Browse Questions

Integrate the function$(\sin^{-1}x)^2$

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

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,

$I=x(\sin^{-1}x)^2-2[-\sin^{-1}x.\sqrt{1-x^2}+x]+c.$

$\;\;\;=x(\sin^{-1}x)^2-2[x-\sin^{-1}x\sqrt{1-x^2}]+c.$