Question

$\int \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$ is equal to

• A. $x-\sqrt{1-x^2}si{n}^{-1}x+c$
• B. $x+\sqrt{1-x^2}si{n}^{-1}x+c$
• C. $x+si{n}^{-1}x+c$
• D. $x-si{n}^{-1}x+c$

Answer 1

The correct option is A $x-\sqrt{1-x^2}si{n}^{-1}x+c$

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

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

$=-t\cos t+\int \cos tdt$ ....integration by parts

$=-t\cos t+\sin t+C$

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

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

Hence, option 'A' is correct.