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

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)$\frac{d}{dx}(\cos^{-1}x)=\frac{-1}{\sqrt{1-x^2}}.$
Given I=$\int x\cos^{-1}xdx.$

Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts

Let u=$\cos^{-1}x.$

On differentiating with respect to x,

du=$\frac{-1}{\sqrt{1-x^2}}dx.$

Let dv=xdv.

Integrting on both sides we get,

v=$\frac{x^2}{2}.$

Now substituting for u,v,du and dv we get,

$\int x\cos^{-1}xdx=\frac{x^2}{2}\cos^{-1}x-\int\frac{x^2}{2}\big(\frac{-1}{\sqrt{1-x^2}}dx.$

$\qquad\qquad=\frac{x^2}{2}\cos^{-1}x-\frac{1}{2}\int\frac{-x^2}{\sqrt{1-x^2}}dx.$----(1)

Let $I_1=\int\frac{-x^2}{\sqrt{1-x^2}}dx.$

Add and subtract 1 to the numerator,

$I_1=\frac{1-x^2-1}{\sqrt{1-x^2}}dx.$