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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Indefinite -- Another one
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[Post New]16 Mar 2008 11:37:56 IST
    Subject: Indefinite -- Another one
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sid.shah.90 (603)

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Evaluate---
 
 
((x-a)/(b-x))dx
 
plz give as many different methods and substitutions.............
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Siddhant Shah


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[Post New]16 Mar 2008 11:46:57 IST
    Subject: Re:Indefinite -- Another one
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sandeepramesh (1247)

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Put x = a^2 \cdot ( sinx )^2 + b^2 \cdot ( cosx )^2 Smile
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[Post New]16 Mar 2008 11:49:28 IST
    Subject: Re:Indefinite -- Another one
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annihilator (343)

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multiply and divide by x-a or b-x then separate the into two terms in which will be in std form and other one can be done by substion
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[Post New]16 Mar 2008 15:26:39 IST
    Subject: Re:Indefinite -- Another one
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ramkumar_november (1270)

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put   x  = a sin2  +  b cos2      ....
 
 
 
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[Post New]16 Mar 2008 20:14:15 IST
    Subject: Indefinite -- Another one
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varshavallig (800)

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yup , annihilator is right,multiply and divide by x-a and then change it into a std integral.

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[Post New]16 Mar 2008 23:00:23 IST
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Taara (53)

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pls explain. didn't understand any of u guys. how to proceed?
TAARA
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[Post New]16 Mar 2008 23:01:22 IST
    Subject: Re:Indefinite -- Another one
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computer001 (1849)

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easiest is sandeep's method..follow tht
Nitwit Blubber Odment Tweak
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[Post New]17 Mar 2008 12:24:39 IST
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computer001 (1849)

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y do i get rated 4 sayin follow some1 else's?? method
Nitwit Blubber Odment Tweak
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[Post New]17 Mar 2008 12:32:02 IST
    Subject: Re:Indefinite -- Another one
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sandeepramesh (1247)

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bcos what u said is the most intelligent thing u have said Huh? Huh?
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[Post New]17 Mar 2008 12:35:07 IST
    Subject: Re:Indefinite -- Another one
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layman (162)

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Both the above methods are right I think.

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[Post New]17 Mar 2008 12:36:13 IST
    Subject: Re:Indefinite -- Another one
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ramkumar_november (1270)

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  I   =   \int\sqrt{\frac{x-a}{b-x}}\;dx
 
PUT \;\;x=\;acos^2\theta\;+bsin^2\theta
 
differentiating we get.........
 
dx\;=\;(b-a)sin2\theta\,d\theta
 
x-a\,=\,(b-a)sin^2\theta
 
b-x\,=\,(b-a)cos^2\theta
 
so   \sqrt{\frac{x-a}{b-x}}\;=\;tan\theta
 
 
substituting in the integral. we get......
 
   I  =  \int(b-a)sin2\theta\,tan\theta\,d\theta
 
   I  =  2(b-a)\int\,sin^2\theta\,d\theta
 
  I  =  2(b-a)\bigg(\frac{\theta}{2}\,-\,\frac{cos2\theta}{4}\bigg)
 
where\;\;\theta\;=\;sin^{-1}\sqrt{\frac{x-a}{b-a}}
 
clear with my proof???
 
 
 
 
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[Post New]17 Mar 2008 12:47:27 IST
    Subject: Re:Indefinite -- Another one
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computer001 (1849)

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@ madness:

Bow Down


edited:
wat was what 4??
n this is jus to give madness the funny feeling tht i repect him
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[Post New]17 Mar 2008 12:49:16 IST
    Subject: Re:Indefinite -- Another one
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layman (162)

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What was that for??

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