E-StoreIndefnite easier
4 Aug 2007 09:09:01 IST
Indefnite easier
Here is what you got to do when integration has been asked of a particular type of expression :
**
f(ax+b)
f(ax+b) put (ax+b)=t
**px+q/ax2+bx+c
px+q/ax2+bx+c take px+q=l(differentiation of ax2+bx+c) + m
Find values of l and m and substitute them in the exp.
and divide the whole by (ax2+bx+c) and now you can do
its integration easily as it is divided in two parts.
**P(x)/ax2+bx+c
P(x)/ax2+bx+cIn such cases we divide the numerator by denominator
and then express as
Q(x) + [R(x)/ax2+bx+c]
Here Q(x) = Quotient after dividing.
and R(x) = Remainder after dividing.
This becomes Q(x).dx + R(x)/ax2+bx+c .dx
Q(x).dx + R(x)/ax2+bx+c .dx**px+q/(ax2+bx+c)1/2
px+q/(ax2+bx+c)1/2Take px+q = l(Diff. of Denominator) + m
Find values of "l" and "m" and solve.
After this take x common and make the coefficient of x2
as unity.
Now add and substract the square of half of the
coefficient of x.
Now apply formula to get the answer.
** 1/aSinx+bCosx , 1/a+bSinx , 1/a+bCosx.dx
1/aSinx+bCosx , 1/a+bSinx , 1/a+bCosx.dxWe proceed as follows.
Take Sinx = 2 Tan x/2 / 1+tan2x/2 or
Take Cosx = 1-tan2x/2 / 1+tan2x/2
Replace 1+tan2x/2 in numerator by sec2x/2.
put tan x/2 = t as 1/2Sec2x/2.dx=dt
And solve.
**1/aSinx+bCosx
1/aSinx+bCosxWe take a=rCos@ and b=rSin@
as r=(a2+b2)1/2.@=tan-1(b/a)
Now solve.
**aSinx+bCosx/cSinx+dCosx
aSinx+bCosx/cSinx+dCosxNumerator=l(Diff. of Denominator) + m(denominator)
Get value of l and m and solve.
**aSinx+bCosx+c/pSinx+qCosx+r
aSinx+bCosx+c/pSinx+qCosx+rHere c and r = Integers(constants)
Num=l(den.)+m(Diff. of Den.)+ n
Find values of l,m,n and then take the form as:
aSinx+bCosx+c/pSinx+qCosx+r=l.dx + mDiff. of Den/Den. + n1/pSinx+qCosx+r
l.dx + mDiff. of Den/Den. + n1/pSinx+qCosx+rNow solve.
**f'(x)/f(x).dx = log{f(x)}
f'(x)/f(x).dx = log{f(x)}**ex{f(x)+f'(x)}.dx = ex{f(x)} + c
ex{f(x)+f'(x)}.dx = ex{f(x)} + c**(px+q)(ax2+bx+c)1/2.dx px+q=l(Diff. of ax2+bx+c)+m Find l and m and solve. **h(x)/P(Q)1/2.dx *P and Q are linear.put Q =t2. Gor example of a form. 1/(ax+b)(cx+d)1/2.dx put (cx+d)=t2. *P is quadratic and Q is linear 1/(ax2+bx+c)(dx+e)1/2.dx put (dx+e)=t2. *P is linear and Q is Quadratic. 1/(ax+b)(cx2+dx+e)1/2.dx put (ax+b)=1/t *P and Q both are Quadratic. 1/(ax2+b)(cx2+d)1/2.dx put x=1/t and then c+dt2 = u2. Hope you find it useful. Rate me if useful. Cheers!!!!!!!!!!!
Comments (19)
joy francis
Blazing goIITian
Joined: 19 Feb 2007 20:50:19 IST
Posts: 1802
4 Aug 2007 09:24:51 IST
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thank you
6 Aug 2007 17:28:36 IST
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gudd work mann.........
its really useful.......and i have rated u........
its really useful.......and i have rated u........
6 Aug 2007 22:04:07 IST
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its great bhai!!!!
thank u very much!!!
keep on posting such type 4 us !!!!!
thanks once again
thank u very much!!!
keep on posting such type 4 us !!!!!
thanks once again
7 Aug 2007 16:32:00 IST
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0 people liked this
these are basic but hats off for u'r patience if u typed it
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