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Home  >>  CBSE XII  >>  Math  >>  Integrals

Integrate the function\[\sqrt{1-4x^2}\]

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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 $I=\int \sqrt{(1-4x^2}dx$
 
$=\int \sqrt{4(\frac{1}{4}-x^2})dx$
 
$=\int 2\sqrt{\frac{1}{4}-x^2}dx$
 
$=2\int \sqrt{\frac{1}{4}-x^2}dx$
 
Clearly this is of the fun $\int \sqrt{a^2-x^2},$ where,
 
$a=\frac{1}{2}$
 
Hence $I=2\bigg[\frac{x}{2}\sqrt{1-4x^2}+\frac{(1/2)^2}{2} \sin^{-1}(\frac{x}{1/2}+c\bigg]$
 
$=\frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}\sin ^{-1}(2x)+c.$

 

 

answered Feb 7, 2013 by meena.p
 
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