# Integrate the function$\frac{x\cos^{-1}x}{\sqrt{1-x^2}}$

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

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)Method of substitution $\int f(x)dx,$ if we substitute f(x) as t,then $f'(x)dx=dt$ hence the integral function becomes $\int t.dt$
• (iii)$\frac{d}{dx}\cos^{-1}x)=\frac{-1}{\sqrt{1-x^2}}.$
Given $I=\frac{x\cos^{-1}x}{\sqrt{1-x^2}}dx.$

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

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

Differentiating with respect to x we get

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

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

This can be integrated by the method of substitution

Let $(1-x^2)=t.$

On differentiating we get

-2xdx=dt.

$\Rightarrow xdx=\frac{-dt}{2}.$

Now substituting t and dt

$\int dv=\frac{-1}{2}\int\frac{dt}{\sqrt t}.$

On integrating we get

v=$\frac{-1}{2}(\sqrt t).$

Substituting for t we get

v=$\frac{-1}{2}\frac{\sqrt{1-x^2}}{1/2}=-\sqrt{1-x^2}.$

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

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

On canceling the common terms,

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

edited Jul 21, 2013