Answer
$(x-\dfrac{1}{2} )\sin^{-1}\sqrt x+\dfrac{1}{2} \sqrt{x(1-x)}+C$
Work Step by Step
Suppose $u=\sqrt x\implies du=\dfrac{1}{2\sqrt x}dx=\dfrac{1}{2u}dx ; dx=2udu$
So, $I=\int\sin^{-1} \space u(2u \space du)=2 \int u\sin^{-1}u \space du$
Now apply Formula 106, which states that
$$\int x^n\sin^{-1}axdx=\dfrac{x^{n+1}}{n+1}\sin^{-1} (ax)-\frac{a}{n+1}\int\frac{x^{n+1}dx}{\sqrt{1-a^2x^2}}$$
For $a=1$ and $n=1$:
$I=u^2\sin^{-1} \space u-\int\dfrac{u^2du}{\sqrt{1-u^2}}$
Now, $I=u^2\sin^{-1}u-(\dfrac{1}{2}\sin^{-1}u-\dfrac{1}{2}u\sqrt{1-u^2} )+C\\=(x-\dfrac{1}{2} )\sin^{-1}\sqrt x+\dfrac{1}{2} \sqrt{x(1-x)}+C$
