M1-Allens Made Maths Theory + Exercise [II]

JEE-Mathematics INDEFINITE INTEGR ATION If f & F are function of x such that F' (x) = f(x) then the function F is called a PRIMITIVE OR ANTIDERIVATIVE OR INTEGRAL of f(x) w.r.t. x and is written symbolically as  f(x) dx  F(x)  c  d {F (x )  c} f(x ) , where c is called the constant of integration. dx 1 . GEOMETRICAL INTERPRETATION OF INDEFINITE INTEGR AL :  f(x)dx  F (x)  c  y (say ) , represents a family of curves. The different values of c will correspond to different members of this family and these members can be obtained by shifting any one of the curves parallel to itself. This is the geometrical interpretation of indefinite integral. Let f(x) = 2x. Then f(x)dx  x2  c. For different values Y of c, we get different integrals. But these integrals are y= x2+3 very similar geometrically. Thus, y = x2 + c, where c is arbitrary constant, represents P3 y= x2+2 a family of integrals. By assigning different values to c, we get different members of the family. These together P2 y= x2+1 constitute the indefinite integral. In this case, each integral represents a parabola with its axis along y-axis. P1 y= x2 If the line x = a intersects the parabolas y = x2, y = x2 +1, y = x2 + 2, y = x2 – 1, y = x2 – 2 at P , P , P , P , P etc., P0 y= x2–1 0 1 2 –1 –2 X' P–1 y= x2–2 X dy respectively, then at these points equals 2a. This P–2 y= x2–3 dx indicates that the tangents to the curves at these points are parallel. Thus, 2 x d x = x 2 + c = f( x ) + c (say), P–3 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 implies that the tangents to all the curves x= a f(x) + c, c  R, at the points of intersection of the Y' curves by the line x = a, (a  R) , are parallel. 2. STANDARD RESULTS : E ( i ) (ax  b)n dx  (ax  b)n1  c ; n  1 (ii)  dx b  1 n ax  b c a(n 1) ax  a ( i i i ) eaxbdx  1eaxb  c ( i v ) apxq dx  1 apxq  c, (a  0) p na a (v)  sin(ax  b)dx   1 cos(ax  b)  c (vi)  cos(ax  b)dx  1 sin(ax  b)  c a a (vii)  tan(ax  b)dx  1  n| sec(ax  b)| c (viii)  cot(ax  b)dx  1 n| sin(ax  b)| c a a (ix)  sec2 (ax  b)dx  1 tan(ax  b)  c (x)  cos ec2 (ax  b)dx   1 cot(ax  b)  c a a (xi)  cosec (ax  b ). cot(ax  b)dx   1 cosec (ax  b)  c a (xi i) sec (ax + b).tan(ax + b)dx = 1 sec(ax  b)  c a 1

JEE-Mathematics (xiii)  sec x dx  n sec x  tan x c = n tan    x   c  4 2  (xiv)  cosec x dx  n cosec x  cot x c = n tan x c = n | cosec x  cot x| + c 2 (x v) dx  sin 1 x  c (xvi ) dx  1 tan 1 x c a2  x2 a a2  x2 a a (x vi i) dx  1 sec1 x  c (xvi ii) dx   n  x  x2  a2  +c x x2  a2 a a x2  a2   ( x i x ) dx   n  x  x2  a2  +c ( x x ) dx  1 n ax c x2  a2   a2  x2 2a ax ( x x i ) dx  1 n x a c ( x x i i ) a2  x2 dx  x a2  x2  a2 sin 1 x  c x2  a2 2a xa 2 2a x2  a2 dx  x x2  a2  a2 n x  x2  a2  c 22  (xxiii) x2  a2 dx  x a2   (xxiv) x2  a2 22 n x x2  a2 c (xx v) eax .sin bx dx  eax (a sin bx  b cos bx)  c  e ax sin  bx  tan 1 b   c a2  b2 a2  b2  a  ( x x v i ) eax.cos bx dx  eax (a cos bx  b sin bx)  c  e ax cos  bx  tan 1 b   c a2  b2 a2  b2  a  3 . TECHNIQUES OF INTEGR ATION : (a) Substitution or change of independent variable : If (x) is a continuous differentiable function, then to evaluate integrals of the form  f((x)) '(x)dx , we substitute (x) = t and '(x)dx = dt. Hence I   f((x)) '(x)dx reduces to  f(t)dt . NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 ( i ) Fundamental deductions of method of substitution :  f(x)n f '(x)dx f '(x) OR  [f(x)]n dx put f(x) = t & proceed. Illustration 1 : Evaluate cos3 x dx sin2 x  sin x Solution : I =  (1  sin2 x) cos x dx =  1  sin x cos x dx sin x(1  sin x) sin x Put sinx = t  cosx dx = dt  I = 1 t dt  n| t| t  c = n| sin x|  sin x  c Ans. t E Illustration 2 : Evaluate x2 1 dx x4  3x2  1 tan1  x  1  x  2

JEE-Mathematics Solution : The given integral can be written as 1  1  dx x2  I =  12   1  x  x   1 tan 1  x  x   Let  x  1 = t. Differentiating we get 1  1  dx = dt  x  x2  dt Hence I =  t2  1 tan1 t dt du  n| u| c u Now make one more substitution tan–1t = u. Then t2  1 = du and I = Returning to t, and then to x, we have I= n| tan 1 t| c  n tan 1  x  1   c Ans.  x  Do yourself -1 : x2 ( i i ) Evaluate :  cos3 x dx ( i ) dx Evaluate : 9 16x6 (ii) Standard substitutions : dx  a2  x2 or a2  x2 dx ; put x = a tan or x = a cot   dx a2  x2 dx ; put x = a sin or x = a cos  a2  x2 or NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 dx  x2  a2 or x2  a2 dx ; put x = a sec or x = a cosec   a  x dx ; put x = a cos2 ax  x  dx or  (x  )(  x) ; put x =  cos2  +  sin2  x  x  dx or  (x  )(x  ) ; put x =  sec2  –  tan2  x   dx ; put x –  = t2 or x –  = t2. (x  )(x  ) Illustration 3 : Evaluate  dx (x  a)(b  x) Solution : Put x = acos2 + bsin2, the given integral becomes E 3

JEE-Mathematics I 2(b  a) sin  cos d 1  (a cos2   b sin2   a )(b  a cos2   b sin2  2  2(b  a) sin  cos d ba 2d  2  c  2 sin1  x  a   c  b  a   b  a  b  a sin cos  = = Ans. Illustration 4 : Evaluate  1 x1 1 . dx xx Solution : Put x = cos2  dx = –2sin cos d  I 1  cos  . 1 ( 2 sin  cos )d   2 tan  tan  d 1  cos  cos2 2   4  sin2 ( / 2) d  2  1  cos  d  2n| sec   tan | 2  c cos  cos   2n 1  1  x  2 cos1 x  c x Do yourself -2 : (i) Evaluate :  x  3 dx (ii) Evaluate :  dx 2x x x2  4 (b) Integration by part :  u.v dx  u v dx    du . v dx  dx where u & v are differentiable functions and  dx  are commonly designated as first & second function respectively. Note : While using integration by parts, choose u & v such that (i)  vdx & (ii)   du . v dx  dx are simple to integrate.  dx  This is generally obtained by choosing first function as the function which comes first in the word ILATE, where; I-Inverse function, L-Logarithmic function, A-Algebraic function, T-Trigonometric function & E-Exponential function. Illustration 5 : Evaluate :  cos x dx NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 Solution : Consider I =  cos xdx Let x  t 1 then dx  dt 2x i.e. dx  2 xdt or dx = 2t dt so I   cos t.2tdt taking t as first function, then integrate it by part  I =2    dt  cos tdt  dt  = 2 t sin t   1 . sin tdt  = 2t sin t  cos t  c  dt    t cos tdt     I = 2  x sin x  cos x  c Ans. Illustration 6 : Evaluate :  1  x x dx E sin 4

JEE-Mathematics Solution : Let I =  x dx =  (1  x(1  sin x) x ) dx sin sin x)(1  sin 1  x = x(1  sin x) =  x(1  sin x) dx   x sec2 xdx   x sec x tan xdx cos2 x  1  sin2 x dx x  sec2 xdx   dx  sec2 xdx  dx    dx   =    – x  sec x tan xdx    dx  sec x ta n xdx  dx    dx       = x tan x   tan xdx   x sec x   sec xdx  = x tan x  n| sec x|  x sec x  n| sec x  tan x|  c = x tan x  sec x   n (sec x  tan x)  c = x(1  sin x)  n|1  sin x|  c Ans. sec x cos x Do yourself -3 : ( i i ) Evaluate : x3 sin(x2 )dx ( i ) Evaluate :  xe x dx Two classic integrands : ( i ) ex [f(x)  f '(x)]dx  ex.f(x)  c  1x  2  1  x2  Illustration 7 : Evaluate e x dx  1x 2 ex (1  2x  x2 ) dx = ex 1  2x  dx  ex c  1  x2 (1  x2 )2  (1  x2 ) (1  x2 )2  1  x2 Solution :   ex  dx = Ans.  Illustration 8 : ex  x4  2  The value of  (1  x2 )5 /2  dx is equal to - ex (x  1) ex (1  x  x2 ) ex (1  x) (D) none of these (A) (1  x2 )3 / 2 (B) (1  x2 )3 / 2 (C) (1  x2 )3 / 2 Ans. (D) NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 Solution :  Let I = e x  x4  2  dx = ex  1  1  2x2    x2 )5 /   x2 (1  x2 )5 / 2  .dx (1 2 (1  )1 / 2 = e x  1 x  x  12 x2  dx  x2  (1  x2 )3 / 2 (1  x2 )5 /  (1  )1 / 2 2 (1  x2 )3 / 2 = ex xex c = ex {1  x2  x} (1  x2 )1 / 2  c (1  x2 )3 / 2 (1  x2 )3 / 2 Do yourself -4 : ( i ) ex  t a n 1 x  1  dx   ( i i ) Evaluate : xe x2 sin x2  cos x2 dx Evaluate :   x2  1 (ii)  [f(x)  xf '(x)] dx  x f(x)  c Illustration 9 : Evaluate  x  sin x dx 1  cos x E 5

JEE-Mathematics  Solution : I =  x  sin x dx =   x  sin x  dx =   x 1 sec 2 x  tan x  dx x Ans. 1  cos x  2 cos2 x   2 2 2  = x tan + c   2 2 Do yourself -5 : (ii) Evaluate :  (nx  1)dx  ( i ) Evaluate : tan(ex )  xe x sec2 (ex ) dx (c) Integration of rational function : P(x) ( i ) Rational function is defined as the ratio of two polynomials in the form Q(x) , where P(x) and Q(x) are polynomials in x and Q(x)  0. If the degree of P(x) is less than the degree of Q(x), then the rational function is called proper, otherwise, it is called improper. The improper rational function P(x) can be reduced to the proper rational functions by long division process. Thus, if Q(x) is improper, then P(x) = T(x) + P1 (x) , where T(x) is a polynomial in x and P1 (x) is proper rational function. Q(x) Q(x) Q(x) It is always possible to write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition. After this, the integration can be carried out easily using the already known methods. S. No. Form of the rational function Form of the partial fraction px2  qx  r ABC xa + xb + xc 1. (x  a) (x  b)(x  c) ABC px2  qx  r x  a + (x  a)2 + x  b A Bx C 2. (x  a )2 (x  b) x  a + x2  bx  c px2  qx  r NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 3. (x  a ) (x2  bx  c) where x2 + bx + c cannot be factorised further f(x) A Bx C Dx  E 4. (x  a )(x2  bx  c)2 x  a + x2  bx  c + (x2  bx  c)2 where f(x) is a polynomial of degree less than 5. Illustration 10 : Evaluate  ( x  2 x  5 ) dx )( x Solution : x AB (x  2)(x  5 ) x  2 x  5 or x = A(x + 5) + B(x – 2). by comparing the coefficients, we get A = 2/7 and B = 5/7 so that  (x x  5) dx  2  dx  5  dx  2 n (x  2)  5 n x 5 c Ans.  2)(x 7 x 2 7 x 5 7 7 E 6

x4 JEE-Mathematics Illustration 11 : Evaluate  (x  2)(x2  1) dx Ans. Solution : x4 3x2  4 (x  2)(x2  1)  (x  2)  (x  2)(x2  1) 12  x 3x2  4  16 5 5 Now, (x  2)(x2  1) 5(x  2) x2  1 12  x So, x4  x2 16 5 5 (x  2)(x2  1) 5(x  2) x2 1 Now,   2)  16 2)  1 x  2  dx  5(x  55  x2 1  (x   = x2  2 x  2 tan 1 x  16 n x  2  1 n(x2  1)  c 25 5 10 Do yourself - 6 : (i) Evaluate :  (x 3x 2 3) dx x2 1  1)(x  ( i i ) Evaluate :  (x  1)(x  2) dx (ii)  ax2 dx  ,  ax2 dx , ax2  bx  c dx  bx c  bx  c Express ax2 + bx + c in the form of perfect square & then apply the standard results. px  q px  q (iii)  ax2  bx  c dx, dx ax2  bx  c Express px + q =  (differential coefficient of denominator ) + m. NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 dx Illustration 12 : Evaluate  2x2  x 1 Solution : I   2x2 dx 1 1 dx 1 dx 1 1 x 1   2  x2  x  1  2  x2  x   22 2 16 16 2 1 dx 1 dx  =  2 (x  1 / 4)2  9 / 16 2 (x  1 / 4)2  (3 / 4)2 = 1 . 1 4) log x 1/4 3/4 c using, x2 dx  1 log x a  c  2 2(3 / x 1/4 3/4  a2 2a xa   1 x 1 / 2 c 1 2x 1 c Ans. = log  log 3 x 1 3 2(x 1) 3x  2 7 Illustration 13 : Evaluate  4x2  4x  5 dx E

JEE-Mathematics Solution : Express 3x + 2 = (d.c. of 4x2 + 4x + 5) + m or, 3x + 2 = (8x + 4) + m Comparing the coefficients, we get 8 = 3 and 4 + m = 2   = 3/8 and m = 2 – 4 = 1/2  I = 3  8x 4 dx  1  4x2 dx  5 8 4x2  4x 5 2  4x = 3 log 1 dx = 3 log 4x2 1 1  1  4x2  4x  5  x5 8  4x 5  8 tan  x  2   c Ans. 88 x2 4 Do yourself -7 : 5x  4 dx (ii) Evaluate :  x2  3x  2 dx ( i ) Evaluate :  x2  x  1 ( i v ) x2 1 dx OR  x4 x2 1 dx where K is any constant. Integrals of the form x4  Kx2 1  Kx2 1 Divide Nr & Dr by x2 & proceed. Note : Sometimes it is useful to write the integral as a sum of two related integrals, which can be evaluated by making suitable substitutions e.g. 2x2 dx  x2  1 dx  x2 1 dx 2 dx  x2  1 dx  x2 1dx      * * x4 1 x4 1 x4 1 x4 1 x4 1 x4 1 These integrals can be called as Algebric Twins. 4 Illustration 14 : Evaluate : dx sin4 x  cos4 x 1 sin2 x  cos2 x sin4 x  cos4 x dx = 4 dx Solution :  I = 4 sin4 x  cos4 x NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 (tan2 x  1) cos2 x dx  4 (tan2 x  1) sec2 x dx (tan4 x  1) cos4 x (tan4 x  1)  = 4 Now, put tanx = t  sec2x dx = dt 1  t2 1 / t2 1 1  t4 dt  4 dt   I = 4 t2  1 / t2 Now, put t – 1/t = z  1  1  dt = dz t2   I = 4 dz  4 tan1 z 2 2 tan 1 t  1 / t = 2 2 tan 1  tan x  1/ tan x   c z2  2 2 2 2  2    Ans. 1 Illustration 15 : Evaluate : dx x4  5x2  1 8 E

JEE-Mathematics 12 Solution : I = 2 dx x4  5x2 1 I1 1  x2 dx  1 1  x2 dx 1 1 1 / x2 1 1 1 / x2 2 x4  5x2 1 2 x2  5  1 / x2 dx  2 dx     = x2  5 1 / x2 x4  5x2 1 2 {dividing Nr and Dr by x2} (1  1 / x2 ) (1  1 / x2 )dx 1 dt 1 du  1  1  2  t2   72  2  u2   3 2 dx  2 (x  1 / x)2  7 2 (x  1 / x)2  3 11 where t = x – and u = x  xx I = 1. 1  tan 1 t   1 . 1  ta n 1 u   c 2 7  7  2 3  3      1  1 tan 1  x 1/ x   1 tan 1  x 1/ x   c 2  7  7  3  3  =      Ans. Do yourself -8 : ( i ) x2  1 dx 1 Evaluate : x4  x2 1 ( i i ) Evaluate :  1  x4 dx (d) Manipulating integrands : ( i ) dx , n N , take xn common & put 1 + x–n = t. x(xn 1) (ii)  dx , n  N , take xn common & put 1 + x–n = tn (n 1)  x2 xn  1 n NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 (i i i ) dx , take xn common and put 1 + x–n = tn. x n (1  x n )1 / n dx Illu stration 16 : Evaluate : xn (1  xn )1/ n Solution :  dx dx Let I = x n (1  x n )1 / n = x n 1 1  1 1/n xn   Put 1 1 = tn, then 1 dx  tn 1 dt xn x n 1 n 1 tn 1 dt tn2 dt tn 1 c 1 1 1  n c   n 1 xn  t n 1  I = –    Ans. Do yourself -9 : dx dx dx ( i ) Evaluate : ( i i ) Evaluate : ( i i i ) Evaluate : x3 (x3  1)1 / 3 x(x2 1) x2 (x3  1)2 / 3 E9

JEE-Mathematics (e) Integration of trigonometric functions : dx dx dx ( i )  a  b sin2 x OR  a  b cos2 x OR a sin2 x  b sin x cos x  c cos2 x Divide Nr & Dr by cos2 x & put tan x = t . Illustration 17 dx Solution : : Evaluate :  2  sin2 x Divide numerator and denominator by cos2x  sec2 xdx sec2 xdx I = 2 sec2 x  tan2 x = 2  3 tan2 x Let 3 tan x  t  3 sec2 x dx  dt I 1 dt 1 . 1 tan 1 t c = 1 1  3 tan x  3  32 2   S o tan  c Ans. 2  t2 Ans. 6  2 Illustration 18 : dx Evaluate :  (2 sin x  3 cos x)2 Solution : Divide numerator and denominator by cos2x sec2 xdx  I =  (2 tan x  3)2 Let 2 tan x  3  t ,  2sec2xdx = dt I  1  dt = 1 c  1  3) c 2 t2 2t  2(2 tan x Do yourself -10 : dx dx ( i i ) Evaluate :  3 sin2 x  sin x cos x  1 ( i ) Evaluate :  1  4 sin2 x (ii) dx OR dx OR dx  a  b sin x  a  b cos x  a  b sin x  c cos x x Convert sines & cosines into their respective tangents of half the angles & put tan  t 2 In this case sin x = 2t ,cos x  1  t2 , x= 2tan–1t; dx = 2dt NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 1  t2 1  t2 1  t2 dx Illustration 19 : Evaluate :  3 sin x  4 cos x dx dx sec2 x dx = 2 3 sin x  4 cos x 4  6 tan x  4 tan2 x Solution :   I = x = 22  tan x  1  3  2 2 x   4   tan2 2      tan2 x  1 tan2  1   2  2 x  1 sec2 x dx  dt let tan = t, 22 2 2dt 1  dt 1 dt so I =  4  6t  4t2  =  2  3 t 2 2 1   t 2  2 25  3 16   t  4   10 E

1 1 5   t  3  1 1  2 tan x JEE-Mathematics . 4  4  5 4  2 tan = n c = n 2 +c Ans. 2  5  5  3  x 2  4  4   t  4  2 Do yourself -11 : dx dx ( i i ) Evaluate :  1  4 sin x  3 cos x ( i ) Evaluate :  3  sin x (iii)  a cos x  b sin x  c dx p cos x  q sin x  r Express Numerator (Nr) = (Dr) + m d (Dr) + n & proceed. dx Illustration 20 : Evaluate :  sin 2  3 cos  3 d   2 cos   Solution : Write the Numerator = (denominator) + m(d.c. of denominator) + n  2 + 3 cos  = (sin + 2cos + 3) + m(cos – 2sin) + n. Comparing the coefficients of sin, cos and constant terms, we get 3 + n = 2, 2 + m = 3,  – 2m = 0   = 6/5, m = 3/5 and n = –8/5 Hence I = 6 d  3  cos   2 sin  3 d  8  sin   d   3 5 5 sin   2 cos   5 2 cos 6   3 n sin   2 cos   3 8 d 55 5 I3 sin   2 cos   3 = – where I3 = In I3, put   t  sec2  d  2dt tan 2 2 dt dt 1 –1  t  1  = tan–1  tan  / 2  1  2 (t  1)2  22 2.  2   2  t2  2t  5 2  I3 = 2 = t a n NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 Hence I = 6  3 n sin   2 cos   3 8 tan 1  tan  / 2  1   c Ans. 55 5  2  Do yourself -12 : (i) Evaluate :  sin sin x x dx (ii) Evaluate  3 sin x  2 cos x dx x cos 3 cos x  2 sin x ( i v ) sin m x cosn xdx Case-I : When m & n  natural numbers. * If one of them is odd, then substitute for the term of even power. * If both are odd, substitute either of the term. * If both are even, use trigonometric identities to convert integrand into cosines of multiple angles. Case-II : m + n is a negative even integer. * In this case the best substitution is tanx = t. E 11

JEE-Mathematics Illustration 21 : Evaluate  sin3 x cos5 x dx Solution : Put cos x = t; – sin x dx = dt. so that I   (1  t2 ).t5 dt = (t7  t5 )dt  t8 t6  cos8 x cos6 x c   86 8 6 Alternate : Put sin x = t; cos x dx = dt    so that I  t3 (1  t2 )2 dt  t3  2t5  t7 dt sin4 x 2 sin6 x sin8 x   c 468 Note : This problem can also be handled by successive reduction or by trignometric identities. Illustration 22 : Evaluate  sin2 x cos4 x dx  1  cos 2x  cos 2 x  1 2 1  2   2  8   sin2 x cos4 xdx 1  Solution :   dx    cos 2x cos2 2x  2 cos 2x  1 dx   1 cos2 2x  2 cos 2x  1  cos3 2x  2 cos2 2x  cos 2x dx 8  1 1  cos 6x 3 cos 2 x 1  cos 4x  8   cos3 2x  cos2 2x  cos 2x  1 dx     4  2  cos 2 x  1  dx 8  1 sin 6x  3 sin 2x   1 x  sin 4x  sin 2x  x  c   6 2  16 64 16 8 32   sin 6 x  sin 4 x  1 sin 2x  x  c 192 64 64 16 sin x Illustration 23 : Evaluate dx cos9 /2 x NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 sin1/2 x dx  Let I  Solution : dx  sin 1 / 2 x cos9 / 2 x cos9 /2 x Here m + n = 1  9  4 (negative even integer). 22 Divide Numerator & Denominator by cos4x.  I  tan x sec4 x dx  tan x (1  tan2 x) sec2 xdx  t (1  t2 )dt (using tan x = t)  2 t3 / 2  2 t7 / 2  c  2 tan3 / 2 x  2 tan7 / 2 x  c 37 3 7 Do yourself -13 : sin2 x sin xdx (i i i ) Evaluate : sin2 x cos5 x dx ( i ) dx Evaluate : cos4 x ( i i ) Evaluate : cos5 /2 x 12 E

JEE-Mathematics (f) Integration of Irrational functions : (i)  (ax  dx x  q & dx ;put px  q  t2 b) p (ax2  bx  c) px  q  dx 1  dx 1 (ii) ,put ax  b  ; , put x = (ax  b) px2  qx  r t (ax2  bx  c) px2  qx  r t x 2  (x2  3x  3) .dx Illustration 24 : Evaluate x 1 x 2 Put x + 1 = t2  dx = 2tdt Solution : Let, I =  (x2  3x  3) .dx x 1 (t2 1)  2 .(2t)dt = 2 t2  1 dt  2 1  1 / t2 dt    I = {(t2  1)2  3(t2  1)  3} t2 t4  t2  1 t2  1  1 / t2 1  1 / t2 du  w h e r e u  t  1   = 2  t  (t  1 / t)2  ( 3 )2 .dt  2 u2  ( 3 )2  = 2 tan 1  u   c = 2 tan 1  t2 1   c  2 tan 1  x   c Ans. 3  3    3  x  Ans.   3  3t   3 (  1)   dx (x 1) Illustration 25 : Evaluate x2  x 1 Solution : Let, I=  dx 1  dx = –1/t2 dt (x 1) put x – 1 = x2  x 1 t I=  1 / t2dt dt 2 1/t 1  1  1 1 = –  3t2  3t 1  t  1  t NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 1 dt =  1 log (t  1 / 2)  (t  1 / 2)2  1 / 12  c 12 3 =–  2  3  t  1 /12 1 log  1 1  12  x 1  1 2 1 3   2   1 2 =    c  x 1 12 dx Illustration 26 : Evaluate (1  x2 ) 1  x2 Solution : dx Let, I  (1  x2 ) 1  x2 1 1 Put x = , So that dx = dt t t2 1 / t2 dt tdt   I= (1  1 / t2 ) 1  1 / t2  (t2  1) t2  1 again let, t2 = u. So that 2t dt = du. E 13

JEE-Mathematics = 1  du dx where both P and Q are linear so that 2 1) (u  u  1 which reduces to the form  P Q we put u – 1 = z2 so that du = 2z dz 1 2zdz dz I 2   (z2  1  1) z2   (z2  2) I 1 tan 1  z   c 2  2    I 1 tan 1  u  1   c  1  t2  1   c  1  1  x2  c Ans.   2 tan 1  2  2 tan 1  2x   2 2   Do yourself -14 :  (x  3) x dx dx x 1 (i) Evaluate ( i i ) Evaluate : x2 1  x2 Miscelleneous Illustrations : Illustration 27 : Evaluate  cos4 xdx 3 sin3 x{sin5 x  cos5 x}5 cos4 x cos4 x 3 dx = cot4 xcosec2 xdx   I = (1 + cot5 x)3 / 5 Solution : dx = 3 sin3 x{sin5 x  cos5 x}5 sin6 x{1  cot5 x}5 Put 1 + cot5x = t 5cot4x cosec2xdx = –dt =  1 dt  1 t2 / 5 c   1 1  cot5 2 / 5 c Ans.  5 t3 / 5 2 2 x Ans. (C,D) dx (B) n|cotx – tanx|+ c E (D) tan–1(–2cot2x) + c Illustration 28 :  cos6 x  sin6 x is equal to - (A) n|tanx – cotx|+ c (C) tan–1(tanx – cotx) + c dx sec6 x (1  tan2 x)2 sec2 xdx NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 cos6 x  sin6 x = dx = 1  tan6 x 1  tan6 x Solution :   Let I = If tan x = p, then sec2 x dx = dp (1  p2 )2 dp (1  p2 ) p2 1  1  1  p6  p4  p2  1 dp = p2     I = dp  1  1 p2  p 2  p2 = dk  tan 1 (k )  c  where p1  k, 1  1  dp  dk  k2 1  p  p2      = tan–1  p  1   c  tan 1 (tan x  cot x)  c = tan–1(–2cot2x) +c  p  Illustration 29 : Evaluate : 2 sin 2x  cos x  6  cos2 x  4 sin x dx Solution :  2 sin 2x  cos x dx (4 sin x  1) cos x dx  (4 sin x  1) cos x dx  cos2 x  4 sin 6  (1  sin2 x)  4 sin x sin2 x  4 sin x  5  I= 6 x = 14

JEE-Mathematics Put sin x = t, so that cos x dx = dt. (4t 1)dt ...... (i)  I =  (t2  4t  5) Now, let (4t – 1) = (2t – 4) + µ Comparing coefficients of like powers of t, we get 2 = 4, –4 + µ = –1 ...... (ii)  = 2, µ = 7 2(2t  4)  7 {using (i) and (ii)}  I =  t2  4t  5 dt 2t  4 dt = 2log t2  4t  5  7 dt  4t  t2  4t  5 t2  4t  4  4  5   = 2 t2 5 dt  7 = 2log t2  4t  5  7 dt = 2log|t2 – 4t + 5|+7 . tan–1 (t – 2) + c (t  2)2  (1)2 = 2log|sin2x – 4sinx + 5| + 7 tan–1(sinx – 2) + c. Ans. Illustration 30 : The value of  3  x . s in 1  1 3  x  dx, is equal to - 3  x  6  1 3  x  2  x   4   3  3   (A) c o s 1   2 9  x2 . cos1  2x   c    (B) 1 3  co s 1  x 2  2 9  x 2 . sin 1  x   2 x   c 4   3  3         (C) 1 3  sin 1 x 2 2 9  x 2 . sin 1  x   2 x   c 4   3  3         (D) none of these Solution : Here, I = 3  x . sin 1  1 3  x  dx 3  x  6  NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 Put x = 3cos2   dx = –6sin2d = 3  3 co s 2 . sin 1  1 3  3 cos 2 (–6 sin 2)d 3  3 co s 2  6  = sin  .sin1 (sin ).(6 sin 2)d = – 6 .(2 sin2 )d cos  2  – 6  (1  cos 2)d     cos 2 d = = – 6  2   2  sin 2 1.  sin 2    – 32  sin 2 cos 2   2  2  d   2 4  = – 6  2     =  6    c  1 3   x  2  x   4    3  3   = co s 1    2 9  x2 . c o s 1  2 x   c Ans. (A)   tan    x   4  Illustration 31 : Evaluate : dx cos2 x tan3 x  tan2 x  tan x E 15

JEE-Mathematics tan    x  (1  tan2 x)dx  4  Solution : dx =  I = cos2 x tan3 x  tan2 x  tan x (1  tan x)2 cos2 x tan3 x  tan2 x  tan x  1  1 x  sec2 xdx tan2  I =  1  1  tan x  2  tan x  tan x  1  tan x let, y tan x  1  1  2y dy   se c 2 x  1 .sec2 x dx  tan2 tan x x  I=  ( 2 y dy dy y2  1).y = –2  1  y2 = – 2tan–1 y + c = – 2 t a n –1  tan x  1  1  Ans.  c  tan x  ANSWERS FOR DO YOURSELF 1: (i) 1 tan 1  4x3   c (ii) sin x  1 sin3 x  c 36  3  3   2 : (i) 3  x  sin 1 3  x  c (ii) n  x  1   x2  x   c x 2  2  1  3 : ( i ) xex – ex + c (ii)  1 x2 cos(x2 )  sin(x2 )  c 2 4 : ( i ) ex tan–1 + c 5 : ( i ) x tan(ex) + c (ii) 1 ex2 sin(x2 )  c 2 (ii) xnx + c 6 : ( i )  1 n| x  1|  7 n| x  3| c ( i i ) n| x  2|  3  c 22 x 2 7 : (i) 3 tan 1  2x 1   c (ii) 5 x2  4 x  1  6n  x  2   x2  4x  1   c 2  3     NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65  x 2  1  1   x2 1   1 n x2  2x 1  x  22 tan 1  2x  2 x2   8 : (i) tan 1    c (ii)    9 : (i) 2x  1  10 : (i)  1 n  x 2 1   c  1  1 1/3  c 1 1  1 2/3  c 2  x2  x3 2 x3   (ii)  (iii)     1 tan1 5 tan x  c 2 tan 1  8 tan x  1   c   5 (ii) 15  15  1 tan 1  3 tan x / 2  1   c 1 n 6  tan x / 2  2  c   2 6 6  tan x / 2  2 11 : (i) 2  22  (ii) 12 : (i) 13 : (i) 1 x  1 n sin x  cos x  c (ii) 12 5 n 3 cos x  2 sin x C 22 x 1 tan3 x  c 13 13 3 (ii) 2 tan3 /2 x  c (i ii ) 1 sin3 x  2 sin5 x  1 sin7 x  c . 3 357 14 : (i) 2 x  1  3 n x 1 2  c (ii)  1 1  x2  c 2 x 1 2 2 16 E

JEE-Mathematics EXERCISE - 01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1. If f(x) =  2 sin x sin 2 x dx , x  0 then lim f'(x) is equal to- x3 x0 (A) 0 (B) 1 (C) 2 (D) 1/2 2 . 4 sin x cos x cos 3x dx is equal to - 22 (A) cos x  1 cos2x  1 cos 3x c (B) cos x  1 cos 2x  1 cos 3x  c 2 3 23 (C) cos x  1 cos 2x  1 cos 3x  c (D) cos x  1 cos2x  1 cos 3x  c 23 23 3.  8x 13 dx is equal to - 4x 7 1 1 (A) 6 (8x + 11) 4x 7 + c (B) 6 (8x + 13) 4x 7 + c 1 1 (C) 6 (8x + 9) 4x 7 + c (D) 6 (8x + 15) 4x 7 + c 4 .  1 cos8 x  sin8 x x  dx equals -   2 sin2 x cos2    (A)  sin 2x c sin 2x cos 2x (D)  cos 2 x  c (B)  c (C)  c 2 2 2 2 x  5 . Primitive of 3 x4  1 4 w.r.t. x is - 1 1 1 1 (A) 3 1  1 3 c (B)  3 1  1 3 c (C) 4 1  1 3  c (D)  4 1  1 3 c 4 x4 1  4 x4 1  3 x4  1  3 x4   1 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 6 . (1  2x  3x2  4x3 ......) dx (|x| < 1) - (A) (1 + x)–1 + c (B) (1 – x)–1 + c (C) (1 + x)–2 + c (D) none of these 7.  x dx is equal to -  1  x2  1  x2 3  (A) 1 ln 1  1  x2  c (B) 2 1  1  x2  c 2 (D) none of these  (C) 2 1  1  x2  c 8. x n x dx equals - 1  n x E 2 1  n x (ln x  2)  c 2 1  n x (ln x  2)  c (A) 3 (B) 3 1 1  n x (ln x  2)  c (D) 2 1  n x (3n x  2)  c (C) 3 17

JEE-Mathematics x4  1 dx  A  n x  B + c, where c is the constant of integration then : 9 . If   x x2  1 2 1  x2 (A) A = 1; B = –1 (B) A = –1; B = 1 (C) A = 1; B = 1 (D) A = –1 ; B = –1 1 0 .  x 3/2 dx equals -  1  x5   2 x5 (B) 2 x  c 21 c 5 1  x5 c (A) 5 1  x5 (C) 5 1  x5 (D) none of these 1 1 . sin x.cos x.cos 2x.cos 4 x.cos 8x.cos16x dx equals - (A) sin 16 x  c (B)  cos 32 x  c (C) cos 32x  c (D)  cos 32x c 1024 1024 1096 1096 1 2 . Identify the correct expression (A) x nx dx  x2n| x| x2  c (B) x n x dx  xex  c (C) x ex dx  xex  c x (D) dx  1 tan 1  x   c a2  x2 a  a  1 3 . x.  n x  1  x2 dx equals - 1  x2 x . n2 x  1  x2 x c 2 1  x2  (A) 1  x2 n x  1  x2  x  c  (B)  x .n2 x  1  x2 x c 2 1  x2  (C)   (D) 1  x2 n x  1  x2  x  c 14. If dx = a n(1 + x2) + btan–1x + 1  n|x + 2| + C then- 5  (x 2)(x2  1) 12 12 12 12 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 (A) a = – 10 , b = – 5 (B) a = 10 , b = – 5 (C) a = – 10 , b = 5 (D) a = 10 , b = 5  x  12 1 5 . dx equals - x4  2x2 1 x3 x  (B) x5  x3  x  3  c x5  4x3  3x  3  c (A) x c 3 x2 1 3 x2 1 3 x2 1  (C) (D) None of these z x2 4 16. dx equals - x4  24x2 16  (A)  (B) 1 tan 1  x2  4   1 cot 1  x2  4   4x  c  x  c   4 4    (C)  (D) 1 1  4 x2  4  1 co t 1  x2  4   x  c  x  c  cot   4 4   18 E

JEE-Mathematics 1 7 . x4  4 dx equals- x2 4  x2  x4 (A) 4  x2  x 4  c (B) 4  x2  x4  c (C) 4  x2  x 4  c 4  x2  x4 x 2 (D)  c 2x x9 1 8 .  (x2  4)6 dx is equal to - (A) 1  4  1 5 c (B) 1  4  1 5 +c 5x  x2  + 5  x2  1 (D) 1 (1 + 4x–2)–5 + c (C) 10x (1 + 4x2)–5+c 40 19. dx = a tan–1  b tan x  + c, then-  2  If  5  4 cos x 21 21 (A) a = 3 , b = – 3 (B) a = 3 , b = 3 21 21 (C) a = – 3 , b = 3 (D) a = – 3 , b = – 3 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 2 0 . Primitive of 1  2 tan x sec x  tan x  w.r.t.x is - (A) n sec x  n sec x  tan x  c (B) n sec x  tan x  n sec x  c (C) 2n x x c (D) n 1  tan x(sec x  tan x)  c sec  tan 2 2 2 1 .  sin 2x dx equals - (A)  cos 2 x  c sin2 x (C)  cos2 x c (D) cos 2x  c 2 (B)  c 2 2 2 2 2 . dx NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 1 1  equals- 2x2  x3  (A) n 2 x2  1  2n| x| c (B) n 2 x2  1  2n| x|  c (C) n 2 x2  1  n(x2 )  n2  c (D) n 1  1 c 2x2 2 3 . If  e3x cos 4 x dx  e3x (A sin 4 x  B cos 4 x)  c , then - (A) 4A = 3B (B) 2A = 3B (C) 3A = 4B (D) 4A + 3B = 1 CHECK YOUR GRASP AANNSSWWEERR KKEEYY EXERCISE-1 Que. 1 2 3 4 5 6 7 8 9 10 Ans. B B A B B B B A C A Que. 11 12 13 14 15 16 17 18 19 20 Ans. B C A C D A A D B A,B,D Que. 21 22 23 Ans. A,B,C B,C,D C,D E 19

JEE-Mathematics BRAIN TEASERS EXERCISE - 02 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) z b g1 . cot x  tan x dx equals - 2 cos x  sin x (A) sec –1 sin x  cos x   c (B) sec–1 sin x  cos x   c (D) n sin x  cos x   sin 2x  c (C) n sin x  cos x   sin 2x  c z2 . sin x  4 sin 3x  6 sin 5x  3 sin 7x dx equals - sin 2x  3 sin 4x  3 sin 6x (A) –2sinx+c (B) 2sinx+c (C) –2cosx+c (D) 2cosx+c 3 . 1  x7 equals - dx x(1  x7 ) (A) n| x|  2 n 1  x7 c (B) n| x|  2 n 1  x7  c 7 4 (C) n| x|  2 n 1  x7 c (D) n| x|  2 n 1  x7  c 7 4 4.  x 3 dx is equal to - 1 x2 1 1  x2 (2 + x2) + c 1 1  x2 (x2 – 1) + c (A) 3 (B) 3 (C) 1 (1 + x2)3/2 + c 1 1  x2 (x2 – 2) + c 3 (D) 3 5 .  sin2 ( n x) dx is equal to - NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 (A) x + 2sin(2nx) + cos(2nx)) + c (B) x + 2sin(2nx) – cos(2nx)) + c 10 (5 10 (5 (C) x – 2sin(2nx) – cos(2nx)) + c (D) x – 2sin(2nx) + cos(2nx)) + c 10 (5 10 (5 6. x2  cos2 x .cosec2x dx is equal to -  1 x2 (A) cotx + tan–1x + c (B) cotx – tan–1x + c (C) –cotx – tan–1x + c (D) tan–1x – cotx + c 7 .  x2  3  equals - ex  dx ,  x  3 2  (A) ex . x  c (B) ex  2  6   c (C) e x 1  6   c (D) ex. 3  c x 3      x 3 x 3 x 3 20 E

JEE-Mathematics 8 .  e tan1 x (1  x  x2 ). d(cot1 x ) is equal to - (A) – e tan1 x + c (B) e tan1 x + c (C) –x. e tan1 x + c (D) x. e tan1 x + c 9 .  e x (1  n.x n 1  x2n ) dx is equal to - (1  xn ) 1  x2n (A) ex 1  xn +c (B) ex 1  xn +c (C) –ex 1  xn +c (D) –ex 1  xn +c 1  xn 1  xn 1  xn 1  xn 1 0 . ex4 (x  x3  2x5 ) e x2 dx is equal to - (A) 1 xex2 . ex4  c (B) 1 x2 ex4  c (C) 1 ex2 . ex4  c (D) 1 x2ex2. ex4  c 2 2 2 2 1 1 . Primitive of 3x4 1 w.r.t. x is -  x4  x  1 2 x x x 1 c  x 1 c (A) c (B)  c (C) x4  x 1 (D) x4  x 1 x4  x 1 x4  x 1  dx 1 2 . x4 [x (x5  1)]1 / 3 equals - 3 x5 2/3 c 3 x5 2/3 c 3 x5 2/3 c 3  x5  1 2 / 3 c 2 10 4 5   1 1 1   (A)   (B)   (C)   (D)  x5   x5   x5  x5 1 3 . sin x dx is equal to - sin 4x NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 (A) 1 n 1  2 sin x + 1 n 1  sin x + c (B) 1 n 1  2 sin x – 1 n 1  sin x + c 2 2 1  2 sin x 8 1  sin x 2 2 1  2 sin x 8 1  sin x (C) 1 n 1  2 sin x + 1 n 1  sin x + c (D) 1 n 1  2 sin x – 1 n 1  sin x + c 4 2 1  2 sin x 8 1  sin x 4 2 1  2 sin x 8 1  sin x 1 4 . The value of integral d can be expressed as irrational function of tan as - cos3  sin 2  (A) 2 tan2   5 tan   c  (B) 2 tan2   5 tan   c 5 5  (C) 2 tan2   5 tan   c  (D) 2 tan2   5 tan   c 5 5 15. If  3 sin x 2 cos x dx = ax + bn[2sinx + 3cosx| + c, then - 3 cos x 2 sin x 12 15 17 6 12 15 17 1 (A) a = – 13 , b = 39 (B) a = 13 , b = 13 (C) a = 13 , b = – 39 (D) a = – 13 , b = – 192 E 21

JEE-Mathematics 16.  x 1 dx is equal to - x x 1 (A) n x  x2  1 – tan–1x + c (B) n x  x2  1 – tan–1x + c (D) n x  x2  1 – sec–1x + c (C) n x  x2  1 – sec–1x + c 17.  dx is equal to - (1  x ) x  x2 2( x 1) 2(1  x ) 2( x 1) 2(1  x ) (A) 1  x + c (B) 1  x + c (C) + c (D) x 1 + c x 1 18. Let f'(x) = 3x2.sin 1 1 1 = 0, then which of the following is/are not correct. x – xcos x , x  0, f(0) = 0, f    (A) f(x) is continuous at x = 0 (B) f(x) is non-differentiable at x = 0 (C) f'(x) is discontinuous at x = 0 (D) f'(x) is differentiable at x = 0 1 9 . 1 n x  1 dx equals - x2 1 x 1 (A) 1 n2 x 1 c (B) 1 n2 x 1 c (C) 1 n2 x 1 c (D) 1 n2 x 1 c 2 x 1 4 x 1 2 x 1 4 x 1 2 0 . dx equals, where x   1 ,1 - x  x2  2 (A) 2 sin 1 x  c (B) sin 1 (2 x  1)  c (D) cos1 2 x  x2  c (C) c  cos1 (2 x  1)  (C) tan1 tan2 x  c  sin 2x NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 2 1 . dx is equal to - sin4 x  cos4 x  (A) cot1 cot2 x  c  (B)  cot1 tan2 x  c (D)  tan1 cos 2 x   c BRAIN TEASERS AANNSSWWEERR KKEEYY EXERCISE-2 Que. 1 2 3 4 5 6 7 8 9 10 Ans. A B C D C C C C B C Que. 11 12 13 14 15 16 17 18 19 20 Ans. B B D C C D A B,C,D B,D A,B,C,D Que. 21 Ans. A,B,C,D 22 E

JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS FILL IN THE BLANKS 1 .If4ex  6ex dx = Ax + B log(9e2x – 4) + C, then A = ......., B = ....... and C = ...... 9ex  4ex 2 . If the graph of the antiderivative F(x) of f(x) = log(logx) + (logx)–2 passes through (e, 1998 – e) then the term independent of x in F(x) is ....... 3 . Let F(x) be the antiderivative of f(x) = 3cosx – 2sinx whose graph passes through the point (/2, 1). Then F(/2) = ....... 4 . Let f be a function satisfying f\"(x) = x–3/2, f'(4) = 2 and f(0) = 0. Then f(784) is equal to ........ MATCH THE COLUMN Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE statement in Column-II. 1 . The antiderivative of Column-I Column-II (A) f(x) = 1 is (p) 1 ta n –1  a tan x +c ab  b 2  (a2  b2 ) (a2  b2 ) cos x 1 (q) a2 1  tan–1  tan x  + c,  = cos–1 b (B) f(x) = a2 sin2 x  b2 cos2 x is sin  sin   a (C) f(x) = 1 is (r) 1 ta n –1  a tan x  + c ab  b  a cos x  b sin x (D) f(x) = 1 is ; (a2 > b2) (s) 1 log tan 1  x  tan 1 a  +c a2  b2 cos2 x a2  b2 2  b  2 . f(x) dx when Column-I Column-II NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 1 (p) c– 1 sin–1 a (A) f(x) = (a2  x2 )3 /2 a | x| x2 (q) a2 sin–1 x x a2  x2 + c (B) f(x) = 2 a – a2  x2 2 1 x (C) f(x) = (r) c – (x2  a2 )3 /2 1 a2 x2  a2 (D) f(x) = x x x2  a2 (s) + c a2 x2  a2 ASSERTION & REASON In each of the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it . Of the statements mark the correct answer as (A) Statement-I is True, Statement-II is True ; Statement-II is a correct explantion for Statement-I (B) Statement-I is True, Statement-II is True ; Statement-II is NOT a correct explantion for Statement-I (C) Statement-I is True, Statement-II is False. (D) Statement-I is False, Statement-II is True. E 23

JEE-Mathematics f1 (x) f2 (x ) f3 (x ) 1 . If D(x) = a2 b2 c2 , where f1, f2, f3 are differentiable function and a2, b2, c2, a3, b3, c3 are constants. a3 b3 c3  f1 (x )dx  f2 (x)dx  f3 (x)dx Statement - I :  D(x) dx = a2 b2 c2 + c a3 b3 c3 Because Statement - II : Integration of sum of several function is equal to sum of integration of individual functions. (A) A (B) B (C) C (D) D dx 2 . Statement - I : If a > 0 and b2 – 4ac < 0, then the values of integral  ax2  bx c will be of the type µ tan–1 x  A + c. where A, B, C, µ are constants. B Because Statement - II : If a > 0, b2 – 4ac < 0, then ax2 + bx + c can be written as sum of two squares. (A) A (B) B (C) C (D) D 3 . If y is a function of x such that y(x – y)2 = x. Statement - I : dx = 1 log[(x – y)2 – 1] 2  x3y Because Statement - II : dx = log(x – 3y) + c.  x3y (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 In calculating a number of integrals we had to use the method of integration by parts several times in succession. The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts  u(x) v(x)dx = u(x) v1(x) – u'(x)v2(x) + u\"(x) v3(x) – ..... + (–1)n–1 un–1(x) vn(x) – (–1)n–1 un(x) vn(x) dx   where v1(x) = v(x)dx, v2(x) = v1(x) dx ...., vn(x) = vn–1(x) dx NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration by parts is especially useful when calculating Pn(x) Q(x) dx, where Pn(x), is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n + 1 times. If1 . (x3 – 2x2 + 3x – 1)cos2x dx = sin 2x u(x) + cos 2 x v(x) + c, then - 48 (A) u(x) = x3 – 4x2 + 3x (B) u(x) = 2x3 – 4x2 + 3x (C) v(x) = 3x2 – 4x + 3 (D) v(x) = 6x2 – 8x 2 . If e2x .x4 dx  e2x f(x )  C then f(x) is equal to - 2 (A)  x 4  2x3  3x2  3x  3 1 (B) x4 – x3 + 2x2 – 3x + 2  2  2 (C) x4 – 2x3 + 3x2 – 3x + 3 (D) x4 – 2x3 + 2x2 – 3x 3 2 + 2 E 24

JEE-Mathematics Comprehension # 2 Integrals of class of functions following a definite pattern can be found by the method of reduction and recursion. Reduction formulas make it possible to reduce an integral dependent on the index n > 0, called the order of the integral, to an integral of the same type with a smaller index. Integration by parts helps us to derive reduction formulas. (Add a constant in each question) 1 . dx then 12n . 1 If In = (x2  a2 )n In+1 + 2n a2 In is equal to - x 11 1x 11 (A) (x2  a2 )n (B) 2na2 (x2  a2 )n 1 (C) 2na2 . (x2  a2 )n (D) 2na2 . (x2  a2 )  sinn x n 1 2. If I = dx then I+ I is equal to- n, –m cosm x m 1 n–2, 2–m n, –m sinn1 x 1 sinn1 x 1 sinn1 x n 1 sinn1 x (A) cosm 1 x (B) (m 1) cosm1 x (C) (n 1) cosm1 x (D) m 1 cosm1 x 3 . If un = xn dx , then (n + 1)au + (2n + 1)bu + ncu is equal to - ax2 2bx  c n+1 n n–1 (A) xn–1 ax2  2bx  c x n 2 xn (D) xn ax2  2bx  c (B) (C) ax2  2bx  c ax2 2bx  c NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3  Fill in the Blanks 3 35 2. 1998 3. 1 4. 2240 1 . 2 , 36 , any real value  Match the Column 1 . (A) p; (B) r; (C)  s; (D)  q 2 . (A) s; (B) q,; (C)  r; (D)  p  Assertion & Reason 1 . (A) 2 . (A) 3 . (C)  Comprehension Based Questions Comprehension # 1 : 1 . (B) 2 . (C) Comprehension # 2 : 1. (C) 2 . (B) 3 . (D) E 25

JEE-Mathematics EXERCISE - 04 [A] CONCEPTUAL SUBJECTIVE EXERCISE Evaluate the following Indefinite integrals :  dx  5x4  4x5 2. dx 3 . tan x. tan 2x. tan 3x dx 1 . sin(x  a) sin(x  b) (x5  x  1)2 4 .  x2 1  n(x2  1)  2  n x      x4  dx  5 . Int eg r at e 1 f '(x) w.r.t. x4 , wh er e f( x ) = tan 1 x   n 1  x   n 1  x 2 6 . cos ecx  cot x . sec x dx  (ax2  b) dx  x2 cos ecx  cot x 1  2 sec x 8. dx 7. x c2 x2  (ax2  b)2 (x sin x  cos x)2 9. cos 2. n cos   sin  d 1 0 .  xx   ex  n x dx  xn x cos sin   e   x    1 1 . (x2  1)3 /2 dx x3  3x  2 13. 3 x2  1 dx  dx 1 2 . dx [JEE 99] (x2  1)3 [JEE 84] (x2  1)2 (x  1) 1 4 . x2 (x4 1)3 / 4  dx 1 6 . (sin x)11 / 3 (cos x)1 / 3 dx 1 7. cos x  sin x dx 7  9 sin 2x 1 5 . sin2 x  sin 2x  cos2 x 1 9 .  tan x  co t x  dx 18. dx [JEE 89] 1  tan x  (cos 2x)1 / 2 1 [JEE 87] 21. dx [JEE 92] 2 0 . dx 3 x4 x sin x CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1 . cos ec(b  a). n sin(x  b)  c 2.  x 1 c sin (x  a) x5  x 1 3.  –  n(sec x)  1  n(sec 2 x )  1  n(sec 3 x )  + c 4. (x2  1) x2 1   3 n 1  1  2 3  9x3 . 2 x2   5 .  n 1  x4  c 6 . sin 1  1 sec2 x   c sin 1  ax 2 b   k sin x  x cos x  c NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65  2 2   cx  x sin x  cos x 7.   8. 9. 1 (sin2)n  cos   sin   – 1 n(sec2) + c 10.  x x   e x  c 11. arc secx – nx + c 2  cos   sin    e  x   2   x2  1   1 2 . 3 tan 1 x  1 n(1  x )  1 n(1  x2 )  x  c x 14. – 1  1 1/4  c x4 22 4 1  x2 1 3 . C – (x2 1)2   1 n tan x c 3(1  4 tan2 x) c 1  n (4  3 sin x 3 cos x)  c 15. 2 tan x  2 16.  17. 24 (4  3 sin x  3 cos x) 8 (tan x)8 / 3 1 8 . 1 n(cos + sinx) + x + 1 19. 2 tan–1  tan x  cot x  +c 2   4 8 (sin2x + cos2x)  2 1 2 1  tan2 x  20. log  2  – log(cotx + cot2 x 1 ) + c 1  tan2 x  2  2 1 . 3 x2/3 – 12 x7/12 + 2x1/2 – 12 x5/12 + 3x1/3 + 6x1/6 – 12x1/12 + 12log|x1/12 + 1| – 4x1/4 + c 27 5 26 E

EXERCISE - 04 [B] JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE 1 . cos 8x  cos7x dx  2. x  x2  2 dx 3. sin(x  a) dx  cot x dx 1  2 cos 5x sin(x  a) 4 . (1  sin x)(sec x  1) 5 . 1  x dx dx dx dx 1 x sec x  cos ecx sin x sin(2x  )  6. 7.  n cos x  cos 2x 9 .  ex 2  x2 8.  dx dx sin2 x (1  x ) 1  x2 1 0 0 x   x 1 0 . Let 6 2 4 0   x 2  = 2x  x2   x  R and f(x) is a differentiable function satisfying, 5    , 5 x  x2  3  3  1   f(xy) = f(x) + x2 (y2 – 1) + x (y – 1) ;  x, y  R and f(1) = 3 . Evaluate x2  x   dx f(x) 1 1 . cot x  tan x dx 12. sin 1 x 1  3 sin 2x dx ax f(x)dx 1 3 . Let f(x) is a quadratic function such that f(0) = 1 and  x2 (x 1)3 is a rational function, find the value of f'(0) 14. ecosx (x sin3 x  cos x) dx x  sin2 x 15.  (7x 10  x2 )3 /2 dx BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1 (2 sin 3x  3 sin 2x)  c 1 x 3/2 2 6 x2  2  1. 2.    3 c 1/2 x  x2  2 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65  3 .  cos x  a. 1 n tan x  1 sec2 x  tan x  c cos a.arc cos  cos a   sin n sin x  sin2 x  sin2 a  c 4 . 2 24 2 2 5 . x 1  x  2 1  x  arc cos x  c 6. 1 sin x  cos x  1 n tan  x    c 2  2  2   8  7.  1   n cot x  cot   cot2 x  2 cot  cot x 1   c sin    8 . cos 2x  x  cot x. n e cos x  cos 2x  c 9. ex 1  x  c sin x 1x HF KI FG JI1 0 . 3x – n x2  x  1  3 tan 1 2x  1  c 11. tan 1  2 sin 2x x   c H K3  sin x  cos  x ax + c 13. 3 1 5 . 2(7x 20) + c 1 2 . (a + x) arc tan – a 9 7x 10  x2 1 4 . C – ecosx(x + cosecx) E 27

JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [A] 1.  cos 2 x 1 dx = [AIEEE-2002] cos 2 x 1 (4) –x – cotx + C (1) tanx – x + C (2) x + tanx + C (3) x – tanx + C (log x) [AIEEE-2002] 2 .  x2 dx 1 1 1 (4) log(x + 1) + C (1) (logx + 1) + C (2) – (logx + 1) + C (3) (logx – 1) + C 2 x x 3 . If sin x dx =Ax + B logsin(x –) + C then values of (A, B) is - [AIEEE-2004] sin(x ) (1) (sin, cos) (2) (cos, sin) (3) (–sin, cos) (4) (–cos, sin) 4. dx is equal to- [AIEEE-2004]  cos x sin x (1) 1 log tan  x    +C (2) 1 log cot  x  +C 2  2 8  2  2  (3) 1 log tan  x  3  +C (4) 1 log tan  x  3  +C 2  2 8  2  2 8  5 .  (log x 1) 2 dx is equals to - [AIEEE-2005] 1  (log x )2    log x x xex x (1) (log x)2 1 + C (2) x2 1 + C (3) 1  x2 + C (4) (log x)2 1 + C 6.  dx equals- [AIEEE-2007] NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 cos x  3 sin x (1) 1 logtan  x    + C (2) 1 logtan  x    + C 2  2 12  2  2 12  (3) logtan  x    + C (4) logtan  x    +C  2 12   2 12  7. The value of  sin x dx is - 2    [AIEEE-2008]  4  sin x  (1) x + log cos  x    +c (2) x – log sin  x    +c (3) x + log sin  x    +c (4) x – log cos  x    +c  4   4   4   4  8 . If the integral 5 tan x dx = x + a ln|sin x – 2 cos x| + k then a is equal to : [AIEEE-2012] tan x 2 E (1) 2 (2) –1 (3) –2 (4) 1 28

9 . If  ƒ(x)dx  (x) , then  x5 ƒ(x3 )dx is equal to : JEE-Mathematics [JEE (Main)-2013] 1  3 3 2 3  1 x3(x3 )  3 x3(x3 )dx  C 3   3 (1) x  ( x )  x  ( x ) dx  C (2) 1 x3(x3 )  x2(x3 )dx  C 1  x 3 (x3 )  x3 (x3 )dx   C 3 3(4) (3) NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Qu e. 1 2 3 4 5 6 7 8 9 Ans 3 2 2 4 4 1 3 1 3 E 29

JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [B] 1 . sin 1  2x 2  Evaluate :  4x2  8x  13  dx . [JEE 2001 (Mains) 5M out of 100] 2 . (x3m x2m xm )(2x2m 3xm 1 where 0 For any natural number m, evaluate + + + + 6) m dx x > z3 . x2 1 dx is equal to - [JEE 2002 5M out of 60] [JEE 2006, (3M, –1M) out of 184] x3 2x4  2x2 1 (A) 2x4  2x2 1  c (B) 2x4  2x2 1  c (C) 2x4  2x2 1  c (D) 2x4  2x2 1  c x2 x3 x 2x2 x (f f...f) (x). Then xn2 g(x)dx equals. [JEE 2007, 3M] Let f(x) = (1  xn )1 / n f occurs n times 4 . for n  2 and g(x) = (A) 1 (1  n x n 1 1 K (B) 1 (1  nx n 1 1 K (C) 1 (1  n x n 1 1 K (D) 1 (1  nx n 1 1 K n n n n ) ) ) ) n(n 1) n 1 n(n  1) n 1 5 . Let F(x) be an indefinite integral of sin2x. [JEE 2007, 3M] Statement-1 : The function F(x) satisfies F(x + ) = F(x) for all real x. because Statement-2 : sin2(x + ) = sin2x for all real x. (A) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. ex ex  6 . Let I= e4x  e2x dx , J = e 4 x  e 2 x  1 dx . [JEE 2008, 3M, –1M] 1 Then, for an arbitrary constant c, the value of J – I equals (A) 1  e4x  e2x 1   c (B) 1  e4x  e2x 1   c 2 log  e4x  e2x 1  2 log  e2x  e2x 1      (C) 1  e2x  ex 1   c (D) 1  e4x  e2x 1   c NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 2 log  e2x  ex 1  2 log  e4x  e2x 1      7. The integral sec2 x equals (for some arbitrary constant K) [JEE 2012, 3M, –1M]  dx (sec x  tan x)9 /2 (A)  1 x 11 / 2 1  1 sec x  tan x 2   K (B) sec 1 x 11 / 2 1  1 sec x  tan x 2   K tan 11 7  tan 11 7  sec x   x   (C)  1 x 11 / 2 1  1 sec x  tan x 2   K (D) sec 1 x 11 / 2 1  1 sec x  tan x 2   K tan 11 7  tan 1 1 7  sec  x   x  PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] 1 . (x  1) tan1 2(x  1)  3  n(4x2  8x  13)  c 2. m 1 34 6.  2 x3m  3 x2m  6 x m m 3. D 4. A 5. D C 6 (m 1) C 7. C 30 E

JEE-Mathematics INDEFINITE INTEGR ATION If f & F are function of x such that F' (x) = f(x) then the function F is called a PRIMITIVE OR ANTIDERIVATIVE OR INTEGRAL of f(x) w.r.t. x and is written symbolically as  f(x) dx  F(x)  c  d {F (x )  c} f(x ) , where c is called the constant of integration. dx 1 . GEOMETRICAL INTERPRETATION OF INDEFINITE INTEGR AL :  f(x)dx  F (x)  c  y (say ) , represents a family of curves. The different values of c will correspond to different members of this family and these members can be obtained by shifting any one of the curves parallel to itself. This is the geometrical interpretation of indefinite integral. Let f(x) = 2x. Then f(x)dx  x2  c. For different values Y of c, we get different integrals. But these integrals are y= x2+3 very similar geometrically. Thus, y = x2 + c, where c is arbitrary constant, represents P3 y= x2+2 a family of integrals. By assigning different values to c, we get different members of the family. These together P2 y= x2+1 constitute the indefinite integral. In this case, each integral represents a parabola with its axis along y-axis. P1 y= x2 If the line x = a intersects the parabolas y = x2, y = x2 +1, y = x2 + 2, y = x2 – 1, y = x2 – 2 at P , P , P , P , P etc., P0 y= x2–1 0 1 2 –1 –2 X' P–1 y= x2–2 X dy respectively, then at these points equals 2a. This P–2 y= x2–3 dx indicates that the tangents to the curves at these points are parallel. Thus, 2 x d x = x 2 + c = f( x ) + c (say), P–3 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 implies that the tangents to all the curves x= a f(x) + c, c  R, at the points of intersection of the Y' curves by the line x = a, (a  R) , are parallel. 2. STANDARD RESULTS : E ( i ) (ax  b)n dx  (ax  b)n1  c ; n  1 (ii)  dx b  1 n ax  b c a(n 1) ax  a ( i i i ) eaxbdx  1eaxb  c ( i v ) apxq dx  1 apxq  c, (a  0) p na a (v)  sin(ax  b)dx   1 cos(ax  b)  c (vi)  cos(ax  b)dx  1 sin(ax  b)  c a a (vii)  tan(ax  b)dx  1  n| sec(ax  b)| c (viii)  cot(ax  b)dx  1 n| sin(ax  b)| c a a (ix)  sec2 (ax  b)dx  1 tan(ax  b)  c (x)  cos ec2 (ax  b)dx   1 cot(ax  b)  c a a (xi)  cosec (ax  b ). cot(ax  b)dx   1 cosec (ax  b)  c a (xi i) sec (ax + b).tan(ax + b)dx = 1 sec(ax  b)  c a 1

JEE-Mathematics (xiii)  sec x dx  n sec x  tan x c = n tan    x   c  4 2  (xiv)  cosec x dx  n cosec x  cot x c = n tan x c = n | cosec x  cot x| + c 2 (x v) dx  sin 1 x  c (xvi ) dx  1 tan 1 x c a2  x2 a a2  x2 a a (x vi i) dx  1 sec1 x  c (xvi ii) dx   n  x  x2  a2  +c x x2  a2 a a x2  a2   ( x i x ) dx   n  x  x2  a2  +c ( x x ) dx  1 n ax c x2  a2   a2  x2 2a ax ( x x i ) dx  1 n x a c ( x x i i ) a2  x2 dx  x a2  x2  a2 sin 1 x  c x2  a2 2a xa 2 2a x2  a2 dx  x x2  a2  a2 n x  x2  a2  c 22  (xxiii) x2  a2 dx  x a2   (xxiv) x2  a2 22 n x x2  a2 c (xx v) eax .sin bx dx  eax (a sin bx  b cos bx)  c  e ax sin  bx  tan 1 b   c a2  b2 a2  b2  a  ( x x v i ) eax.cos bx dx  eax (a cos bx  b sin bx)  c  e ax cos  bx  tan 1 b   c a2  b2 a2  b2  a  3 . TECHNIQUES OF INTEGR ATION : (a) Substitution or change of independent variable : If (x) is a continuous differentiable function, then to evaluate integrals of the form  f((x)) '(x)dx , we substitute (x) = t and '(x)dx = dt. Hence I   f((x)) '(x)dx reduces to  f(t)dt . NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 ( i ) Fundamental deductions of method of substitution :  f(x)n f '(x)dx f '(x) OR  [f(x)]n dx put f(x) = t & proceed. Illustration 1 : Evaluate cos3 x dx sin2 x  sin x Solution : I =  (1  sin2 x) cos x dx =  1  sin x cos x dx sin x(1  sin x) sin x Put sinx = t  cosx dx = dt  I = 1 t dt  n| t| t  c = n| sin x|  sin x  c Ans. t E Illustration 2 : Evaluate x2 1 dx x4  3x2  1 tan1  x  1  x  2

JEE-Mathematics Solution : The given integral can be written as 1  1  dx x2  I =  12   1  x  x   1 tan 1  x  x   Let  x  1 = t. Differentiating we get 1  1  dx = dt  x  x2  dt Hence I =  t2  1 tan1 t dt du  n| u| c u Now make one more substitution tan–1t = u. Then t2  1 = du and I = Returning to t, and then to x, we have I= n| tan 1 t| c  n tan 1  x  1   c Ans.  x  Do yourself -1 : x2 ( i i ) Evaluate :  cos3 x dx ( i ) dx Evaluate : 9 16x6 (ii) Standard substitutions : dx  a2  x2 or a2  x2 dx ; put x = a tan or x = a cot   dx a2  x2 dx ; put x = a sin or x = a cos  a2  x2 or NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 dx  x2  a2 or x2  a2 dx ; put x = a sec or x = a cosec   a  x dx ; put x = a cos2 ax  x  dx or  (x  )(  x) ; put x =  cos2  +  sin2  x  x  dx or  (x  )(x  ) ; put x =  sec2  –  tan2  x   dx ; put x –  = t2 or x –  = t2. (x  )(x  ) Illustration 3 : Evaluate  dx (x  a)(b  x) Solution : Put x = acos2 + bsin2, the given integral becomes E 3

JEE-Mathematics I 2(b  a) sin  cos d 1  (a cos2   b sin2   a )(b  a cos2   b sin2  2  2(b  a) sin  cos d ba 2d  2  c  2 sin1  x  a   c  b  a   b  a  b  a sin cos  = = Ans. Illustration 4 : Evaluate  1 x1 1 . dx xx Solution : Put x = cos2  dx = –2sin cos d  I 1  cos  . 1 ( 2 sin  cos )d   2 tan  tan  d 1  cos  cos2 2   4  sin2 ( / 2) d  2  1  cos  d  2n| sec   tan | 2  c cos  cos   2n 1  1  x  2 cos1 x  c x Do yourself -2 : (i) Evaluate :  x  3 dx (ii) Evaluate :  dx 2x x x2  4 (b) Integration by part :  u.v dx  u v dx    du . v dx  dx where u & v are differentiable functions and  dx  are commonly designated as first & second function respectively. Note : While using integration by parts, choose u & v such that (i)  vdx & (ii)   du . v dx  dx are simple to integrate.  dx  This is generally obtained by choosing first function as the function which comes first in the word ILATE, where; I-Inverse function, L-Logarithmic function, A-Algebraic function, T-Trigonometric function & E-Exponential function. Illustration 5 : Evaluate :  cos x dx NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 Solution : Consider I =  cos xdx Let x  t 1 then dx  dt 2x i.e. dx  2 xdt or dx = 2t dt so I   cos t.2tdt taking t as first function, then integrate it by part  I =2    dt  cos tdt  dt  = 2 t sin t   1 . sin tdt  = 2t sin t  cos t  c  dt    t cos tdt     I = 2  x sin x  cos x  c Ans. Illustration 6 : Evaluate :  1  x x dx E sin 4

JEE-Mathematics Solution : Let I =  x dx =  (1  x(1  sin x) x ) dx sin sin x)(1  sin 1  x = x(1  sin x) =  x(1  sin x) dx   x sec2 xdx   x sec x tan xdx cos2 x  1  sin2 x dx x  sec2 xdx   dx  sec2 xdx  dx    dx   =    – x  sec x tan xdx    dx  sec x ta n xdx  dx    dx       = x tan x   tan xdx   x sec x   sec xdx  = x tan x  n| sec x|  x sec x  n| sec x  tan x|  c = x tan x  sec x   n (sec x  tan x)  c = x(1  sin x)  n|1  sin x|  c Ans. sec x cos x Do yourself -3 : ( i i ) Evaluate : x3 sin(x2 )dx ( i ) Evaluate :  xe x dx Two classic integrands : ( i ) ex [f(x)  f '(x)]dx  ex.f(x)  c  1x  2  1  x2  Illustration 7 : Evaluate e x dx  1x 2 ex (1  2x  x2 ) dx = ex 1  2x  dx  ex c  1  x2 (1  x2 )2  (1  x2 ) (1  x2 )2  1  x2 Solution :   ex  dx = Ans.  Illustration 8 : ex  x4  2  The value of  (1  x2 )5 /2  dx is equal to - ex (x  1) ex (1  x  x2 ) ex (1  x) (D) none of these (A) (1  x2 )3 / 2 (B) (1  x2 )3 / 2 (C) (1  x2 )3 / 2 Ans. (D) NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 Solution :  Let I = e x  x4  2  dx = ex  1  1  2x2    x2 )5 /   x2 (1  x2 )5 / 2  .dx (1 2 (1  )1 / 2 = e x  1 x  x  12 x2  dx  x2  (1  x2 )3 / 2 (1  x2 )5 /  (1  )1 / 2 2 (1  x2 )3 / 2 = ex xex c = ex {1  x2  x} (1  x2 )1 / 2  c (1  x2 )3 / 2 (1  x2 )3 / 2 Do yourself -4 : ( i ) ex  t a n 1 x  1  dx   ( i i ) Evaluate : xe x2 sin x2  cos x2 dx Evaluate :   x2  1 (ii)  [f(x)  xf '(x)] dx  x f(x)  c Illustration 9 : Evaluate  x  sin x dx 1  cos x E 5

JEE-Mathematics  Solution : I =  x  sin x dx =   x  sin x  dx =   x 1 sec 2 x  tan x  dx x Ans. 1  cos x  2 cos2 x   2 2 2  = x tan + c   2 2 Do yourself -5 : (ii) Evaluate :  (nx  1)dx  ( i ) Evaluate : tan(ex )  xe x sec2 (ex ) dx (c) Integration of rational function : P(x) ( i ) Rational function is defined as the ratio of two polynomials in the form Q(x) , where P(x) and Q(x) are polynomials in x and Q(x)  0. If the degree of P(x) is less than the degree of Q(x), then the rational function is called proper, otherwise, it is called improper. The improper rational function P(x) can be reduced to the proper rational functions by long division process. Thus, if Q(x) is improper, then P(x) = T(x) + P1 (x) , where T(x) is a polynomial in x and P1 (x) is proper rational function. Q(x) Q(x) Q(x) It is always possible to write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition. After this, the integration can be carried out easily using the already known methods. S. No. Form of the rational function Form of the partial fraction px2  qx  r ABC xa + xb + xc 1. (x  a) (x  b)(x  c) ABC px2  qx  r x  a + (x  a)2 + x  b A Bx C 2. (x  a )2 (x  b) x  a + x2  bx  c px2  qx  r NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 3. (x  a ) (x2  bx  c) where x2 + bx + c cannot be factorised further f(x) A Bx C Dx  E 4. (x  a )(x2  bx  c)2 x  a + x2  bx  c + (x2  bx  c)2 where f(x) is a polynomial of degree less than 5. Illustration 10 : Evaluate  ( x  2 x  5 ) dx )( x Solution : x AB (x  2)(x  5 ) x  2 x  5 or x = A(x + 5) + B(x – 2). by comparing the coefficients, we get A = 2/7 and B = 5/7 so that  (x x  5) dx  2  dx  5  dx  2 n (x  2)  5 n x 5 c Ans.  2)(x 7 x 2 7 x 5 7 7 E 6

x4 JEE-Mathematics Illustration 11 : Evaluate  (x  2)(x2  1) dx Ans. Solution : x4 3x2  4 (x  2)(x2  1)  (x  2)  (x  2)(x2  1) 12  x 3x2  4  16 5 5 Now, (x  2)(x2  1) 5(x  2) x2  1 12  x So, x4  x2 16 5 5 (x  2)(x2  1) 5(x  2) x2 1 Now,   2)  16 2)  1 x  2  dx  5(x  55  x2 1  (x   = x2  2 x  2 tan 1 x  16 n x  2  1 n(x2  1)  c 25 5 10 Do yourself - 6 : (i) Evaluate :  (x 3x 2 3) dx x2 1  1)(x  ( i i ) Evaluate :  (x  1)(x  2) dx (ii)  ax2 dx  ,  ax2 dx , ax2  bx  c dx  bx c  bx  c Express ax2 + bx + c in the form of perfect square & then apply the standard results. px  q px  q (iii)  ax2  bx  c dx, dx ax2  bx  c Express px + q =  (differential coefficient of denominator ) + m. NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 dx Illustration 12 : Evaluate  2x2  x 1 Solution : I   2x2 dx 1 1 dx 1 dx 1 1 x 1   2  x2  x  1  2  x2  x   22 2 16 16 2 1 dx 1 dx  =  2 (x  1 / 4)2  9 / 16 2 (x  1 / 4)2  (3 / 4)2 = 1 . 1 4) log x 1/4 3/4 c using, x2 dx  1 log x a  c  2 2(3 / x 1/4 3/4  a2 2a xa   1 x 1 / 2 c 1 2x 1 c Ans. = log  log 3 x 1 3 2(x 1) 3x  2 7 Illustration 13 : Evaluate  4x2  4x  5 dx E

JEE-Mathematics Solution : Express 3x + 2 = (d.c. of 4x2 + 4x + 5) + m or, 3x + 2 = (8x + 4) + m Comparing the coefficients, we get 8 = 3 and 4 + m = 2   = 3/8 and m = 2 – 4 = 1/2  I = 3  8x 4 dx  1  4x2 dx  5 8 4x2  4x 5 2  4x = 3 log 1 dx = 3 log 4x2 1 1  1  4x2  4x  5  x5 8  4x 5  8 tan  x  2   c Ans. 88 x2 4 Do yourself -7 : 5x  4 dx (ii) Evaluate :  x2  3x  2 dx ( i ) Evaluate :  x2  x  1 ( i v ) x2 1 dx OR  x4 x2 1 dx where K is any constant. Integrals of the form x4  Kx2 1  Kx2 1 Divide Nr & Dr by x2 & proceed. Note : Sometimes it is useful to write the integral as a sum of two related integrals, which can be evaluated by making suitable substitutions e.g. 2x2 dx  x2  1 dx  x2 1 dx 2 dx  x2  1 dx  x2 1dx      * * x4 1 x4 1 x4 1 x4 1 x4 1 x4 1 These integrals can be called as Algebric Twins. 4 Illustration 14 : Evaluate : dx sin4 x  cos4 x 1 sin2 x  cos2 x sin4 x  cos4 x dx = 4 dx Solution :  I = 4 sin4 x  cos4 x NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 (tan2 x  1) cos2 x dx  4 (tan2 x  1) sec2 x dx (tan4 x  1) cos4 x (tan4 x  1)  = 4 Now, put tanx = t  sec2x dx = dt 1  t2 1 / t2 1 1  t4 dt  4 dt   I = 4 t2  1 / t2 Now, put t – 1/t = z  1  1  dt = dz t2   I = 4 dz  4 tan1 z 2 2 tan 1 t  1 / t = 2 2 tan 1  tan x  1/ tan x   c z2  2 2 2 2  2    Ans. 1 Illustration 15 : Evaluate : dx x4  5x2  1 8 E

JEE-Mathematics 12 Solution : I = 2 dx x4  5x2 1 I1 1  x2 dx  1 1  x2 dx 1 1 1 / x2 1 1 1 / x2 2 x4  5x2 1 2 x2  5  1 / x2 dx  2 dx     = x2  5 1 / x2 x4  5x2 1 2 {dividing Nr and Dr by x2} (1  1 / x2 ) (1  1 / x2 )dx 1 dt 1 du  1  1  2  t2   72  2  u2   3 2 dx  2 (x  1 / x)2  7 2 (x  1 / x)2  3 11 where t = x – and u = x  xx I = 1. 1  tan 1 t   1 . 1  ta n 1 u   c 2 7  7  2 3  3      1  1 tan 1  x 1/ x   1 tan 1  x 1/ x   c 2  7  7  3  3  =      Ans. Do yourself -8 : ( i ) x2  1 dx 1 Evaluate : x4  x2 1 ( i i ) Evaluate :  1  x4 dx (d) Manipulating integrands : ( i ) dx , n N , take xn common & put 1 + x–n = t. x(xn 1) (ii)  dx , n  N , take xn common & put 1 + x–n = tn (n 1)  x2 xn  1 n NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 (i i i ) dx , take xn common and put 1 + x–n = tn. x n (1  x n )1 / n dx Illu stration 16 : Evaluate : xn (1  xn )1/ n Solution :  dx dx Let I = x n (1  x n )1 / n = x n 1 1  1 1/n xn   Put 1 1 = tn, then 1 dx  tn 1 dt xn x n 1 n 1 tn 1 dt tn2 dt tn 1 c 1 1 1  n c   n 1 xn  t n 1  I = –    Ans. Do yourself -9 : dx dx dx ( i ) Evaluate : ( i i ) Evaluate : ( i i i ) Evaluate : x3 (x3  1)1 / 3 x(x2 1) x2 (x3  1)2 / 3 E9

JEE-Mathematics (e) Integration of trigonometric functions : dx dx dx ( i )  a  b sin2 x OR  a  b cos2 x OR a sin2 x  b sin x cos x  c cos2 x Divide Nr & Dr by cos2 x & put tan x = t . Illustration 17 dx Solution : : Evaluate :  2  sin2 x Divide numerator and denominator by cos2x  sec2 xdx sec2 xdx I = 2 sec2 x  tan2 x = 2  3 tan2 x Let 3 tan x  t  3 sec2 x dx  dt I 1 dt 1 . 1 tan 1 t c = 1 1  3 tan x  3  32 2   S o tan  c Ans. 2  t2 Ans. 6  2 Illustration 18 : dx Evaluate :  (2 sin x  3 cos x)2 Solution : Divide numerator and denominator by cos2x sec2 xdx  I =  (2 tan x  3)2 Let 2 tan x  3  t ,  2sec2xdx = dt I  1  dt = 1 c  1  3) c 2 t2 2t  2(2 tan x Do yourself -10 : dx dx ( i i ) Evaluate :  3 sin2 x  sin x cos x  1 ( i ) Evaluate :  1  4 sin2 x (ii) dx OR dx OR dx  a  b sin x  a  b cos x  a  b sin x  c cos x x Convert sines & cosines into their respective tangents of half the angles & put tan  t 2 In this case sin x = 2t ,cos x  1  t2 , x= 2tan–1t; dx = 2dt NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 1  t2 1  t2 1  t2 dx Illustration 19 : Evaluate :  3 sin x  4 cos x dx dx sec2 x dx = 2 3 sin x  4 cos x 4  6 tan x  4 tan2 x Solution :   I = x = 22  tan x  1  3  2 2 x   4   tan2 2      tan2 x  1 tan2  1   2  2 x  1 sec2 x dx  dt let tan = t, 22 2 2dt 1  dt 1 dt so I =  4  6t  4t2  =  2  3 t 2 2 1   t 2  2 25  3 16   t  4   10 E

1 1 5   t  3  1 1  2 tan x JEE-Mathematics . 4  4  5 4  2 tan = n c = n 2 +c Ans. 2  5  5  3  x 2  4  4   t  4  2 Do yourself -11 : dx dx ( i i ) Evaluate :  1  4 sin x  3 cos x ( i ) Evaluate :  3  sin x (iii)  a cos x  b sin x  c dx p cos x  q sin x  r Express Numerator (Nr) = (Dr) + m d (Dr) + n & proceed. dx Illustration 20 : Evaluate :  sin 2  3 cos  3 d   2 cos   Solution : Write the Numerator = (denominator) + m(d.c. of denominator) + n  2 + 3 cos  = (sin + 2cos + 3) + m(cos – 2sin) + n. Comparing the coefficients of sin, cos and constant terms, we get 3 + n = 2, 2 + m = 3,  – 2m = 0   = 6/5, m = 3/5 and n = –8/5 Hence I = 6 d  3  cos   2 sin  3 d  8  sin   d   3 5 5 sin   2 cos   5 2 cos 6   3 n sin   2 cos   3 8 d 55 5 I3 sin   2 cos   3 = – where I3 = In I3, put   t  sec2  d  2dt tan 2 2 dt dt 1 –1  t  1  = tan–1  tan  / 2  1  2 (t  1)2  22 2.  2   2  t2  2t  5 2  I3 = 2 = t a n NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 Hence I = 6  3 n sin   2 cos   3 8 tan 1  tan  / 2  1   c Ans. 55 5  2  Do yourself -12 : (i) Evaluate :  sin sin x x dx (ii) Evaluate  3 sin x  2 cos x dx x cos 3 cos x  2 sin x ( i v ) sin m x cosn xdx Case-I : When m & n  natural numbers. * If one of them is odd, then substitute for the term of even power. * If both are odd, substitute either of the term. * If both are even, use trigonometric identities to convert integrand into cosines of multiple angles. Case-II : m + n is a negative even integer. * In this case the best substitution is tanx = t. E 11

JEE-Mathematics Illustration 21 : Evaluate  sin3 x cos5 x dx Solution : Put cos x = t; – sin x dx = dt. so that I   (1  t2 ).t5 dt = (t7  t5 )dt  t8 t6  cos8 x cos6 x c   86 8 6 Alternate : Put sin x = t; cos x dx = dt    so that I  t3 (1  t2 )2 dt  t3  2t5  t7 dt sin4 x 2 sin6 x sin8 x   c 468 Note : This problem can also be handled by successive reduction or by trignometric identities. Illustration 22 : Evaluate  sin2 x cos4 x dx  1  cos 2x  cos 2 x  1 2 1  2   2  8   sin2 x cos4 xdx 1  Solution :   dx    cos 2x cos2 2x  2 cos 2x  1 dx   1 cos2 2x  2 cos 2x  1  cos3 2x  2 cos2 2x  cos 2x dx 8  1 1  cos 6x 3 cos 2 x 1  cos 4x  8   cos3 2x  cos2 2x  cos 2x  1 dx     4  2  cos 2 x  1  dx 8  1 sin 6x  3 sin 2x   1 x  sin 4x  sin 2x  x  c   6 2  16 64 16 8 32   sin 6 x  sin 4 x  1 sin 2x  x  c 192 64 64 16 sin x Illustration 23 : Evaluate dx cos9 /2 x NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 sin1/2 x dx  Let I  Solution : dx  sin 1 / 2 x cos9 / 2 x cos9 /2 x Here m + n = 1  9  4 (negative even integer). 22 Divide Numerator & Denominator by cos4x.  I  tan x sec4 x dx  tan x (1  tan2 x) sec2 xdx  t (1  t2 )dt (using tan x = t)  2 t3 / 2  2 t7 / 2  c  2 tan3 / 2 x  2 tan7 / 2 x  c 37 3 7 Do yourself -13 : sin2 x sin xdx (i i i ) Evaluate : sin2 x cos5 x dx ( i ) dx Evaluate : cos4 x ( i i ) Evaluate : cos5 /2 x 12 E

JEE-Mathematics (f) Integration of Irrational functions : (i)  (ax  dx x  q & dx ;put px  q  t2 b) p (ax2  bx  c) px  q  dx 1  dx 1 (ii) ,put ax  b  ; , put x = (ax  b) px2  qx  r t (ax2  bx  c) px2  qx  r t x 2  (x2  3x  3) .dx Illustration 24 : Evaluate x 1 x 2 Put x + 1 = t2  dx = 2tdt Solution : Let, I =  (x2  3x  3) .dx x 1 (t2 1)  2 .(2t)dt = 2 t2  1 dt  2 1  1 / t2 dt    I = {(t2  1)2  3(t2  1)  3} t2 t4  t2  1 t2  1  1 / t2 1  1 / t2 du  w h e r e u  t  1   = 2  t  (t  1 / t)2  ( 3 )2 .dt  2 u2  ( 3 )2  = 2 tan 1  u   c = 2 tan 1  t2 1   c  2 tan 1  x   c Ans. 3  3    3  x  Ans.   3  3t   3 (  1)   dx (x 1) Illustration 25 : Evaluate x2  x 1 Solution : Let, I=  dx 1  dx = –1/t2 dt (x 1) put x – 1 = x2  x 1 t I=  1 / t2dt dt 2 1/t 1  1  1 1 = –  3t2  3t 1  t  1  t NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 1 dt =  1 log (t  1 / 2)  (t  1 / 2)2  1 / 12  c 12 3 =–  2  3  t  1 /12 1 log  1 1  12  x 1  1 2 1 3   2   1 2 =    c  x 1 12 dx Illustration 26 : Evaluate (1  x2 ) 1  x2 Solution : dx Let, I  (1  x2 ) 1  x2 1 1 Put x = , So that dx = dt t t2 1 / t2 dt tdt   I= (1  1 / t2 ) 1  1 / t2  (t2  1) t2  1 again let, t2 = u. So that 2t dt = du. E 13

JEE-Mathematics = 1  du dx where both P and Q are linear so that 2 1) (u  u  1 which reduces to the form  P Q we put u – 1 = z2 so that du = 2z dz 1 2zdz dz I 2   (z2  1  1) z2   (z2  2) I 1 tan 1  z   c 2  2    I 1 tan 1  u  1   c  1  t2  1   c  1  1  x2  c Ans.   2 tan 1  2  2 tan 1  2x   2 2   Do yourself -14 :  (x  3) x dx dx x 1 (i) Evaluate ( i i ) Evaluate : x2 1  x2 Miscelleneous Illustrations : Illustration 27 : Evaluate  cos4 xdx 3 sin3 x{sin5 x  cos5 x}5 cos4 x cos4 x 3 dx = cot4 xcosec2 xdx   I = (1 + cot5 x)3 / 5 Solution : dx = 3 sin3 x{sin5 x  cos5 x}5 sin6 x{1  cot5 x}5 Put 1 + cot5x = t 5cot4x cosec2xdx = –dt =  1 dt  1 t2 / 5 c   1 1  cot5 2 / 5 c Ans.  5 t3 / 5 2 2 x Ans. (C,D) dx (B) n|cotx – tanx|+ c E (D) tan–1(–2cot2x) + c Illustration 28 :  cos6 x  sin6 x is equal to - (A) n|tanx – cotx|+ c (C) tan–1(tanx – cotx) + c dx sec6 x (1  tan2 x)2 sec2 xdx NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 cos6 x  sin6 x = dx = 1  tan6 x 1  tan6 x Solution :   Let I = If tan x = p, then sec2 x dx = dp (1  p2 )2 dp (1  p2 ) p2 1  1  1  p6  p4  p2  1 dp = p2     I = dp  1  1 p2  p 2  p2 = dk  tan 1 (k )  c  where p1  k, 1  1  dp  dk  k2 1  p  p2      = tan–1  p  1   c  tan 1 (tan x  cot x)  c = tan–1(–2cot2x) +c  p  Illustration 29 : Evaluate : 2 sin 2x  cos x  6  cos2 x  4 sin x dx Solution :  2 sin 2x  cos x dx (4 sin x  1) cos x dx  (4 sin x  1) cos x dx  cos2 x  4 sin 6  (1  sin2 x)  4 sin x sin2 x  4 sin x  5  I= 6 x = 14

JEE-Mathematics Put sin x = t, so that cos x dx = dt. (4t 1)dt ...... (i)  I =  (t2  4t  5) Now, let (4t – 1) = (2t – 4) + µ Comparing coefficients of like powers of t, we get 2 = 4, –4 + µ = –1 ...... (ii)  = 2, µ = 7 2(2t  4)  7 {using (i) and (ii)}  I =  t2  4t  5 dt 2t  4 dt = 2log t2  4t  5  7 dt  4t  t2  4t  5 t2  4t  4  4  5   = 2 t2 5 dt  7 = 2log t2  4t  5  7 dt = 2log|t2 – 4t + 5|+7 . tan–1 (t – 2) + c (t  2)2  (1)2 = 2log|sin2x – 4sinx + 5| + 7 tan–1(sinx – 2) + c. Ans. Illustration 30 : The value of  3  x . s in 1  1 3  x  dx, is equal to - 3  x  6  1 3  x  2  x   4   3  3   (A) c o s 1   2 9  x2 . cos1  2x   c    (B) 1 3  co s 1  x 2  2 9  x 2 . sin 1  x   2 x   c 4   3  3         (C) 1 3  sin 1 x 2 2 9  x 2 . sin 1  x   2 x   c 4   3  3         (D) none of these Solution : Here, I = 3  x . sin 1  1 3  x  dx 3  x  6  NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 Put x = 3cos2   dx = –6sin2d = 3  3 co s 2 . sin 1  1 3  3 cos 2 (–6 sin 2)d 3  3 co s 2  6  = sin  .sin1 (sin ).(6 sin 2)d = – 6 .(2 sin2 )d cos  2  – 6  (1  cos 2)d     cos 2 d = = – 6  2   2  sin 2 1.  sin 2    – 32  sin 2 cos 2   2  2  d   2 4  = – 6  2     =  6    c  1 3   x  2  x   4    3  3   = co s 1    2 9  x2 . c o s 1  2 x   c Ans. (A)   tan    x   4  Illustration 31 : Evaluate : dx cos2 x tan3 x  tan2 x  tan x E 15

JEE-Mathematics tan    x  (1  tan2 x)dx  4  Solution : dx =  I = cos2 x tan3 x  tan2 x  tan x (1  tan x)2 cos2 x tan3 x  tan2 x  tan x  1  1 x  sec2 xdx tan2  I =  1  1  tan x  2  tan x  tan x  1  tan x let, y tan x  1  1  2y dy   se c 2 x  1 .sec2 x dx  tan2 tan x x  I=  ( 2 y dy dy y2  1).y = –2  1  y2 = – 2tan–1 y + c = – 2 t a n –1  tan x  1  1  Ans.  c  tan x  ANSWERS FOR DO YOURSELF 1: (i) 1 tan 1  4x3   c (ii) sin x  1 sin3 x  c 36  3  3   2 : (i) 3  x  sin 1 3  x  c (ii) n  x  1   x2  x   c x 2  2  1  3 : ( i ) xex – ex + c (ii)  1 x2 cos(x2 )  sin(x2 )  c 2 4 : ( i ) ex tan–1 + c 5 : ( i ) x tan(ex) + c (ii) 1 ex2 sin(x2 )  c 2 (ii) xnx + c 6 : ( i )  1 n| x  1|  7 n| x  3| c ( i i ) n| x  2|  3  c 22 x 2 7 : (i) 3 tan 1  2x 1   c (ii) 5 x2  4 x  1  6n  x  2   x2  4x  1   c 2  3     NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65  x 2  1  1   x2 1   1 n x2  2x 1  x  22 tan 1  2x  2 x2   8 : (i) tan 1    c (ii)    9 : (i) 2x  1  10 : (i)  1 n  x 2 1   c  1  1 1/3  c 1 1  1 2/3  c 2  x2  x3 2 x3   (ii)  (iii)     1 tan1 5 tan x  c 2 tan 1  8 tan x  1   c   5 (ii) 15  15  1 tan 1  3 tan x / 2  1   c 1 n 6  tan x / 2  2  c   2 6 6  tan x / 2  2 11 : (i) 2  22  (ii) 12 : (i) 13 : (i) 1 x  1 n sin x  cos x  c (ii) 12 5 n 3 cos x  2 sin x C 22 x 1 tan3 x  c 13 13 3 (ii) 2 tan3 /2 x  c (i ii ) 1 sin3 x  2 sin5 x  1 sin7 x  c . 3 357 14 : (i) 2 x  1  3 n x 1 2  c (ii)  1 1  x2  c 2 x 1 2 2 16 E

JEE-Mathematics EXERCISE - 01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1. If f(x) =  2 sin x sin 2 x dx , x  0 then lim f'(x) is equal to- x3 x0 (A) 0 (B) 1 (C) 2 (D) 1/2 2 . 4 sin x cos x cos 3x dx is equal to - 22 (A) cos x  1 cos2x  1 cos 3x c (B) cos x  1 cos 2x  1 cos 3x  c 2 3 23 (C) cos x  1 cos 2x  1 cos 3x  c (D) cos x  1 cos2x  1 cos 3x  c 23 23 3.  8x 13 dx is equal to - 4x 7 1 1 (A) 6 (8x + 11) 4x 7 + c (B) 6 (8x + 13) 4x 7 + c 1 1 (C) 6 (8x + 9) 4x 7 + c (D) 6 (8x + 15) 4x 7 + c 4 .  1 cos8 x  sin8 x x  dx equals -   2 sin2 x cos2    (A)  sin 2x c sin 2x cos 2x (D)  cos 2 x  c (B)  c (C)  c 2 2 2 2 x  5 . Primitive of 3 x4  1 4 w.r.t. x is - 1 1 1 1 (A) 3 1  1 3 c (B)  3 1  1 3 c (C) 4 1  1 3  c (D)  4 1  1 3 c 4 x4 1  4 x4 1  3 x4  1  3 x4   1 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 6 . (1  2x  3x2  4x3 ......) dx (|x| < 1) - (A) (1 + x)–1 + c (B) (1 – x)–1 + c (C) (1 + x)–2 + c (D) none of these 7.  x dx is equal to -  1  x2  1  x2 3  (A) 1 ln 1  1  x2  c (B) 2 1  1  x2  c 2 (D) none of these  (C) 2 1  1  x2  c 8. x n x dx equals - 1  n x E 2 1  n x (ln x  2)  c 2 1  n x (ln x  2)  c (A) 3 (B) 3 1 1  n x (ln x  2)  c (D) 2 1  n x (3n x  2)  c (C) 3 17

JEE-Mathematics x4  1 dx  A  n x  B + c, where c is the constant of integration then : 9 . If   x x2  1 2 1  x2 (A) A = 1; B = –1 (B) A = –1; B = 1 (C) A = 1; B = 1 (D) A = –1 ; B = –1 1 0 .  x 3/2 dx equals -  1  x5   2 x5 (B) 2 x  c 21 c 5 1  x5 c (A) 5 1  x5 (C) 5 1  x5 (D) none of these 1 1 . sin x.cos x.cos 2x.cos 4 x.cos 8x.cos16x dx equals - (A) sin 16 x  c (B)  cos 32 x  c (C) cos 32x  c (D)  cos 32x c 1024 1024 1096 1096 1 2 . Identify the correct expression (A) x nx dx  x2n| x| x2  c (B) x n x dx  xex  c (C) x ex dx  xex  c x (D) dx  1 tan 1  x   c a2  x2 a  a  1 3 . x.  n x  1  x2 dx equals - 1  x2 x . n2 x  1  x2 x c 2 1  x2  (A) 1  x2 n x  1  x2  x  c  (B)  x .n2 x  1  x2 x c 2 1  x2  (C)   (D) 1  x2 n x  1  x2  x  c 14. If dx = a n(1 + x2) + btan–1x + 1  n|x + 2| + C then- 5  (x 2)(x2  1) 12 12 12 12 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 (A) a = – 10 , b = – 5 (B) a = 10 , b = – 5 (C) a = – 10 , b = 5 (D) a = 10 , b = 5  x  12 1 5 . dx equals - x4  2x2 1 x3 x  (B) x5  x3  x  3  c x5  4x3  3x  3  c (A) x c 3 x2 1 3 x2 1 3 x2 1  (C) (D) None of these z x2 4 16. dx equals - x4  24x2 16  (A)  (B) 1 tan 1  x2  4   1 cot 1  x2  4   4x  c  x  c   4 4    (C)  (D) 1 1  4 x2  4  1 co t 1  x2  4   x  c  x  c  cot   4 4   18 E

JEE-Mathematics 1 7 . x4  4 dx equals- x2 4  x2  x4 (A) 4  x2  x 4  c (B) 4  x2  x4  c (C) 4  x2  x 4  c 4  x2  x4 x 2 (D)  c 2x x9 1 8 .  (x2  4)6 dx is equal to - (A) 1  4  1 5 c (B) 1  4  1 5 +c 5x  x2  + 5  x2  1 (D) 1 (1 + 4x–2)–5 + c (C) 10x (1 + 4x2)–5+c 40 19. dx = a tan–1  b tan x  + c, then-  2  If  5  4 cos x 21 21 (A) a = 3 , b = – 3 (B) a = 3 , b = 3 21 21 (C) a = – 3 , b = 3 (D) a = – 3 , b = – 3 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 2 0 . Primitive of 1  2 tan x sec x  tan x  w.r.t.x is - (A) n sec x  n sec x  tan x  c (B) n sec x  tan x  n sec x  c (C) 2n x x c (D) n 1  tan x(sec x  tan x)  c sec  tan 2 2 2 1 .  sin 2x dx equals - (A)  cos 2 x  c sin2 x (C)  cos2 x c (D) cos 2x  c 2 (B)  c 2 2 2 2 2 . dx NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 1 1  equals- 2x2  x3  (A) n 2 x2  1  2n| x| c (B) n 2 x2  1  2n| x|  c (C) n 2 x2  1  n(x2 )  n2  c (D) n 1  1 c 2x2 2 3 . If  e3x cos 4 x dx  e3x (A sin 4 x  B cos 4 x)  c , then - (A) 4A = 3B (B) 2A = 3B (C) 3A = 4B (D) 4A + 3B = 1 CHECK YOUR GRASP AANNSSWWEERR KKEEYY EXERCISE-1 Que. 1 2 3 4 5 6 7 8 9 10 Ans. B B A B B B B A C A Que. 11 12 13 14 15 16 17 18 19 20 Ans. B C A C D A A D B A,B,D Que. 21 22 23 Ans. A,B,C B,C,D C,D E 19

JEE-Mathematics BRAIN TEASERS EXERCISE - 02 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) z b g1 . cot x  tan x dx equals - 2 cos x  sin x (A) sec –1 sin x  cos x   c (B) sec–1 sin x  cos x   c (D) n sin x  cos x   sin 2x  c (C) n sin x  cos x   sin 2x  c z2 . sin x  4 sin 3x  6 sin 5x  3 sin 7x dx equals - sin 2x  3 sin 4x  3 sin 6x (A) –2sinx+c (B) 2sinx+c (C) –2cosx+c (D) 2cosx+c 3 . 1  x7 equals - dx x(1  x7 ) (A) n| x|  2 n 1  x7 c (B) n| x|  2 n 1  x7  c 7 4 (C) n| x|  2 n 1  x7 c (D) n| x|  2 n 1  x7  c 7 4 4.  x 3 dx is equal to - 1 x2 1 1  x2 (2 + x2) + c 1 1  x2 (x2 – 1) + c (A) 3 (B) 3 (C) 1 (1 + x2)3/2 + c 1 1  x2 (x2 – 2) + c 3 (D) 3 5 .  sin2 ( n x) dx is equal to - NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 (A) x + 2sin(2nx) + cos(2nx)) + c (B) x + 2sin(2nx) – cos(2nx)) + c 10 (5 10 (5 (C) x – 2sin(2nx) – cos(2nx)) + c (D) x – 2sin(2nx) + cos(2nx)) + c 10 (5 10 (5 6. x2  cos2 x .cosec2x dx is equal to -  1 x2 (A) cotx + tan–1x + c (B) cotx – tan–1x + c (C) –cotx – tan–1x + c (D) tan–1x – cotx + c 7 .  x2  3  equals - ex  dx ,  x  3 2  (A) ex . x  c (B) ex  2  6   c (C) e x 1  6   c (D) ex. 3  c x 3      x 3 x 3 x 3 20 E