Question

$\int {\left[xsi{n}^{-1}x{\left\{\sqrt{(1-x^2}\right\}}^{-1}\right]}^{}$ dx

Answer 1

$\int \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx[Here{1}^{st}Function={\sin }^{-1}x{2}^{nd}Function=\frac{x}{\sqrt{1-x^2}}]$

$\Rightarrow$ Using integration by part $5$

$\int I.IIdx=I\int IIdx-\int \left(\frac{dI}{dx}\int IIdx\right)dx$

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

Now Solving, Here $\int \frac{x}{\sqrt{1-x^2}}dx=\int \frac{-dt}{2\sqrt{t}}$ Putting $1-x^2=t$

$-2xdx=dt$

$xdx=\frac{-dt}{2}$

$=\frac{-1}{2}\frac{t}{1/2}=-\sqrt{t}$

$=-\sqrt{1-x^2}$

Hence, $\int \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx={\sin }^{-1}x[-\sqrt{1-x^2}]-\int \frac{1}{\sqrt{1-x^2}}\times -\sqrt{1-x^2}dx$

$=-\sqrt{1-x^2}{\sin }^{-1}x+\frac{x^2}{2}+C$ where $C=$ constant of Integration.