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$iii)$ 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

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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 $iv)$.

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  • $\begingroup$ I am getting a negative sign at the end. Can you please check that. $\endgroup$ – mathnoob123 Aug 4 '16 at 18:00
  • $\begingroup$ $$ \begin{align} 2\int_0^t\sqrt{1-x^2}\:dx&-t\sqrt{1-t^2}=-\int_0^t\frac{1}{\sqrt{1-x^2}}\:dx \end{align} $$ $$ \begin{align} \sin^{-1}t=-\int_0^t\frac{1}{\sqrt{1-x^2}}\:dx \end{align} $$ $\endgroup$ – mathnoob123 Aug 4 '16 at 18:08
  • $\begingroup$ You can even see from your example.$$ \begin{align} sin^{-1}=-sin^{-1} -2\int_0^t\frac{1}{\sqrt{}}\ \end{align} $$$$ \begin{align} 2sin^{-1}=-2\int_0^t\frac{1}{\sqrt{}}\ \end{align} $$$$ \begin{align} sin^{-1}=-\int_0^t\frac{1}{\sqrt{}}\ \end{align} $$ $\endgroup$ – mathnoob123 Aug 4 '16 at 18:24
  • $\begingroup$ But this statement is true. So why are we getting a different answer$$ \begin{align} sin^{-1}=\int_0^t\frac{1}{\sqrt{1-x^2}}\ \end{align} $$ $\endgroup$ – mathnoob123 Aug 4 '16 at 18:33
  • $\begingroup$ @FaiqRaees A typo in my answer. Edited. Deeply sorry. $\endgroup$ – Olivier Oloa Aug 4 '16 at 18:34

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