Hi **Stonehambey**.

Originally Posted by

**Stonehambey** $\displaystyle \sin^{-1} t = 2\left[ t\sqrt{1-t^2} + \int^{t}_{0}\frac{x^2}{\sqrt{1-x^2}}\,dx \right] - t\sqrt{1-t^2}$

$\displaystyle =2\int^{t}_{0}\frac{x^2}{\sqrt{1-x^2}}\, dx + t\sqrt{1-t^2}$

...and it's here I get stuck. I've looked at it for a while now and can't see how use the above information to derive the required result.

$\displaystyle \int_0^t\frac{x^2}{\sqrt{1-x^2}}\, dx$

$\displaystyle =\ -\int_0^t\frac{-x^2}{\sqrt{1-x^2}}\, dx$

$\displaystyle =\ -\int_0^t\left(\frac{1-x^2}{\sqrt{1-x^2}}-\frac1{\sqrt{1-x^2}}\right)\, dx$

$\displaystyle =\ -\int_0^t\left(\sqrt{1-x^2}-\frac1{\sqrt{1-x^2}}\right)\, dx$

$\displaystyle =\ \int_0^t\frac{dx}{\sqrt{1-x^2}}-\int_0^t\sqrt{1-x^2}\,dx$

$\displaystyle =\ \int_0^t\frac{dx}{\sqrt{1-x^2}}-t\sqrt{1-t^2}-\int_0^t\frac{x^2}{\sqrt{1-x^2}}\, dx$

$\displaystyle \therefore\ 2\int_0^t\frac{x^2}{\sqrt{1-x^2}}\, dx+t\sqrt{1-t^2}\ =\ \int_0^t\frac{dx}{\sqrt{1-x^2}}$