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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\]

$\begin{array}{1 1} -[\cos^{-1}x\sqrt{1-x^2}+x]+c \\ [\cos^{-1}x\sqrt{1+x^2}+x]+c \\-[\cos^{-1}x\sqrt{1-x^2}-x]+c \\ [\cos^{-1}x\sqrt{1+x^2}-x]+c \end{array} $

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Toolbox:
  • (i)When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts\[\int udv=uv-\int vdu\]
  • (ii)Method of substitution $\int f(x)dx,$ if we substitute f(x) as t,then $f'(x)dx=dt$ hence the integral function becomes $\int t.dt$
  • (iii)$\frac{d}{dx}\cos^{-1}x)=\frac{-1}{\sqrt{1-x^2}}.$
Given $ I=\frac{x\cos^{-1}x}{\sqrt{1-x^2}}dx.$
 
Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts
 
Let $u=\cos{-1}x.$
 
Differentiating with respect to x we get
 
du=$\frac{-1}{\sqrt{1-x^2}}dx.$
 
Let $dv=\frac{x}{\sqrt{1-x^2}}dx.$
 
This can be integrated by the method of substitution
 
Let $(1-x^2)=t.$
 
On differentiating we get
 
-2xdx=dt.
 
$\Rightarrow xdx=\frac{-dt}{2}.$
 
Now substituting t and dt
 
$\int dv=\frac{-1}{2}\int\frac{dt}{\sqrt t}.$
 
On integrating we get
 
v=$\frac{-1}{2}(\sqrt t).$
 
Substituting for t we get
 
v=$\frac{-1}{2}\frac{\sqrt{1-x^2}}{1/2}=-\sqrt{1-x^2}.$
 
Now substituting for u,v,du and dv we get,
 
$I=(\cos^{-1}x).\sqrt{1-x^2}-\int\sqrt{1-x^2}.\frac{-1}{\sqrt{1-x^2}}dx.$
 
On canceling the common terms,
 
$I=-[\cos^{-1}x\sqrt{1-x^2}+x]+c.$

 

answered Feb 10, 2013 by sreemathi.v
edited Jul 21, 2013 by balaji.thirumalai
 
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