Question

Evaluate $\int \frac{\arcsin \sqrt{x}}{\sqrt{1-x}}dx$ $=$

• A. $2[\sqrt{x}-\sqrt{1-x}$ arc $\sin \sqrt{x}]+c$
• B. $2[\sqrt{x}+\sqrt{1-x}$ arc $\sin \sqrt{x}]+c$
• C. $2[\sqrt{x}+\sqrt{1-x}$ arc $\cos \sqrt{x}]+c$
• D. $2[\sqrt{x}-\sqrt{1-x}$ arc $\cos \sqrt{x}]+c$

Answer 1

The correct option is A $2[\sqrt{x}-\sqrt{1-x}$ arc $\sin \sqrt{x}]+c$

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

Put $x={\sin }^2\theta$ $\Rightarrow dx=2\sin \theta \cos \theta d\theta .$

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

$=\int \frac{{\sin }^{-1}\left(\sin \theta \right)}{\sqrt{1-{\sin }^2\theta }}\cdot d\theta \left(2\sin \theta \right)\left(\cos \theta \right)$

$=2\int \theta \sin \theta d\theta$

$=2\left[\theta \right(-\cos \theta )+\int \cos \theta d\theta .]$

$=2\left[\theta \right(-\cos \theta )+\sin \theta ]+c$

$=2[\sin \theta -\theta \cos \theta ]+c$

$=2[\sqrt{x}-({\sin }^{-1}\sqrt{x}\left)\sqrt{1-x}\right]+c$

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