Integration giving $\arcsin x$ by a unique method: $\sin^{-1}t=\int_0^t\frac{dx}{\sqrt{1-x^2}}$ $iii)$ Hence show that$$
\begin{align}
\sin^{-1}t&=2\int_0^t\sqrt{1-x^2}\:dx-t\sqrt{1-t^2}
\end{align}
$$ $iv)$ By using integration by parts, show that$$
\begin{align}
\int_0^t\sqrt{1-x^2}\:dx&=t\sqrt{1-t^2}+\int_0^t\frac{x^2}{\sqrt{1-x^2}}\:dx
\end{align}
$$ $v)$ By using parts $iii)$ and $iv)$, prove that, show that$$
\begin{align}
\sin^{-1}t=\int_0^t\frac{dx}{\sqrt{1-x^2}}
\end{align}
$$ 
I am just having trouble at the last part. I tried substituting the integral but that method results in endless loop

To conclude, one may observe that
$$
\begin{align}
\int_0^t\frac{x^2}{\sqrt{1-x^2}}dx&=-\int_0^t\frac{(1-x^2)-1}{\sqrt{1-x^2}}dx
\\\\&=-\int_0^t\frac{1-x^2}{\sqrt{1-x^2}}dx+\int_0^t\frac{1}{\sqrt{1-x^2}}dx
\\\\&=-\int_0^t\sqrt{1-x^2}dx+\int_0^t\frac{1}{\sqrt{1-x^2}}dx
\end{align}
$$ then use $iii)$ and $iv)$.
