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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\frac{1}{x\sqrt{ax-x^2}}[Hint:Put\;x=\frac{a}{t}]\]

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Toolbox:
  • (i)If f(x)=t, then f'(x)=dt.Henc $int f(x)dx=\int t.dt$
  • (ii)$\int x^n dx=\frac{x^{n+1}}{n+1}+c$
Given $I=\frac{1}{x\sqrt{ax-x^2}}$
 
Let us put $x=\frac{a}{t}$ then differentiating then w.r.t x $dx=-\frac{a}{t^2}$
 
Therefore $I=\large\frac{1}{\frac{a}{t}\sqrt{a(\frac{a}{t})-(\frac{a}{t})^2}} \times (\frac{-a}{t^2})$
 
On simplifying we get
 
$-\int \large\frac{1}{at} \times \frac{1}{\sqrt {\frac{1}{t}-\frac{1}{t^2}}}dt$
 
$=\frac{-1}{a}\int \large\frac{1}{\sqrt {\frac{t^2}{t}-\frac{t^2}{t^2}}}dt$
 
$=\frac{-1}{a}\int \frac{1}{\sqrt {t-1}}dt$
 
On integrating we get,
 
$-\frac{1}{a} \bigg[\frac{(t-1)^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\bigg]+c$
 
$-\frac{1}{a} \bigg[\frac{(t-1)^{\frac{1}{2}}}{\frac{1}{2}}\bigg]+c$
 
$=\frac{-2}{a}[\sqrt {t-1}]+c$
 
Substituting for t we get,
 
$I=\frac{-2}{9}\bigg[\sqrt{\frac{a}{x}-1}\bigg]+c$
 
$I=\frac{-2}{9}\bigg[\sqrt{\frac{a-x}{x}}\bigg]+c$

 

 

answered Feb 15, 2013 by meena.p
 
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