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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[(\sin^{-1}x)^2\]

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

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1 Answer

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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)$\frac{d}{dx}(x^2)=2x.$
  • (iii)$\sin xdx=-\cos x+c.$
Given $I=\int(\sin^{-1}x)^2dx.$
 
Let us substitute $\sin^{-1}x=t.$
 
Since $\sin^{-1}x=t.\Rightarrow x=\sin t.$
 
On differentiating with respect to x we get
 
$dx=\cos t.dt.$
 
On substituting in I we get
 
$I=\int t^2\cos t dt.$
 
Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts
 
Let u=$t^2$,on differentiating we get
 
du=2tdt.
 
Let dv=$\cos tdt$.
 
On integrating we get,
 
v=sin t.
 
Now substituting for u,v,du and dv we get,
 
$i=t^2\sin t.-\int\sin t.2tdt.$
 
$I=t^2\sin t-2\int t\sin t.dt$-------(1)
 
Again $\int t\sin t.dt$ is of the form $\int udv.$
 
Let us have this as $I_1$.
 
Therefore $I_1=\int t\sin tdt.$
 
Let u=t.On differentiating with respect to t
 
du=dt.
 
Let dv=sint dt,on integrting we get
 
v=-cos t.
 
Therefore $I_1=-t\cos t-\int(-\cos t)dt.$
 
$\qquad\qquad=-t\cos t+\int\cos t.dt.$
 
On integrating we get,
 
$I_1=-t\cos t+\sin t.$
 
Now substituting we get,
 
$I=t^2\sin t-[-2t\cos t+2\sin t+c].$
 
$\;\;\;=t^2\sin t+2t\cos t-2\sin t+c.$
 
But we know $\cos t=\sqrt{1-\sin^2t}$,hence
 
$I=t^2\sin t+2t\sqrt{1-\sin^2t}-2\sin t+c.$
 
Now substituting back for t we get,
 
$I=x(\sin^{-1}x)^2-2[-\sin^{-1}x.\sqrt{1-x^2}+x]+c.$
 
$\;\;\;=x(\sin^{-1}x)^2-2[x-\sin^{-1}x\sqrt{1-x^2}]+c.$

 

answered Feb 10, 2013 by sreemathi.v
 
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