proof of derivative of arcsin(x)

**Stonehambey** · May 26th 2009, 12:45 PM

Hi guys

I have this problem I'm a little stuck on. It's a 5 part question which leads you through a proof of $\frac{d}{dx}\left( \sin^{-1} x\right)$

In the previous two parts of the question, I was asked to show that

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

and that

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

which I did. It's the final part which I'm stuck on. It says the following

"Use your answers to the above two parts to prove that"

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

I thought it would be simple enough, I proceeded as follows

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

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

...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.

(oh, I can already derive the result using other methods, it's this particular question which is bugging me)

Thanks

**TheAbstractionist** · May 26th 2009, 01:36 PM

Hi Stonehambey.

Originally Posted by Stonehambey

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

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

...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.

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

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

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

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

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

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

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

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