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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\sqrt{1+3x-x^2}\]

$\begin{array}{1 1}\large \frac{(2x-3)(\sqrt{1+3x-x^2})}{4} +\frac{13 }{8}\sin^{-1}(\large \frac{2x-3}{\sqrt {13}})+c \\\large \frac{(2x-3)(\sqrt{1+3x-x^2})}{4} -\frac{13 }{8}\cos^{-1}(\large \frac{2x-3}{\sqrt {13}})+c \\ \large \frac{(3x+2)(\sqrt{1+3x-x^2})}{4} +\frac{13 }{8}\cos^{-1}(\large \frac{2x-3}{\sqrt {13}})+c \\ \large \frac{(3-2x)(\sqrt{1+3x-x^2})}{4} -\frac{13 }{8}\sin^{-1}(\large \frac{2x-3}{\sqrt {13}})+c \end{array} $

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1 Answer

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  • $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac {a^2}{2}\sin^{-1}(\frac{x}{a})+c$
Given $I=\sqrt{1+3x-x^2}dx$
 
Consider $1+3x-x^2=-(x^2-3x-1)$
 
This can split as $-[(x^2-3x+\frac{9}{4})-\frac{9}{4}-1]$
 
                         $=-[(x-\frac{3}{2})^2-\frac {9}{4}-1]$
 
                         $=-[(x-\frac{3}{2})^2-(\frac {\sqrt {13}}{2})^2]$
 
                          $I=[(\frac{\sqrt {13}}{2})^2-(x-\frac{3}{2})^2]$
 
         Hence $I=\int \sqrt{(\frac{\sqrt {13}}{2})^2-(x-\frac{3}{2})^2} dx$
 
Clearly This is of the form $\int \sqrt{a^2-x^2} dx$
 
                        $=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}(\frac{x}{a})+c$
 
Here in the place a we have $\frac{\sqrt{13}}{2}$ and in the place of x we have (x-3/2).
 
Therefore $\int \sqrt{1+3x-x^2}dx=\frac{(x-\frac{3}{2})}{2}\sqrt{1+3x-x^2}+\frac{(\frac{\sqrt {13}}{2})^2}{2}\sin^{-1}(\frac{x-3/2}{\frac{\sqrt {13}}{2}})+c$
 
On simplifying we get,
 
$\frac{2x-3}{4} \sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}(\frac{2x-3}{\sqrt {13}})+c$
answered Feb 7, 2013 by meena.p
edited Apr 11 by meena.p
 
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