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Home  >>  CBSE XII  >>  Math  >>  Integrals
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$\Large \int \normalsize\frac{dx}{\sqrt{9x-4x^2}}$ equals

\begin{array}{1 1}(A)\;\frac{1}{9}\sin^{-1}\bigg(\frac{9x-8}{8}\bigg)+C \\ (B)\;\frac{1}{2}\sin^{-1}\bigg(\frac{8x-9}{9}\bigg)+C\\(C)\;\frac{1}{3}\sin^{-1}\bigg(\frac{9x-8}{8}\bigg)+C \\ (D)\;\frac{1}{2}\sin^{-1}\bigg(\frac{9x-8}{9}\bigg)+C\end{array}

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1 Answer

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  • $\int\frac{dx}{a^2-x^2}=\sin^{-1}\big(\frac{x}{a}\big)+c.$
Given $I=\int\frac{dx}{\sqrt{9x-4x^2}}.$
 
$9x-4x^2=-(4x^2-9x)$
 
$\;\;\;\;\qquad\;\;=-4(x^2-\frac{9}{4}x)$.
 
$\;\;\;\;\qquad\;\;=-4(x-\frac{9}{8})^2-\frac{81}{64}$
 
$\;\;\;\;\qquad\;\;=4\big(\frac{9}{8}\big)^2-\big(x-\frac{9}{8}\big)^2$
 
Therefore $I=\frac{1}{2}\int\frac{dx}{\sqrt{\big(\frac{9}{8}\big)^2-\big(x-\frac{9}{8}\big)^2}}$
 
On integrating we get,
 
$\;\;\;\;\qquad\;\;=\frac{1}{2}\sin^{-1}\big(\frac{x-9/8}{9/8}\big)+c.$
 
$\;\;\;\;\qquad\;\;=\frac{1}{2}\sin^{-1}\frac{(8x-9)}{9}+c.$
 
Hence the correct answer is B.

 

answered Feb 5, 2013 by sreemathi.v
 
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