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

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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)$\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt (1-x^2}.$
  • (iii)$\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(x/a)+c.$
  • (iv)$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}(x/a)+c.$
Given $I=\int x\sin^{-1}xdx.$
 
Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts where \[\int udv=uv-\int vdu\]
 
Let $u=\sin^{-1}x.$
 
On differentiating with respect to x,
 
$du=2\frac{1}{\sqrt{1-x^2}}dx.$
 
let dv=x dx.
 
On integrating on both sides we get
 
$v=\frac{x^2}{2}.$
 
Now substituting for u,v,du and dv we get,
 
$\int x\sin^{-1}xdx=\sin^{-1}x\frac{x^2}{2}-\int\frac{x^2}{2}.\frac{1}{\sqrt{1-x^2}}dx.$
 
This can be written as
 
$I=\frac{x^2\sin^{-1}x}{2}-\frac{1}{2}\int\frac{x^2}{\sqrt{1-x^2}}dx.$----(1)
 
Let us add 1 and subtract 1 to the numerator
 
$\;\;\;=-\int\frac{x^2-1+1}{\sqrt{1-x^2}}dx.$
 
We can separate the terms as,
 
$\;\;\;=\int[\frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}]dx.$
 
But we know $\frac{1-x^2}{\sqrt{1-x^2}}=(1-x^2)^{1-1/2}=(1-x^2)^{1/2}.$
 
Hence $\int\sqrt{1-x^2}dx-\int\frac{1}{\sqrt{1-x^2}}dx.$
 
On integrating we get,
 
$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}(x/a)+c.$
 
Here a=1,so by substituting for a we get
 
$\int\sqrt{1-x^2}dx=\frac{x}{2}\sqrt{1-x62}+\frac{1}{2}\sin^{-1}x+c.$
 
And \int\frac{1}{\sqrt{1-x^2}dx=\sin^{-1}x.$
 
So $\int\frac{x^2}{\sqrt{1-x^2}}dx=\frac{x}{2}\sqrt{1^2-x^2}+\frac{1}{2}\sin^{-1}x-\sin^{-1}x+c.$
 
Now substituting this in eq(1) we get,
 
$I=\frac{x^2\sin^{-1}x}{2}+\frac{1}{2}[x/2\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x-\sin^{-1}x]+c.$
 
$\;\;\;=\frac{x^2\sin^{-1}x}{2}+\frac{1}{2}[x/2\sqrt{1-x^2}-\frac{1}{2}\sin^{-1}x]+c.$
 
Multiplying by 1/2 we get
 
$\;\;\;=\frac{x^2\sin^{-1}x}{2}+x/4\sqrt{1-x^2}+\frac{1}{4}\sin^{-1}x+c.$
 
Taking $\sin^{-1}x$ as the common factor we get
 
$\;\;\;=\frac{1}{4}(2x^2-1)\sin^{-1}x+\frac{x}{4}\sqrt{1-x^2}+c.$

 

answered Feb 8, 2013 by sreemathi.v
 
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