Question

Integrate:

$\int \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$.

Answer 1

Consider the given integral.

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

Let $f\left(x\right)={\sin }^{-1}x$ and $g\left(x\right)=\frac{x}{\sqrt{1-x^2}}$.

We know that,

$\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-\int (f^{\prime }\left(x\right)\int g\left(x\right)dx)dx$

Therefore,

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

$I=I_1+I_2$

Let $1-x^2=t$. Therefore,

$dx=\frac{dt}{-2x}$

Now,

$I_1=\int \frac{x}{\sqrt{1-x^2}}dx$

$I_1=\int \frac{x}{\sqrt{t}}\frac{dt}{-2x}$

$I_1=\int \frac{1}{-2\sqrt{t}}dt=-\frac{1}{2}\int t^{-1/2}dt=-\frac{1}{2}\left(\frac{t^{1/2}}{1/2}\right)$

$I_1=-t^{1/2}$

$I_1=-\sqrt{t}$

$I_1=-\sqrt{1-x^2}$

Therefore,

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

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

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

$I=-\sqrt{1-x^2}{\sin }^{-1}x+\int 1\cdot dx$

$I=-\sqrt{1-x^2}{\sin }^{-1}x+x+C$

$I=x-\sqrt{1-x^2}{\sin }^{-1}x+C$

Hence, this is the required value of the integral.