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 $iv)$.
• \begin{align} 2\int_0^t\sqrt{1-x^2}\:dx&-t\sqrt{1-t^2}=-\int_0^t\frac{1}{\sqrt{1-x^2}}\:dx \end{align} \begin{align} \sin^{-1}t=-\int_0^t\frac{1}{\sqrt{1-x^2}}\:dx \end{align} – mathnoob123 Aug 4 '16 at 18:08
• You can even see from your example.\begin{align} sin^{-1}=-sin^{-1} -2\int_0^t\frac{1}{\sqrt{}}\ \end{align}\begin{align} 2sin^{-1}=-2\int_0^t\frac{1}{\sqrt{}}\ \end{align}\begin{align} sin^{-1}=-\int_0^t\frac{1}{\sqrt{}}\ \end{align} – mathnoob123 Aug 4 '16 at 18:24
• But this statement is true. So why are we getting a different answer\begin{align} sin^{-1}=\int_0^t\frac{1}{\sqrt{1-x^2}}\ \end{align} – mathnoob123 Aug 4 '16 at 18:33