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Home  >>  CBSE XII  >>  Math  >>  Integrals

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

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  • $I=\int \sqrt {a^2-x^2} dx=\frac {x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2} \sin {-1}(\frac{x}{a})+c$
Given $I=\int \sqrt{4-x^2}$
 
Clearly this is of the for $\sqrt{a^2-x^2}=\frac{x}{2}\sqrt{a^2=x^2}+\frac{a^2}{2}\sin^{-1}(x/9)$
 
Here $ a^2=4\qquad=>a=2$
 
I=$\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}(\frac{x}{2})+c$
 
   $\qquad=\frac{x}{2}\sqrt{4-x^2}+2 \sin^{-1}(\frac{x}{2})+c$

 

answered Feb 7, 2013 by meena.p
 
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