Evaluate the following integrals:

Let $I=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$

Let $\mathrm{t}=\cos ^{-1} \mathrm{x}$

$\mathrm{dt}=\frac{1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}$

Also,

$\cos t=x$

Thus,

$I=-\int t \cos t d t$

Now let us solve this by 'by parts' method

Using integration by parts,

$I=-t\left(\int \cos t d t-\int \frac{d}{d t} t \int \cos t d t\right)$

Let

$\mathrm{U}=\mathrm{t} ; \mathrm{d} \mathrm{u}=\mathrm{dt}$

$\int \cos t \mathrm{dt}=\mathrm{v} ; \sin \mathrm{t}=\mathrm{dv}$

Thus,

$I=-\left[t \sin t-\int \sin t d t\right]$

$I=-[t \sin t+\cos t]+c$

Substituting

$t=\cos ^{-1} x$

$I=-\left[\cos ^{-1} x \sin t+x\right]+c$

$I=-\left[\cos ^{-1} X \sqrt{1-X^{2}}+X\right]+c$