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# Integrate the function$x\;\sin^{-1}x$

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

$\int\sqrt{1-x^2}dx=\frac{x}{2}\sqrt{1-x62}+\frac{1}{2}\sin^{-1}x+c.$