Indefinite Integration Formulas Handbook PDF

ULTIMATE FORMULA HANDBOOK MATHONGO Basic Formulas  cosec 2 xdx   cot x  dx  x x n  sec x tan xdx  sec x n 1  x dx  n  1 ( n   1)  x  cosec x cot xdx   cosec x dx  sec  1 x 2 x 1 a x dx  ax ,a  0 log e a  e x dx  e x 1 dx  log | x | x   cos dx  sin x 1 1 x 2   sec 2 xd...

ULTIMATE FORMULA HANDBOOK MATHONGO Basic Formulas  cosec 2 xdx   cot x  dx  x x n  sec x tan xdx  sec x n 1  x dx  n  1 ( n   1)  x  cosec x cot xdx   cosec x dx  sec  1 x 2 x 1 a x dx  ax ,a  0 log e a  e x dx  e x 1 dx  log | x | x   cos dx  sin x 1 1 x 2   sec 2 xdx  tan x 1 1 x 2  tan x d x  log sec x  c  cot x dx  lo g sin x  c dx  sin  1 x  sin xdx   cos x  sec x dx  log  1 dx  tan  1 x x  sec x  tan x  c  log tan     c 2 4  cos ec x dx  log cos ec x  cot x  c  tan x 2  c Method of Substitution Type 1  Type 3 Type 4 f ´( x) dx  log f ( x)  c f ( x) Type 2 n   f ( x ) f ´( x ) dx  f ( x ) n 1  c n 1  f  g ( x )  g ´( x ) dx. In this type, we substituteg (x) = t, then, Hence integral reduces to  f (t) dt P( x).(ax  b) n or Working Rule : P ( x) (ax  b)n where P(x) is a polynomial in x and n is a positive rational number. Put z = ax + b Type 5 sin f(x) or cos f(x) then put z = f(x) Type 6 :  s in m x c o s n x d x Working Rule : (i) If power of sin x is odd positive integer, put z = cos x (ii) If power of cos x is odd positive integer, put z = sin x (iii) If powers of both sin x and cos x are odd positive integers, put z = sin x or z = cos x. (iv) If powers of neither cos x nor sin x is odd positive integer, see the sum of powers of sin x and cos x. (a) If the sum of powers is even negative integer, put z = tan x. (b) If the sum of powers (m + n) is even positive integer and m, n are integers, express the integrand as the algebraic sum of sines and cosines of multiple angles. INDEFINITE INTEGRATION ULTIMATE FORMULA HANDBOOK m x sec n xdx or m x sec n xdx : similar can be derived for the other pair Type 7 :  tan.  tan For MATHONGO  cot m xco sec n xdx (i) If power of secx is even positive integer, put z = tanx. (ii) If power of secx is not even positive integer, then see the power of tanx. (a) If power of tanx is odd positive integer, put z = secx. (b) If power of tanx is even positive integer, then put sec2x – 1 in place of tan2x and then substitute z = tanx. (iii) If power of tanx is zero and power of secx is odd positive integer greater than 1, then method of integration by parts is used. SOME STANDARD SUBSTITUTIONS 1 1. 2. a 2  x 2 or 1  a2  x2 2 a  x 1 or  2 a2  x2 x = a sin  or a cos  x = a tan  or a cot  6. x a  x or 7.  x  a   or b  x  4. 2 or x2  a 2 a  x a  x or x  a 2 x = a sec  or a cosec  x  a x  b 8. a x a  x x = a cos 2 x  a  b  x  x = a cos2  + b sin2  1 3. 1 x x = a tan2  or x = a cot2  x or x  a x  b x = a sec2  – b tan2  5. x a  x or a  x x = a sin2  or x = a cos2  x 1 9.  a x  b x x – a = t2 or x – b = t2 SOME STANDARD FORMULAS DERIVED FROM SUBSTITUTION Set-I Set-II 1.  x2 3.  x2 1.  3.  dx  a dx  a  2 dx 2 a x 2 dx 2 1 x  a log  c 2a x  a  2 x  a 2 2.  a2 2.  4.  dx  x  2 1 a  x log  c 2a a  x 1 x  c tan  1 a a  sin  1 x c a  log x  x2  a 2  c dx 2 x  a 2 dx x x2  a2 INDEFINITE INTEGRATION  log x   x2  a 2  c 1 x sec  1 c a a ULTIMATE FORMULA HANDBOOK Set-III 1. 2. 3. MATHONGO a2 x sin  1  c 2 2  a 2  x 2 dx  x 2 a2  x2   x 2  a 2 dx  x 2 x2  a2   x 2  a 2 dx  x 2 x 2  a 2 dx  a2 lo g x  2 x 2 x2  a2  c x2  a2  a2 lo g x  2 x2  a2  c INTEGRATION BY PARTS Integral of product of two functions = (1st function) × (Integral of 2nd function) – Integral of {(differential of 1st function) × Integral of 2nd function} d  f ( x).  g( x) dx  dx In symbols :  f ( x). g ( x) dx  f ( x).  g ( x) dx     dx  or  u.vdx  u  vdx –  u´( vdx)dx where I stands for Inverse circular function L stands for Logarthmic function A stands for Algebraic function T stands for Trigonometrical functions and E stands for Exponential function (ii) If both the functions are trigonometrical, take that function as v whose integral is simpler. (iii) If both the functions are algerbraic take that functions as u whose d c is simpler. Standard Forms derived using By Parts x  f ( x )  f ´  x   dx  e x f ( x )  c (i) e (iii)  [ xf ´( x)  f ( x)]dx  x f ( x )  c (iv) (v) e e ax ax sin  bx  c  dx = eax cos  bx  c  dx = eax (ii) mx e mf (x)  f ´ x  dx  emx f (x)  c b  sin  bx  c  tan1  a  c a2  b2 b  cos  bx  c  tan1  a  c a2  b2 INDEFINITE INTEGRATION ULTIMATE FORMULA HANDBOOK MATHONGO METHOD : INTEGRATION BY PARTIAL FRACTION We have divided this method into 2 types, depending upon the denominator. 1. If denominator has non repeated factors 2. If denominator has repeating factors Type 1 : For non-Repeating roots When denominator can be expressed as non repeating factors i.e. D(x) = (x – 1) (x – 2)... (for linear factors) 2 = (ax + bx + c) (px + qx + r)... (for quadratic factors) Type 2 When repeating factors are present i.e. when denominator is of the form k1 D(x) = (x – ) (x – )k2... {for linear factor} = (ax2 + bx + c)k1 (px2 + qx + c)k {for quadratic (1) If function is linear. N( x) i.e. (2) 2 ( x  a) ( x  b) ( x  c) B1 C3 A B2 C1 C2    = ( x  a)  ( x  b)  2 2 ( x  c) ( x  c) ( x  b) ( x  c)3 3 If function has quadratic factors N( x) i.e of the form 2 (ax  bx  c) ( px2  qx  c)2 = Ax  B 2 ax  bx  c  P1 x  Q1  px 2  qx  r  P2 x  Q2   px 2  qx  r 2  INTEGRATION OF RATIONAL & IRRATIONAL FUNCTIONS Integral of the form  ax dx 2  bx  c ,  dx 2 and ax  bx  c  ax 2  bx  c dx For evaluating such integral we make the coefficients of x2 in ax2 + bx + c as one. Complete the square by adding and subtracting the square of half of the coefficient of x to get the form 2  c b2   b     a  x     a 4 a2   2a     Integrals of the form  ax px  q 2  bx  c dx,  px  q ax 2  bx  c dx and ( px  q) ax 2  bx  c dx For evaluating such integrals we choose suitable constants A and B such that d  px  q  A  ( ax 2  bx  c)   B  dx  INDEFINITE INTEGRATION ULTIMATE FORMULA HANDBOOK px 2  qx  r  Integrals of the Form : MATHONGO ax 2  bx  c dx For evaluating such integrals we choose suitable constants, A, B and C such that  d  (ax2  bx  c   C px2 + qx + r = A (ax2 + bx + c) + B  dx   Integrals of the form : x x2  1 4  kx 2  1 dx For evaluating such integrals, divide the numerator and denominator by x2. Complete the square of 2 2 1 1   denominator to get the form  x    a or  x    a x x   Then the integral can be evaluated by using the method of substitution. Special Integration Type I Type II Type III x x x x2  q 4 Divide numerator & denominator by x2  px2  q dx 4  px2  q x2  r 4 write this in form  px2  q dx 1 2 q  x 2   q  x2   4 dx  a  b cos x or dx x  px  q   express x2 + r as 2 x2 + r = l x  q + m where l + m = 1 &  a  b sin x or q l  m  r dx  a  b cos x  c sin x 2 tan Working Rule :  2 Integration of Trigonometric Functions Type I q Put sin x  x 2 x 2 cos x  2 x 1  tan 2 1  tan 2 1  tan 2 x 2 whichever is needed and then put z = tan x 2 and INDEFINITE INTEGRATION x 2  q  ULTIMATE FORMULA HANDBOOK  Type II MATHONGO sin x dx, a sin x  b co s x  cos x d x , or a sin x  b co s x  p sin x  q co s x dx c s in x  b co s x Step – 1 : Put Numerator = A (dinominator) + B (derivative of denominator.) where a  0, b  0 Step – 2 : Then equate the coefficients of sinx and cosx to find A and B. a sin x  b cos x  c  p sin x  q cos x  r Type 3. (i) Write Numerator =  (Diff. of denominator) + µ (Denominator) + v i.e. a sin x + b cos x + c =  (p cosx – q sinx) + µ (psin x + q cosx + r) + v Type 4.  a sin 1 2 2 x  b cos x dx. 1  a  b sin 2 x dx , 1  a  b cos 2 x dx, 1   a sin x  b cos x (i) Divide numerator and denominator both by cos2x (ii) Replace sec2 x, if any, in denominator by 1 + tan2 x 2 dx,  a  b sin 1 2 x  cos2 x dx (iii) Put tan x = t so that sec2 x dx = dt This substitution reduces the integral in the form Integrals of the form :  at 1 2  bt  c dt dx P Q where P and Q are linear or quadratic expression in x 1. Q is linear and P is linear or quadratic., For evaluating such integrals, put Q = t2 2. Q is quadratic and P is linear., For evaluating such integrals, put P = 3. Both P and Q are pure quadratic., For evaluating such integrals, put x  1 t 1. t Integration of Irrational Functions Types of functions (intergrand) Approach 1.   ax  b  a / n  f  x,    a, b, c, d, , n  R     cx  d   Substitute : 2. f x,  ax  b  a/n ,  ax  b /m  ax  b  tn cx  d ax + b = tp, where p is L.C.M. of m and n. INDEFINITE INTEGRATION ULTIMATE FORMULA HANDBOOK 3.  f  x    a 2  x2 MATHONGO n   Workrule : x  m + p  N, m + p > 1 Workrule : a + bx = tx a2  x2  t 1 4. x m  a  bx  p 1 5. 6. m L1  x  n L 2  x  (i) If n > m; L1 ( x) t L 2 ( x) (ii) If n < m; L 2 ( x) t L1 ( x) (i) If p  I, substitute x = ts where s is L.C.M. of denominator of m & n. (ii) If xm(a + bxn)pdx m 1 is an Integer, substitute a + bxn = ts n is the denominator of fraction p. (iii) m 1  p substitute ax–n + b = ts n s is denominator of rational number p. If INDEFINITE INTEGRATION