Question

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

Answer 1

Let $I=\int xco{s}^{-1}xdx$

Applying integration by parts, by taking $co{s}^{-1}x$ as first function and $x$ as second function
$I=co{s}^{-1}x\int xdx-\int \left\{\left(\frac{d}{dx}co{s}^{-1}x\right)\right\}\int \left(xdx\right)dx$

$=co{s}^{-1}x\frac{x^2}{2}-\int \frac{-1}{\sqrt{1-x^2}}\cdot \frac{x^2}{2}dx$

$=\frac{x^{2}co{s}^{-1}x}{2}-\frac{1}{2}\int \frac{1-x^2-1}{\sqrt{1-x^2}}dx$

$=\frac{x^{2}co{s}^{-1}x}{2}-\frac{1}{2}\int \{\sqrt{1-x^2}+\left(\frac{-1}{\sqrt{1-x^2}}\right)\}dx$

$=\frac{x^{2}co{s}^{-1}x}{2}-\frac{1}{2}\int \sqrt{1-x^2}dx-\frac{1}{2}\int \left(\frac{-1}{\sqrt{1-x^2}}\right)dx$

$=\frac{x^{2}co{s}^{-1}x}{2}-\frac{1}{2}{I}_1-\frac{1}{2}co{s}^{-1}x$ .......(1)
where, $I_1=\int \sqrt{1-x^2}dx$

Applying integration by parts
$\Rightarrow I_1=\sqrt{1-x^2}\int 1dx-\int \frac{d}{dx}\sqrt{1-x^2}\int 1dx$
$\Rightarrow I_1=x\sqrt{1-x^2}-\int \frac{-x^2}{\sqrt{1-x^2}}\cdot dx$
$\Rightarrow I_1=x\sqrt{1-x^2}-\int \frac{-x^2}{\sqrt{1-x^2}}dx$
$\Rightarrow I_1=x\sqrt{1-x^2}-\int \frac{1-x^2-1}{\sqrt{1-x^2}}dx$
$\Rightarrow I_1=x\sqrt{1-x^2}-\{\int \sqrt{1-x^2}dx+\int \frac{-dx}{\sqrt{1-x^2}}\}$
$\Rightarrow I_1=x\sqrt{1-x^2}-\{I_1+co{s}^{-1}x\}$
$\Rightarrow 2I_1=x\sqrt{1-x^2}-co{s}^{-1}x$
$I_1=\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}co{s}^{-1}x$
Substituting in (1), we obtain
$I=\frac{x^{2}co{s}^{-1}x}{2}-\frac{1}{2}(\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}co{s}^{-1}x)-\frac{1}{2}co{s}^{-1}x$
$=\frac{(2x^2-1)}{4}co{s}^{-1}x-\frac{x}{4}\sqrt{1-x^2}+C$