# How to integrate $x/\sqrt{1-x^2}$?

Could somebody please tell me where I made a mistake? I want to integrate

\begin{equation*} \int_a^b\frac{x}{\sqrt{1-x^2}}dx. \end{equation*}

As far as I know the subsitution $u=1-x^2$ works, but I would like to do it exactly the way we did it in class.

I did the following:

\begin{equation*} \int_a^b\frac{x}{\sqrt{1-x^2}}dx=\frac{1}{2}\int_a^b\frac{2x}{\sqrt{1-x^2}}dx. \end{equation*}

Now let $\phi(x)=x^2$. Therefore

\begin{align*} \int\frac{x}{\sqrt{1-x^2}}dx =\frac{1}{2}\int_a^b\frac{2x}{\sqrt{1-x^2}}dx &=\frac{1}{2}\int_a^b\frac{\phi'(x)}{\sqrt{1-\phi(x)^2}}dx \\ &= \int_{\phi(a)}^{\phi(b)}\frac{1}{\sqrt{1-t^2}}dt \\ &=\arcsin(\phi(b))-\arcsin(\phi(a)). \end{align*}

Thank you very much in advance!

• A better substitution would be $u^2=1-x^2$ and $(-u)\,\mathrm du=x\,\mathrm dx$. – Prasun Biswas May 19 '15 at 13:54
• Or $x= \sin(\theta)$ – user222031 May 19 '15 at 13:55

Your mistake: you wrote $1-\phi(x)^2$ instead of $1-\phi(x)$.
(Which will lead you to the antiderivative $-\sqrt{1-\phi(x)}=-\sqrt{1-x^2}$).
The derivative of $\sqrt{f(x)}$ is $$\frac{f'(x)}{2\sqrt{f(x)}}$$
Have you tried what happens with $f(x)=1-x^2$?