Question

If $\int \frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}dx=-k[\sqrt{1-x^2}{\cos }^{-1}x+x]+C$. what will be the value of $k$?

Answer 1

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

$I=\frac{-1}{2}\int \frac{-2x}{\sqrt{1-x^2}}\cdot {\cos }^{-1}xdx$
Taking ${\cos }^{-1}x$ as first function and $\left(\frac{-2x}{\sqrt{1-x^2}}\right)$ as second function and integrating by parts, we obtain $I=\frac{-1}{2}[{\cos }^{-1}x\int \frac{-2x}{\sqrt{1-x^2}}dx-\int \{\left(\frac{d}{dx}{\cos }^{-1}x\right)\int \frac{-2x}{\sqrt{1-x^2}}dx\}dx]$

$=\frac{-1}{2}[{\cos }^{-1}x\cdot 2\sqrt{1-x^2}-\int \frac{-1}{\sqrt{1-x^2}}\cdot 2\sqrt{1-x^2}dx]$

$=\frac{-1}{2}[2\sqrt{1-x^2}{\cos }^{-1}x+\int 2dx]$

$=\frac{-1}{2}[2\sqrt{1-x^2}{\cos }^{-1}x+2x]+C$

$=-[\sqrt{1-x^2}{\cos }^{-1}x+x]+C$