$iii)$ 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 $iv)$.