Question

$\int \frac{2x-\sqrt{{\sin }^{-1}x}}{\sqrt{1-x^2}}dx$ is equal to
(where $C$ is constant of integration)

• A. $-3(1-x^2{)}^{\frac{1}{2}}-2({\sin }^{-1}x{)}^{\frac{3}{2}}+C$
• B. $-2(1-x^2{)}^{\frac{1}{2}}-\frac{2}{3}({\sin }^{-1}x{)}^{\frac{3}{2}}+C$
• C. $-\frac{1}{2}(1-x^2{)}^{\frac{1}{2}}-\frac{2}{3}({\sin }^{-1}x{)}^{\frac{3}{2}}+C$
• D. $-2(1-x^2{)}^{\frac{1}{2}}-3({\sin }^{-1}x{)}^{\frac{3}{2}}+C$

Answer 1

The correct option is B $-2(1-x^2{)}^{\frac{1}{2}}-\frac{2}{3}({\sin }^{-1}x{)}^{\frac{3}{2}}+C$

$I=\int \frac{2x-\sqrt{{\sin }^{-1}x}}{\sqrt{1-x^2}}dx=\int \frac{2x}{\sqrt{1-x^2}}dx-\int \frac{\sqrt{{\sin }^{-1}x}}{\sqrt{1-x^2}}dx=\int \frac{2x}{\sqrt{1-x^2}}dx-\int \sqrt{{\sin }^{-1}x}\frac{dx}{\sqrt{1-x^2}}$
Let, $1-x^2=t$ and ${\sin }^{-1}x=u$
$\Rightarrow -2xdx=dt$ and $\frac{dx}{\sqrt{1-x^2}}=du\Rightarrow I=\int \frac{-dt}{\sqrt{t}}-\int \sqrt{u}du\Rightarrow I=-2t^{\frac{1}{2}}-\frac{2}{3}{u}^{\frac{3}{2}}+C$
$\Rightarrow I=-2(1-x^2{)}^{\frac{1}{2}}-\frac{2}{3}({\sin }^{-1}x{)}^{\frac{3}{2}}+C$