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

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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)$\frac{d}{dx}(\cos^{-1}x)=\frac{-1}{\sqrt{1-x^2}}.$
Given I=$\int x\cos^{-1}xdx.$
 
Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts
 
Let u=$\cos^{-1}x.$
 
On differentiating with respect to x,
 
du=$\frac{-1}{\sqrt{1-x^2}}dx.$
 
Let dv=xdv.
 
Integrting on both sides we get,
 
v=$\frac{x^2}{2}.$
 
Now substituting for u,v,du and dv we get,
 
$\int x\cos^{-1}xdx=\frac{x^2}{2}\cos^{-1}x-\int\frac{x^2}{2}\big(\frac{-1}{\sqrt{1-x^2}}dx.$
 
$\qquad\qquad=\frac{x^2}{2}\cos^{-1}x-\frac{1}{2}\int\frac{-x^2}{\sqrt{1-x^2}}dx.$----(1)
 
Let $I_1=\int\frac{-x^2}{\sqrt{1-x^2}}dx.$
 
Add and subtract 1 to the numerator,
 
$I_1=\frac{1-x^2-1}{\sqrt{1-x^2}}dx.$

 

answered Feb 10, 2013 by sreemathi.v
 
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