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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the functions\[\frac{1}{\sqrt{7-6x-x^2}}\]

$\begin{array}{1 1} \sin^{-1} \large \frac{(x+3)}{4}+c\\ \sin^{-1} \large \frac{(x-3)}{4}+c \\\cos^{-1} \large \frac{(x+3)}{4}+c \\ \cos^{-1} \large \frac{(x-3)}{4}+c \end{array} $

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  • $\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\big(\frac{x}{a}\big)+c.$
Given $I=\int\frac{dx}{\sqrt{7-6x-x^2}}.$
 
$7-6x-x^2$ can be written as $-(x^2+6x-7).$
 
$\;\;\;=-[(x+3)^2-16].$
 
$\;\;\;=[16-(x+3)^2].$
 
Therefore $I=\int\frac{dx}{\sqrt{16-(x+3)^2}}$.
 
On integrating we get,
 
$\;\;\;=\sin^{-1}\frac{(x+3)}{4}+c.$

 

answered Feb 4, 2013 by sreemathi.v
 
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