Question

If $\int \frac{{\sin }^{-1}\sqrt{x}}{\sqrt{1-x}}dx=2[\sqrt{x}-\sqrt{1-x}f(x\left)\right]+C$
$(C$ is integration constant$)$, then the value of $f\left(0\right)$ is equal to

Answer 1

$I=\int \frac{{\sin }^{-1}\sqrt{x}}{\sqrt{1-x}}dx$
Put $x={\sin }^2\theta$
$\Rightarrow dx=2\sin \theta \cos \theta d\theta$
$\therefore I=2\int \underset{I}{\theta }\cdot \underset{II}{\sin }\theta d\theta =2\left[\theta \right(-\cos \theta )-\int (-\cos \theta )d\theta ]=2[-\theta \cos \theta +\sin \theta ]+C\Rightarrow I=2[\sqrt{x}-\sqrt{1-x}{\sin }^{-1}\sqrt{x}]+C$
$\therefore f\left(x\right)={\sin }^{-1}\sqrt{x}$
$\Rightarrow f\left(0\right)=0$