Indefinite Integration Formulas PDF

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BITSAT WBJEE MHT CET and more... 4.8 50,000+ 2,00,000+ Rating on Google Play Students using daily Questions available With MARKS app you can do all these things for free Solve Chapter-wise PYQ of JEE Main, JEE Advanced, NEET, BITSAT, WBJEE, MHT CET & more Create Unlimited Custom Tests for any exam Attempt Top Questions for JEE Main which can boost your rank Track your exam preparation with Preparation Trackers Complete daily goals, rank up on the leaderboard & compete with other aspirants 4.8 50,000+ 2,00,000+ Rating on Google Play Students using daily Questions available  Indefinite Integration BASIC THEOREMS ON INTEGRATION If f(x), g(x) are two functions of a variable x and k is a constant, then (i)  k f (x) dx  k  f (x) dx (ii)  [f(x)  g( x)]   dx  f ( x) dx  g( x ) dx (iii) d dx  f (x) dx  f(x)  d  (iv)   dx f ( x )  dx  f ( x ) + c SOME STANDARD FORMULAE x n1 1 (i)  x n dx   c (n  1) (ii)  dx  loge x  c n 1 x x ax e x dx  e x  c  a dx  log  c  a x loga e  c (iii)  (iv) ea (v)  sin x dx   cos x  c (vi)  cos x dx  sin x  c (vii)  tan x dx  log | sec x |  c   log | cos x |  c (viii)  cot x dx  log | sin x | c  x (ix)  sec x dx  log | sec x  tan x |  c   log | sec x  tan x |  c = log tan  4  2  + c   x (x)  cosecx dx   log | cosec x  cot x | log | cosecx  cot x |  c = log tan   + c 2 (xi)  sec x tan x dx  sec x  c (xii)  cosec x cot x dx   cosec x  c 2 (xiii)  sec x dx  tan x  c 2 (xiv)  cosec x dx   cot x  c  Indefinite Integration 1 1 x (xv)  dx  tan 1   c x 2  a2 a a 1 1 xa (xvi)  2 2 dx  log  c (x > a) x a 2a xa 1 1 ax (xvii)  2 2 dx  log  c (x < a) a x 2a ax 1 x x (xviii)  dx  sin 1   c   cos 1   c a x 2 2 a   a 1 x (xix)  dx  log | x  x 2  a2 | c  sinh1    c 2 x a 2 a 1 x (xx)  dx  log | x  x 2  a2 |  c  cosh1    c 2 x a 2 a x 2 a2 x (xxi)  a 2  x 2 dx  a  x2  sin 1  c 2 2 a x a2 x (xxii)  x 2  a 2 dx  x 2  a2  sinh 1  c 2 2 a x a2 x (xxiii)  x 2  a 2 dx  x2  a2  cosh 1  c 2 2 a 1 1 x (xxiv)  dx  sec 1  c x x 2  a2 a a ax eax eax   b  (xxv)  e sin bx dx  (a sin bx  b cos bx)  c  sin bx  tan1    c 2 2 a b a2  b2   a  e ax eax   b  (xxvi)  e ax cos bx dx  2 2 (a cos bx  b sin bx )  c  cos bx  tan 1   c a b a 2  b2   a  (xxvii)  eax  b (af(x)  f '(x))dx  eax b f(x)  c 1 (xxviii)  f(ax  b)dx    ax  b  + c a METHOD OF INTEGRATION Integration by Substitution (a) When integrand is the product of two factors such that one is the derivative of the other i.e, I=  f (x) f ' (x) dx In this case we put f(x) = t to convert it into a standard integral. (b) When integrand is a function of function Indefinite Integration i.e.  f [  (x) ]  ' (x) dx Here we put f(x) = t so that f ' (x) dx = dt and in that case the integrand is reduced to  f (t) dt. (c) Integral of a function of the form (ax+b) dx Here put ax + b = t and convert it into standard integral. Obviously if  f (x) dx  (x), then 1  f(ax  b) dx  a  (ax  b) (d) Some standard forms of integrals The following three forms are very useful to write integral directly. [ f ( x )]n 1 (i)  [ f ( x )]n f ' ( x ) dx   c (where n ¹ –1) n1 f ' (x) (ii)  f (x) dx  log [f (x)]  c f ' (x) (iii)  f ( x) dx  2 f ( x )  c dx (e) Integral of the form  a sin x  b cos x Putting a = r cos  and b = r sin  , we get dx 1 I=  r sin ( x  )  r  cos ec (x  ) dx 1 1 x 1 b = log tan (x/2 +  /2) + c =  log tan   tan 1   c r 2 a b 2  2 2 a (f) Standard Substitution Following standard substitution will be useful- Integrand form Substitution 1 (i) a 2  x 2 or x = a sin  a  x2 2 1 (ii) 2 x a 2 or x = a tan  or x = a sinh  x 2  a2 1 (iii) x 2  a 2 or x = a sec  or x = a cos h  x  a2 2 x ax 1 (iv) or or x (a  x ) or x (a  x ) x = a tan2  ax x x ax 1 (v) or or x(a  x ) or x(a  x) x = a sin2  ax x  Indefinite Integration x xa 1 (vi) or or x( x  a) or x( x  a) x = a sec2  xa x ax ax (vii) or x = a cos 2  ax ax x (viii) x or ( x   )(  x ) (b > a) x = a cos2  + b sin2 q (a) Integration by Parts :  d   If u and v are the differentiable functions of x, then  u.v dx  u  vdx    dx (u)    vdx  dx. i.e. Integral of the product of two functions = first function x integral of second function –  [derivative of first) x ( Integral of second) ] (i) How to choose Ist and IInd function : If two functions are of different types take that function as Ist which comes first in the word ILATE, where I stands for inverse circular function, L stands for logrithmic function, A stands for algebric functions, T stands for trigonometric and E for exponential functions. (ii) For the integration of logarthmic or inverse trigonometric functions alone, take unity (1) as the second function x (b) If the integral is of the form  e [f( x)  f ' (x)] dx then by breaking this integral into two integrals, integrate one integral by parts and keep other integral as it is, By doing so, we get -  e x [f(x)  f '(x) ] dx  e x f(x)  c (c) 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 keep other integral as it is, by doing so, we get  [x f ' (x)  f ( x)] dx  x f (x)  c Integration of the Trigonometrical Functions dx dx (i)  a  b sin 2 x , (ii)  a cos 2 x b dx dx  (iii) a cos 2 x  b sin 2 x , (iv)  (a cos x  b sin x ) 2. (For their integration we multiply and divide by sec2x and then put tan x = t.) Some integrals of different expressions of ex ae x (i)  b  ce x dx [put ex = t] 1 (ii)  1 e x dx [multiplying and divide by e–x and put e–x = t] 1 (iii)  1 e x dx [multiplying and divide by e–x and put e–x = t] 1 (iv) e x dx [multiply and divided by ex]  ex Indefinite Integration e x  e x  f ' ( x)  form (v) e x e x dx   f ( x )  ex  1 (vi) e x dx [multiply and divide by e–x/2] 1 2  e x  e x    dx [integrand = tanh2 x] (vii)   e x  e x    2  e2x  1  (viii)   2x  dx [integrand = coth2 x]  e  1 1 1 (ix)  (e x e x 2 ) dx [integrand = 4 sech2x] 1 1 (x)  (e x  e  x )2 dx [integrand = 4 cosech2x] 1 (xi)  (1  e x )(1  e  x ) dx [multiply and divide by ex and put ex = t] 1 (xii)  1 e x dx [multiply and divide by e–x/2] 1 (xiii)  1 ex dx [multiply and divide by e–x/2] 1 (xiv)  e 1 x dx [multiply and divide by e–x/2] 1 (xv)  2e x  1 dx [multiply and divide by 2e  x / 2 ] (xvi)  1  e x dx [integrand = (1 – ex) / 1 e x ] (xvii)  1  e x dx [integrand = (1 + ex) / 1  ex ] (xviii)  e x  1 dx [integrand = (ex – 1) / ex  1 ] ex  a (xix)  dx [integrand = (ex +a) / e2x  a2 ] ex  a x2  1 8.  x 4  kx 2  1 dx (Divide N.r and Dr by x2 then put x ±1/x = t) x2  1 9.  x 4  kx 2  1 dx (Divide N.r and Dr by x2 then put x ±1/x = t)  Indefinite Integration d Nr = A (Dr) + B (Dr) + C dx x2 1 10.  x 4  kx 2  1 dx x2 = {(x2 + 1) + (x2 – 1)} 2 1 1 11.  x 4  kx 2  1 dx 1= {(x2 + 1) – (x2 – 1)} 2 1 1 12.  x 4  a4 dx 1 {(x 2  a2 )  (x 2  a 2 )} 2a 2 1 13.  (ax  b) dx ; Put (x + d) = t2 cx  d 1 1  dx 14. ; Put (px + q) = (px  q) ax 2  bx  c t 1 15.  (ax 2  bx  c) px  q dx ; Put (px + q) = t2 1 16.  dx; Put (x = 1/t) (Ax 2  B) cx 2  D 1 17.  (a sin2 x  b sin x cos x  c.cos2 x) dx 1 2 tan x / 2 18.  (a  b sin x) dx ; put sinx = 1  tan2 x / 2 1  1  tan2 x / 2  19.  dx put cosx =  2  & put tan x/2 = t (a  b cos x)  1  tan x / 2  1 20.  (asin x  bcos x  c) dx P sin x  qcos x  r 21.  a sin x  bcos x  c dx  The only app you need to prepare for JEE Main JEE Adv. BITSAT WBJEE MHT CET and more... 4.8 50,000+ 2,00,000+ Rating on Google Play Students using daily Questions available With MARKS app you can do all these things for free Solve Chapter-wise PYQ of JEE Main, JEE Advanced, NEET, BITSAT, WBJEE, MHT CET & more Create Unlimited Custom Tests for any exam Attempt Top Questions for JEE Main which can boost your rank Track your exam preparation with Preparation Trackers Complete daily goals, rank up on the leaderboard & compete with other aspirants 4.8 50,000+ 2,00,000+ Rating on Google Play Students using daily Questions available