 2000] 60 Solution: (d) I log (a) 2 tan 1 1 1 1 c log | 1 tan | log | tan 2 tan 1 | tan 1 3 6 3 3 2 tan 1 1 1 1 c log | 1 tan | log | tan 2 tan 1 | tan 1 3 6 3 3 (c) 1 tan 2 1 c log 3 3 tan ST (b) (d) None of these Solution: (b) Let I tan tan 3 d 1 tan 3 tan (1 tan 2 ) I 1 tan 3 Putting tan t sec 2 d dt , we have, I Let tan sec 2 1 tan 1t t 3 dt 3 d (1 t)(1 t t t dt 2 ) t A BtC 2 (1 t) (1 t t ) 1 t 1 t t 2 t A(1 t t 2 ) (Bt C) (1 t)......(i) dx 2 1) x c 278 Indefinite Integral Putting t 1 in equation (i), we get, 1 A(1 1 1) A 1 3 t 1 2t 2 (2 t 1) 3 1 t 3 t t 1 dt 3 1 t 6 t t 1 dt 3 1 t 6 t t 1 1 dt 1 2t 1 1 dt dt dt 1 2t 1 3 1 t 6 t t 1 dt 6. t t 1 3 1 t 6 t t 1 dt 2 1 dt 1 1 dt 1 2 2 dt 1 2 2 2 dt 2 3 t 2 4 c 1 log | 1 t | 1 log | t 2 t 1 | 1 tan 1 2 t 1 c 3 3 6 3 ID t 1 1 1 1 2 2 1 2 log | 1 t | log | t t 1 | . tan 3 6 2 3 3 2 1 2 E3 1 3 1 3 Hence, I 60 Comparing the coefficients of t 2 and constant terms on both the sides of equation (i) we get 1 1 and 0 A C C A 0 A B B A 3 3 1 1 1 t 1 t . . 3 1 r 3 1 t t2 (1 t) (1 t t 2 ) 2 tan 1 1 1 1 c. log | 1 tan | log | tan 2 tan 1 | tan 1 3 6 3 3 x (x dx n 1) [Roorkee 1979; Pb. CET 1991; Rajasthan PET 1998; Karnataka CET U Example: 70 1996] 1 xn log n c n x 1 dx Let I x (x n 1) (a) xn 1 c xn (c) 1 xn log n c n x 1 (d) n log xn 1 c xn x n 1 dx x n (x n 1) D YG Solution: (a) (b) n log Putting x t nx n 1 dx dt , we have n I 1 n 1 1 t(t 1) n t t 1 dt, dt 1 (by resolving into partial fractions) 1 t 1 xn 1 c log n c. [log t log (t 1)] c log n t 1 n n x 1 U 5.13 Integration of Trigonometric Functions. (1) Integral of the form sin m x cos n x dx : (i) To evaluate the integrals of the form ST I sin m x cos n x dx , where m and n are rational numbers. (a) Substitute sin x t, if n is odd; (b) Substitute cos x = t, if m is odd; (c) Substitute tan x t, if m n is a negative even integer; and 1 (d) Substitute cot x t, if (n 1) is an integer. 2 m n 2 (e) If m and n are rational numbers and is a negative integer, then substitution 2 cos x t or tan x t is found suitable. Indefinite Integral 279 (ii) Integrals of the form R(sin x, cos x )dx, where R is a rational function of sin x and cos x , are x t, where x . 2 This is the so called universal substitution. Sometimes it is more convenient to make the x substitution cot t for 0 x 2. 2 The above substitution enables us to integrate any function of the form R(sin x , cos x ). 60 transformed into integrals of a rational function by the substitution tan However, in practice, it sometimes leads to extremely complex rational function. In some cases, the integral can be simplified by: E3 R(sin x )cos x dx. (b) Substituting cos x t, if the integral is of the form R(cos x ) sin x dx. (a) Substituting sin x t, if the integral is of the form dt , if the integral is dependent only on tan x. 1 t2 (d) Substituting cos x t , if R( sin x , cos x ) R(sin x , cos x ) ID (c) Substituting tan x = t, i.e., dx (e) Substituting sin x t , if R(sin x , cos x ) R(sin x , cos x ) U (f) Substituting tan x t , if R( sin x , cos x ) R(sin x , cos x ) Important Tips sin mx cos nx dx , sin mx. sin nx dx , cos mx. cos nx dx D YG To evaluate integrals of the form cos mx. sin nx dx , we use the and following trigonometrical identities. sin mx. cos nx 1 1 [sin(m n)x sin(m n)x ] cos mx. sin nx [sin(m n)x sin(m n)x ] 2 2 sin mx. sin nx 1 1 [cos(m n)x cos(m n)x ] cos mx. cos nx [cos(m n)x cos(m n)x ] 2 2 sin 3 x cos 2 x dx cos 2 x cos 3 x c 5 3 U Example: 71 (a) I sin x(1 cos ST Solution: (a) 2 (b) cos 5 cos 3 c 5 3 d sin . cos 3 sin 5 sin 3 c 5 3 (d) sin 5 sin 3 c 5 3 t5 t3 cos 5 x cos 3 x c c. 5 3 5 3 (a) log tan tan 2 c Solution: (c) (c) x ) cos 2 x dx Put cos x t sin x dx dt I (t 2 t 4 ) dt Example: 72 [Pb. CET 1992; EAMCET 1996] d sin cos 3 sec 2 d sin cos (b) log tan 1 tan 2 c 2 sec 2 (1 tan 2 ) d tan d (c) log tan 1 tan 2 c 2 (d)None of these 280 Indefinite Integral t tan Put then it reduces to 1 t2 1 dt t dt t t t2 1 c log tan tan 2 c. 2 2 sin 3 2 x dx 5 2x cos Example: 73 [Karnataka CET 1999] (b) tan 4 x c (a) tan 4 x c Given, I Solution: (d) (c) tan 4 2 x x c sin 3 2 x dx. The given equation may be written as 5 2x cos 1 3 t4 tan 4 2 x t dt c c. 2 8 8 sin nx Solution: (b) where n 1, then In In2 (a) 2 cos(n 1)x (n 1) In sin x dx 2 sin(n 1)x n 1 (b) sin nx In 2 sin(n 2)x dx In In 2 sin x (c) sin nx sin(n 2)x dx sin x 2 sin(n 1)x (n 1) 2 cos n x n 2 cos(n 1)x. sin x dx sin x D YG In I n 2 (2) Reduction formulae for special cases (i) (ii) sin n x dx cos x. sin n 1 x n 1 sin n 2 x dx n n cos n x dx sin x cos n 1 x n 1 cos n 2 x dx n n tan n 1 x tan n 2 x dx n 1 1 (iv) cot n x dx cot n 1 x cot n 2 x dx n 1 1 sec n 2 x. tan x (n 2) sec n 2 x dx (v) sec n x dx (n 1) tan n x dx U (iii) ST (vi) cosec (vii) (viii) (ix) n x dx sin p x cos q x dx (x dx 2 k) n 1 cos ec n 2 x. cot x (n 2) cosec n 2 x dx (n 1) sin p x cos q x dx sin q 1 x. cos p 1 x p 1 sin p 2 x. cos q x dx pq pq sin p 1 x. cos q 1 x p 1 sin p x. cos q 2 x dx pq pq x k (2n 2) (x k ) 2 n 1 (2n 3) dx 2 k (2n 2) (x k )n 1 1 tan 4 2 x c 8 sin 3 2 x 1. dx tan 3 2 x. sec 2 2 x dx. 3 2 x cos 2 2 x cos ID sin x dx , If In Example: 74 U I (d) E3 Put tan 2 x t and 2 sec 2 2 x dx dt 60 log t dt sec 2 d (d) 2 sin n x n Indefinite Integral 281 Important Tips Reduction formulae for I(n, m ) Example: 76 d (a) tan 4 tan 2 log sec c 4 2 (b) (c) tan 4 tan 2 log | sec | c 4 2 (d) None of these cosec 4 x dx Solution: (c) Example: 77 Integrate [Rajasthan PET 2002] cot 3 x c 3 cosec 4 x dx sin 8 (b) tan x tan 3 x c 3 (c) cot x cot 3 x c 3 (d) tan x cosec 2 x (1 cot 2 x )dx cosec 2 x dx cot 2 x cosec 2 x dx cot x x dx tan 3 x c 3 (cot x )3 c. 3 1 sin 8 x sin 6 x sin 4 x sin 2 x 8 28 56 35 x c 6 4 2 27 8 D YG (a) (b) tan 4 tan 2 log sec c 4 2 tan 4 tan 2 tan 4 tan 2 tan 4 log | sec | c. I1 I3 4 2 4 2 4 tan 5 d I5 (a) cot x 1 27 sin 6 x sin 4 x sin 2 x sin 8 x 8 8 6 28 4 56 2 35 x c sin 8 x sin 6 x sin 4 x sin 2 x 8 28 56 35 x c 8 6 4 2 (d) None of these Let cos x i sin x y ; then (c) Solution: (a) [EAMCET 1993] 60 5 E3 Solution: (c) tan ID Example: 75 1 sin n 1 x (n 1) sin n x .. I(n 2, m 2) dx is I(n,m ) m m 1 cos m 1 x (m 1) cos x U 2 cos x y U results) 1 1 1 1 , 2 cos nx y n n 2 i sin x y , 2 i sin nx y n n y y y y (Remember as the standard 8 ST 1 1 1 1 1 Thus 2 8 i 8 sin 8 x y y 8 8 8 y 6 6 28 y 4 4 56 y 2 2 70 y y y y y 2 cos 8 x 16 cos 6 x 56 cos 4 x 112 cos 2 x 70 Thus sin 8 x and sin 8 1 (cos 8 x 8 cos 6 x 28 cos 4 x 56 cos 3 x 35 ), 27 x dx 1 27 sin 6 x sin 4 x sin 2 x sin 8 x 8 8 6 28 4 56 2 35 x c 5.14 Integration of Hyperbolic Functions sinh x dx cos hx c (3) sec h x dx tan h x c (5) sec hx tan hx dx sec hx c (1) 2 cos hx dx sin hx c (4) cosec h x dx cot hx c (6) cosec hx cot h dx cosec hx c (2) 2 282 Indefinite Integral cos 1 dx x [Rajasthan PET 2002] (a) x sec 1 x cos h 1 x c (b) x sec 1 x cos h 1 x c (c) cos h 1 x x sec 1 x c (d) None of these cos Example: 79 1 1 1 dx sec x dx x sec 1 x x cos x sin x x 1 x 2 1 dx equals [Rajasthan PET 1996] sin 2 x (a) cos h 1 (sin x cos x ) c (b) sin h 1 (sin x cos x ) c (c) cos h 1 (sin x cos x ) c (d) sin h 1 (sin x cos x ) c I cos x sin x dx sin 2 x cos x sin x dx 1 sin 2 x 1 sin h 1 x dx equals x.1 dx x sin h 1 x I t 1 II 1 ex a 2x U ex a dx ex a ST (e a ex ex t dx dt a a Put Example: 82 e x (e Example: 83 x dt t 1 2 ex e 2x a t dx 2 dt t2 1 (d) None of these a e 2x a 2 dx ex /a (e x / a)2 1 2 dx 1 (e x / a)2 1 2 dx cos h 1 (e x / a) sec 1 (e x / a) c. dx e x )2 (a) tan hx c Solution: (c) 2 dx 1 (b) cos h 1 (e x / a) sec 1 (e x / a) c a 1 (c) cos h 1 (e x / a) sec 1 (e x / a) c a [Rajasthan PET 1996] ex a dx ex a (a) cos h 1 (e x / a) sec 1 (e x / a) c Solution: (a) cos h 1 t c cos h 1 (sin x cos x ) c.. x dx x sin h 1 x 1 x 2 c. 1 x2 D YG Example: 81 dt 2 U 1 dx (d) x sin h 1 x x 2 1 c (c) x sin h 1 x x 2 1 c sin h (cos x sin x )2 1 (b) x cos h 1 x x 2 1 c (a) x cos h 1 x x 2 1 c Solution: (d) cos x sin x ID Put cos x sin x t (cos x sin x ) dx dt Example: 80 x dx x sec 1 x cos h 1 x c. 60 Solution: (b) Solution: (a) 1 E3 Example: 78 dx 1 x 2 4 e ) tan x 1 tan 2 x (b) (e x 4. dx 1 x 2 4 e ) 1 tan h x c 2 (c) 1 tan h x c 4 (d) None of these sec h x dx 4 tan hx c. 2 1 dx (a) cos h 1 ( 2 cos x ) c (b) cos h 1 ( 2 cos x ) c (c) 1 2 cos 1 ( 2 cos x ) c (d) 1 2 cos h 1 ( 2 cos x ) c Indefinite Integral 283 tan x dx 1 tan x Put I sin x dx cos x sin x 2 2 sin x dx 2 cos 2 x 1 1 dt t 1 2 1 cos h 1 ( 2 cos x c). 2 60 sin x sin hx dx 1 [sin x cos h x cos x sin h x ] c 2 (a) 1 x (e e x ) c 2 (b) (c) 1 [sin x cos h x cos x cos h x ] c 2 (d) None of these E3 Solution: (b) 2 cos x t 2 sin x dx dt 2 Example: 84 2 I sin x sin hx dx sin hx cos x cos hx cos x dx sin hx cos x [cos hx sin x sin hx sin x dx ] I 1 [sin x cos x cos x sin h x ] c. 2 ID Solution: (d) 5.15 Integral of the type f [x, (ax b) m 1 /n1 , (ax b) m 2 /n2...] where f is a rational function U and m 1 , n1 , m 2 , n 2 are Integers To evaluate such type of integral, we transform it into an integral of rational function by D YG putting (ax b) t s , where s is the least common multiple (L.C.M.) of the numbers n1 , n 2. Integrals of the form x m (a bx n ) p dx Case I : If p N (Natural number). We expand the integral with the help of binomial theorem and integrate. Example: 85 Evaluate x (a) 3 x 4 / 3 I x 1/3 (2 x 1 / 2 ) dx 7 7/3 x c 3 (b) x 4 / 3 5 5/3 x c 3 (c) 3 x 4 / 3 3 3 5/3 24 11 / 6 x x c (d) 3 x 4 / 3 x 7 / 3 c 7 5 11 (2 x 1 / 2 )2 dx U Solution: (d) 1/3 Since P is natural number, x 1/3 (4 x 4 x 1 / 2 ) dx (4 x 1 / 3 x 4 / 3 4 x 5 / 6 ) dx ST I 3 24 11 / 6 4 x 4 / 3 x 7 / 3 4. x 11 / 6 x c. c 3x 4 /3 x 7 /3 7 11 4 /3 7/3 11 / 6 Case II: If p I (i.e. negative integer). Write x t k , where k is the L.C.M. of denominator of m and n. Example: 86 Evaluate x 2 / 3 (1 x 2 / 3 )1 dx (a) 3 tan 1 (x 1 / 3 ) c Solution: (a) (b) 3 tan 1 x c (c) 3 tan 1 (x 2 / 3 ) c (d) None of these If we substitute x t 3 (as we know P negative integer) Let x t k , where k is L.C.M. of denominator m and n. x t 3 dx 3 t 2 dt or I 3 t 2 dt dt 3 2 3 tan 1 (t) c I 3 tan 1 (x 1 / 3 ) c. t (1 t 2 ) t 1 2 284 Indefinite Integral Case III: If m 1 is an integer and p fraction put (a bx n ) t k , where k is the denominator of n the fraction p. Evaluate x 2 / 3 (1 x 1 / 3 )1 / 2 dx (b) 2(1 x 1 / 3 )3 / 2 c (a) 2(1 x 1 / 3 )2 / 3 c 1 dx dt 3x 2/3 If we substitute 1 x 1 / 3 t 2 then, Solution: (b) I (c) 2(1 x 2 / 3 )3 / 2 c (d) 2 (1 x 2 / 3 )2 / 3 c 60 Example: 87 t. 6 t dt 6 t 2 dt 2 t 3 c or I 2(1 x 1 / 3 )3 / 2 c. 1 x 11 (1 x 4 )1 / 2 dx 1 5 3 (t t t) c 2 (a) where t 1 (b) 1 x4 ID Evaluate t5 2t 3 1 t 4 2t 3 t c t c (c) 2 4 5 3 3 1 2 (d) None of these U Example: 88 E3 m 1 Case IV: If P is an integer and P fraction. We put (a bx n ) t k. x n , where k is the n denominator of the fraction P. D YG m 1 11 1 1 P 3 Here n 2 4 Solution: (b) If we substitute then 4 1 1 4 t 2 and 5 dx 2 t dt I x x I dx 1 x (1 1 / x 4 )1 / 2 4 13 where t 1 x 2 t dt 8 t x 11 dx (1 x 4 )1 / 2 x 11 dx. x (1 1 / x 4 )1 / 2 2 1 1 4 (t 2 1)2 dt (t 2t 2 1) dt 2 2 1 t 5 2t 3 t c, 2 5 3 1 x4 U 5.16 Integrals using Euler's substitution Integrals of the form f (x ), (ax 2 bx c) dx are calculated with the aid of one of the there Euler ST substitution: (1) ax 2 bx c t x a , if a 0. (2) ax 2 bx c t x c , if c 0. (3) ax 2 bc c (x ) t, if ax 2 bx c a(x )(x ), i.e., if x is real root of (ax 2 bx c). Note : The Euler substitution often lead to rather some calculations, therefore they should be applied only when it is difficult to find another method for calculating the given integral. Example: 89 Evaluate x (a) 2 log e | t | dx x2 x 1 1 3 3 log e | t 1 | log e | t 1 | c 2 2 (t 1) (b) 2 log e | t | 1 3 3 log e | t 1 | log e | t 1 | c 2 2 (t 1) Indefinite Integral 285 (c) 2 log e | t | Where t Solution: (a) 1 3 3 log e | t 1 | log e | t 1 | c (d) None of these 2 2 (t 1) x 2 x 1 1 x Since here c 1, we can apply the second Euler substitution. 2t 1 t2 1 Hence (2t 1)x (t 2 1)x 2 ; x I 2(t 2 t 1)dt 1 and x x 2 x 1 t 1 (t 2 1)2 x dx x2 x 1 E3 dx 60 x 2 x 1 tx 1 2t 2 2t 2 dt t(t 1)(t 1) 2 Using partial fractions, we have, We get A 2, B 1 / 2, C 3 / 2, D 3. Evaluate (a) 1 2 t 1 2 (t 1) 3 (t 1) dt 3 dt dt 2 x dx 2 log e | t | 1 3 3 log e | t 1 | log e | t 1 | c 2 2 (t 1) Where t x 2 x 1 1 x (7 x 10 x 2 )3 2 5 2t c 9 t where t (b) 2 5 2t c 9 t (c) 1 5 2t c 9 t (d) None of these 7 x 10 x 2 x 2 In this case a 0 and c 0. Therefore neither (I) nor (II) Euler substitution is applicable. But the quadratic 7 x 10 x 2 has real roots 2, 5. U Solution: (a) dt D YG Example: 90 t U Hence I 2 ID 2t 2 2t 2 A B C D or (2t 2t 2) A(t 1)(t 1)2 B(t 1)2 C(t 1)(t 1)t Dt. 2 t t 1 ( t 1 ) t(t 1)(t 1) (t 1)2 We use the (III) i.e., 7 x 10 x 2 (x 2)(5 x ) (x 2) t ST where (5 x ) (x 2) t 2 or 5 2t 2 x (1 t 2 ) x 5 2t 2 6 t 3t 5 2t 2 (x 2) t 2 t dt , dx 2 2 2 1 t 1 t ( 1 t 2 )2 1t Hence, I ( ( x dx 7 x 10 x ) x dx 7 x 10 x 2 )3 2 3 5 2t 2 1 t2 6t . (1 t 2 ) 2 dt 3t 2 1 t 2 5 2t c , 9 t 5.17 Some Integrals which can not be Found 3 6 27 2 5 2 5 5 2t 2 2t c dt 2 2 dt 2 9 t 9 t t where t 7 x 10 x 2 x 2 286 Indefinite Integral Any function continuous on interval (a, b) has an antiderivative in that interval. In other words, there exists a function F(x ) such that F ' (x ) f (x ). (ii) (iv) 3 1 x 2 dx e (vii) (x) x 2 (v) (viii) dx sin x dx dx 1 x 3 dx sin x dx x sin( x 2 (iii) ) dx ST U D YG U etc. (xi) e x2 x2 dx 1 x5 1 k 2 sin 2 x dx (vi) E3 dx log x (ix) (xii) cos x dx x cos(x ID (i) 60 However not every antiderivative F(x ), even when it exists is expressible in closed form in terms of elementary functions such as polynomials, trigonometric, logarithmic, exponential etc. function. Then we say that such antiderivatives or integrals "can not be found." Some typical examples are: 2 ) dx (xiii) x tan x dx
