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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the functions $\frac{1}{\sqrt{9-25x^2}}$

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  • $\int\frac{dx}{\sqrt{a^2+x^2}}=\sin^{-1}\big(\frac{x}{a}\big)+c.$
Given:$I=\int \frac{dx}{\sqrt{9-25x^2}}dx.$
 
Let 5x=t.
 
On differentiating we get,
 
-5dx=dt.$\Rightarrow dx=\frac{dt}{5}.$
 
On substituting t and dt we get,
 
Hence $I=\frac{1}{5}\frac{dx}{\sqrt{9-t^2}}=\frac{1}{5}\frac{dx}{\sqrt{3^2-t^2}}.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{5}\sin^{-1}\big(\frac{t}{3}\big)+c.$
 
substituting for t we get,
 
$\;\;\;=\frac{1}{5}\sin^{-1}\big(\frac{5x}{3}\big)+c.$

 

answered Feb 4, 2013 by sreemathi.v
 
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