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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Mar 2008 19:10:52 IST
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q) Evaluate : 1/  (x-a)(x-b) |
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17 Mar 2008 19:18:06 IST
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(x-a)(x-b) = x2 -(a+b)x + ab
Complete the square [ x- (a+b)/2 ]2
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17 Mar 2008 19:21:17 IST
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hope its integration then 1/ (x-a)(x-b) =1/rt(x2-(a+b)x+ab) =1/rt (x-(a+b)/2)2-((a-b)/2)2) its , log (x-(a+b)/2+(rt(x-a)(x-b)) thats it !
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17 Mar 2008 19:34:27 IST
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naidu is corrrect.we have to proceed like that
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17 Mar 2008 19:36:22 IST
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![[tex] \\ \int \frac{1}{\sqrt{(x-a)(x-b)}} \\ \\ \mbox{Consider the denominator ..it is nothing but} \\ \\ \sqrt{(x-(\frac{a+b}{2}))^2 - (\frac{a-b}{2})^2} \\ \\ \mbox{So the given integral simplifies to} \\ \\ \int \frac{1}{\sqrt{(x-(\frac{a+b}{2}))^2 - (\frac{a-b}{2})^2}} \\ \mbox{As per formula we get } \\ \\ \log{[(x-(\frac{a+b}{2}))+\sqrt{(x-(\frac{a+b}{2}))^2 - (\frac{a-b}{2})^2}}]+C[\tex]](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/3/4/e/34ecf7fb4197424560f9719f1565ab08f99c64ba.gif)
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17 Mar 2008 19:56:55 IST
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convert into partial fractions
the ans is 1/(a-b) ln(x-a/x-b)
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21 Mar 2008 12:54:03 IST
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let x = a cos^2 A + b sin^2 A
it becomes easy afterwards
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21 Mar 2008 12:58:39 IST
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u cannot use x = a cos^2 A + b sin^2 A
thts possible only in case of root((a-x)(x-b))
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