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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\frac{\large \sin^{-1}x}{\large \sqrt{1-x^2}}$

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Toolbox:
  • Method of substitution :
  • Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
  • $\Rightarrow $dx=g'(t)dt.
  • Thus $I=\int f(g(t).g'(t))dt.$
Given $I=\int \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx.$
 
Put $\sin^{-1}x$=t.
 
$\frac{1}{\sqrt{1-x^2}}dx=dt.$
 
Now substituting for x and dx we get,
$I=\int tdt.$
 
On integrating we get,
 
$\frac{t^2}{2}+c.$
 
Substituting back for t we get,
 
$\int \frac{\sin^{-1}x}{\sqrt {1-x^2}}dx=\frac{1}{2}(\sin^{-1}x)^2+c.$
 
 
 
 

 

answered Jan 29, 2013 by sreemathi.v
 
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