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

Hint: If $f\left( x \right)$ and $g\left( x \right)$ be two functions then we can write $\int{f\left( x \right)g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)dx-\int{\left( f'\left( x \right)\int{g\left( x \right)dx} \right)dx}}}$. This method is called integration by parts. As in question there is an inverse trigonometric function so first convert into a trigonometric function by assuming ${{\cos }^{-1}}x=\theta $. Then integrate by using integration by parts.

Complete step-by-step answer:
We have to integrate the function $x{{\cos }^{-1}}x$.
That means we have to find the value of $\int{x{{\cos }^{-1}}xdx}$.
${{\cos }^{-1}}x$ is inverse trigonometric function so to convert to trigonometric function putting ${{\cos }^{-1}}x=\theta $.
So we get $x=\cos \theta $.
We know that derivative of $\cos \theta $ is $-\sin \theta $.
So differentiating both side with respect to $x$ we get $dx=-\sin \theta d\theta $.
Hence the integration become $\int{x{{\cos }^{-1}}xdx}=\int{\cos \theta }\left( {{\cos }^{-1}}\left( \cos \theta \right) \right)\left( -\sin \theta \right)d\theta $/
We know that ${{\cos }^{-1}}\left( \cos \theta \right)=\theta $.
Hence $\int{x{{\cos }^{-1}}xdx}=\int{\cos \theta }\left( \theta \right)\left( -\sin \theta \right)d\theta $.
Rearranging it we get $\int{x{{\cos }^{-1}}xdx}=-\int{\theta \left( sin\theta \cos \theta \right)}d\theta $.
Multiply and divide by $2$ we get $\int{x{{\cos }^{-1}}xdx}=-\dfrac{1}{2}\int{\theta \left( 2sin\theta \cos \theta \right)}d\theta $.
Trigonometric identity is $2sin\theta \cos \theta =\sin 2\theta $.
Using the identity we get $\int{x{{\cos }^{-1}}xdx}=-\dfrac{1}{2}\int{\theta \sin 2\theta }d\theta $.
Integrating by parts we get $\int{f\left( x \right)g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)dx-\int{\left( f'\left( x \right)\int{g\left( x \right)dx} \right)dx}}}$.
Putting $f\left( x \right)=\theta $ and $g\left( x \right)=\sin 2\theta $ we get $\int{x{{\cos }^{-1}}xdx}=-\dfrac{1}{2}\left[ \theta \int{\sin 2\theta d\theta -\int{\left( \left( \theta \right)'\int{\sin 2\theta d\theta } \right)d\theta }} \right]$.
Derivative of $\theta $ is $1$. Also we know that $\int{\sin \left( ax \right)dx=}-\dfrac{1}{a}\cos \left( ax \right)$ where $a$ is non-zero constant.
Using this formula we get $\int{x{{\cos }^{-1}}xdx}=-\dfrac{1}{2}\left[ \theta .\left( -\dfrac{1}{2}\cos 2\theta \right)-\int{1.}\left( -\dfrac{1}{2}\cos 2\theta \right)d\theta \right]$.
Taking $-\dfrac{1}{2}$ common we get $\int{x{{\cos }^{-1}}xdx}=\left( -\dfrac{1}{2} \right)\left( -\dfrac{1}{2} \right)\left[ \theta .\left( \cos 2\theta \right)-\int{\cos 2\theta d\theta } \right]$.
Further simplifying we get $\int{x{{\cos }^{-1}}xdx}=\dfrac{1}{4}\left[ \theta .\left( \cos 2\theta \right)-\int{\cos 2\theta d\theta } \right]$.
Since we have $\int{\cos \left( ax \right)dx=}\dfrac{1}{a}\sin \left( ax \right)$ where $a$ is non-zero constant. Also in integration we add a constant term.
Using this formula we get $\int{x{{\cos }^{-1}}xdx}=\dfrac{1}{4}\left[ \theta .\left( \cos 2\theta \right)-\dfrac{1}{2}\sin 2\theta \right]+c$ where $c$ is the arbitrary constant.
Also two trigonometric identities are $\cos 2\theta =2{{\cos }^{2}}\theta -1$ and $\sin 2\theta =2\sin \theta \cos \theta $.
Using these identities we get \[\int{x{{\cos }^{-1}}xdx}=\dfrac{1}{4}\left[ \theta .\left( 2{{\cos }^{2}}\theta -1 \right)-\dfrac{1}{2}.2\sin \theta \cos \theta \right]+c\].
Simplifying the terms and also using the identity $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$ we get \[\int{x{{\cos }^{-1}}xdx}=\dfrac{1}{4}\left[ \theta .\left( 2{{\cos }^{2}}\theta -1 \right)-\sqrt{1-{{\cos }^{2}}\theta }\cos \theta \right]+c\].
Now putting ${{\cos }^{-1}}x=\theta $ and $x=\cos \theta $ we get \[\int{x{{\cos }^{-1}}xdx}=\dfrac{1}{4}\left[ {{\cos }^{-1}}x\left( 2{{x}^{2}}-1 \right)-\sqrt{1-{{x}^{2}}}.x \right]+c\].
Rearranging the terms we get \[\int{x{{\cos }^{-1}}xdx}=\dfrac{1}{4}\left[ \left( 2{{x}^{2}}-1 \right){{\cos }^{-1}}x-x\sqrt{1-{{x}^{2}}} \right]+c\].
Therefore the integration of the function $x{{\cos }^{-1}}x$ is \[\dfrac{1}{4}\left[ \left( 2{{x}^{2}}-1 \right){{\cos }^{-1}}x-x\sqrt{1-{{x}^{2}}} \right]+c\].
This is the required solution.

Note: In this problem while converting inverse trigonometric function to trigonometric function student must be careful and use appropriate substitution. While integrating so many trigonometric identities are used, so students. So students must be aware of those. Students must remember to add one arbitrary constant at the end of the integration as we are dealing with indefinite integrals.