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Integrate the functions\[\frac{1}{\sqrt{8+3x-x^2}}\]

$\begin{array}{1 1} \sin^{-1} \bigg(\frac{2x-3}{\sqrt {41}}\bigg)+c \\ \cos^{-1} \bigg(\frac{2x-3}{\sqrt {41}}\bigg)+c \\ \sin^{-1}\large \bigg(\frac{2x+3}{\sqrt {41}}\bigg)+c \\ \cos^{-1} \bigg(\frac{2x+3}{\sqrt {41}}\bigg)+c \end{array} $

1 Answer

Toolbox:
  • $\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\big(\frac{x}{a}\big)+c.$
Given:$I=\int\frac{dx}{\sqrt{8+3x-x^2}}.$
 
$8+3x-x^2$ can be written as,
 
$\;\;\;=-(x^2-3x-8)$.
 
$\;\;\;==[(x-\frac{3}{2})^2-(\frac{\sqrt {41}}{2})^2]$.
 
$\;\;\;=(\frac{\sqrt {41}}{2})^2-(x-\frac{3}{2})^2$
 
Therefore $I=\frac{dx}{(\frac{\sqrt{41}}{2})^2-(x-\frac{3}{2})^2}.$
 
On integrating we get,
 
$\;\;\;=\sin^{-1}\bigg(\frac{(x-3/2)}{\frac{\sqrt{ 41}}{2}}\bigg)+c.$
 
$\;\;\;=\sin^{-1}\bigg(\frac{2x-3}{\sqrt {41}}\bigg)+c.$

 

answered Feb 4, 2013 by sreemathi.v
 
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