Calculate the integral $ \int_{0}^{1}(ax^2+bx+c)^{-3/{2}}dx $
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How to calculate the integral below:
$$ \int_{0}^{1}(ax^2+bx+c)^{-3/{2}}dx $$
calculusintegration
asked 2011-11-29
user id: user
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What did you try? Where are you stuck? – 2011-11-29
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For $a\ne0$, write $ax^2+bx+c={ a({ (x+{b\over 2a})^2+\underbrace{{c\over a} -{b^2\over 4a^2} }_D) }}$. There are three cases to consider now, depending on whether $D=0$, $D>0$, and $D<0$. Two of these require "inverse trigonometric substitutions". The other leads to a simple integral. – 2011-11-29
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I think it would be a good idea if you did some "warm up" problems with $a$, $b$, and $c$ given particular values. – 2011-11-29
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@Leaozinho - Okay, then complete the square and try a trigonometric substitution. – 2011-11-29
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@Leaozinho - Not getting **what**? Again, try completing the square. If you don't know how to do that, then step away from calculus and learn how to complete the square. Then come back to the problem. – 2011-11-29
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@Leaozinho - Okay, so once you complete the square, try a trigonometric substitution in the two cases that David Mitra pointed out in his comment above. – 2011-11-29