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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

  • (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
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
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,


answered Feb 10, 2013 by sreemathi.v