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Tagged with:       [Post New]posted on 23 Jun 2008 18:17:52 IST    


There are four methods of integration:

 

1) decomposition into sum or difference -------for trigonometric functions

2) integration by parts-------when the expression is a product of 2 terms

3) substitution

4) formulae

 

 
i'm postin a few formulae.......

i'm asuuming that u kno the basic formulae

 

 

 


1)when trigonometric functions are given you reduce them to the sum or difference of trigonometric functions....


 

fro ex: (1- cosx)/(1+cosx) can be reduced to

2cosec2x - 2 cosecx cotx -1 which can b easily integrated

 

 


2)INTEGRATION BY PARTS(very important)


 

 

 u dv = uv - v du

 

it's success  depends upon the choice of u

(1) if integrand contains any non integrable functions directly from the formula, like tan-1x, logx we have to take these non integrable functions as u and the other as dv.

(2) if integrand contains both integrable functions, and one of these is xn ( n is positive integer). then we take u=xn .

(3) for other cases choice of u is yours

 

 


3) SUBSTITUTION :


 


       x3 dx/(1 + x4 )        in this Q  we substitute 1+x4  = t


so

x3dx = dt-----------------substitute this in the given expression and reduce it in terms of "t"

so we get,

 

       x3 dx/(1 + x4 )    =    dt/t  = log t + c  = log(1 + x4) + c

 

 


4) FORMULAE


certain standard integrals
 

 

 (a2 - x2) dx  = (x/2)((a2 - x2) + (a2/2)sin-1(x/a) + c

 (x2 - a2) dx  = (x/2)((x2 - a2) - (a2/2)log[x + (x2 - a2)] + c

 (x2 + a2) dx  = (x/2)((x2 + a2) + (a2/2)log[x + (x2 + a2)] + c

 

dx/[(a2 - x2)   = sin-1(x/a) +c

dx/[(x2 - a2)  = log[x+(x2 - a2)] + c

dx/[(x2 + a2) =  log[x+(x2 + a2)] + c

 

dx/(a2 - x2)    = (1/2a)log [ (a+x)/(a-x)] + c

dx/(x2 - a2)   = (1/2a)log [ (x-a)/(x+a)] + c

dx/(x2 + a2)  = (1/a)tan-1(x/a) + c

 

 

 

 

(eax sin bx ) dx = [eax/a2 + b2][a sinbx - bcosbx] + c

 

axdx = ax/logea   + c

 

f '(x)/f(x)   dx   = log[f(x)] + c

 

f ' (x) /f(x)  dx  = 2 *f(x) + c

 

all these results can be derived using the first three methods



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sahilgupta_iit (438)

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sahilgupta_iit
sahilgupta_iit is offline comment by sahilgupta_iit    (posted on 23 Jun 2008 18:21:10 IST)
couldn't recall where i have seen this. so didn't mention the source
rudra.panda
rudra.panda is offline comment by rudra.panda    (posted on 23 Jun 2008 18:32:27 IST)
it is OK.
tweety
tweety is offline comment by tweety    (posted on 23 Jun 2008 19:18:30 IST)
nice...
ramyani
ramyani is offline comment by ramyani    (posted on 23 Jun 2008 20:27:38 IST)
very nice ! take a salute.
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