Question

Integrate the function.
$\int \frac{xco{s}^{-1}x}{\sqrt{1-x^2}}dx.$

Answer 1

Let $I=\int \frac{xco{s}^{-1}x}{\sqrt{1-x^2}}dx\Rightarrow I=\int co{s}^{-1}x.\frac{x}{\sqrt{1-x^2}}dx$
Consider $co{s}^{-1}x$ as first function and $\frac{x}{\sqrt{1-x^2}}$ as second function and integrating by parts, we get
$I=co{s}^{-1}x\int \frac{x}{\sqrt{1-x^2}}dx-\int \left[\frac{d}{dx}\right(co{s}^{-1}x)\int \frac{x}{\sqrt{1-x^2}}dx]dx$
Let $1-x^2=t^2\Rightarrow -2x=2t\frac{dt}{dx}\Rightarrow dx=-\frac{t}{x}dt$
$\therefore I=co{s}^{-1}x\int \frac{x}{t}\left(\frac{-t}{x}\right)dt-\int \left[\frac{d}{dx}\right(co{s}^{-1}x)\int \frac{x}{t}(\frac{-t}{x}\left)dt\right]dx=co{s}^{-1}x(-t)-\int \left[\frac{(-1)}{\sqrt{1-x^2}}\right(-t\left)dt\right]dx[\because 1-x^2=t^2\Rightarrow t=\sqrt{1-x^2}]=-\sqrt{1-x^2}.co{s}^{-1}x-\int 1dx=-\sqrt{1-x^2}.co{s}^{-1}x-x+C$