Find $ \int\frac{\mathrm{d}x}{(1-x^2)^{3/2}}$ $$
\int\frac{1}{(1-x^2)^{3/2}}\mathrm{d}x=?
$$
I am aware that you can do it with $x=\sin u$ but how do you do it with $x=\tanh u$? I kept ending with with $x^2+1$ in my denominator instead of the correct $x^2-1$.
 A: By letting $x=\sin(u)$, then we have that $dx=\cos(u) du$, and
$$\int\frac{1}{(1-x^2)^{3/2}} dx=\int\frac{\cos(u)}{\cos^3(u)} du
=\int\frac{du}{\cos^2(u)} =\tan(u)+C=\frac{x}{\sqrt{1-x^2}}+C$$
because $\tan(u)=\sin(u)/\cos(u)$ and $\cos(u)=\sqrt{1-x^2}$.
A: If you set $x=\tanh u$, $\;\mathrm d\mkern1mu x=\dfrac1{\cosh^2u}\,\mathrm d\mkern1mu u$, you get
\begin{align*}\int\frac{1}{(1-x^2)^{3/2}}\mathrm{d}x&=\int\cosh^3u\dfrac1{\cosh^2u}\,\mathrm d\mkern1mu u=\int\cosh u\,\mathrm d\mkern1mu u\\
&=\sinh u =\tanh u\cosh u=\frac{\tanh u}{\sqrt{1-\tanh^2 u}}\\
&=\frac x{\sqrt{1-x^2}}.
\end{align*}
Other method: integration by parts:
We'll integrate $I=\displaystyle\int\dfrac{\mathrm d\mkern1mu x}{\sqrt{1-x^2}}$ by parts. Set
\begin{align*}
u&=\dfrac{1}{\sqrt{1-x^2}},&\mathrm{d}x&=x\\
\text{whence}\qquad\mathrm d\mkern1mu u&=\frac x{(1-x^2)^{3/2}},& v&=x.
\end{align*}
We obtain
\begin{align*}
I&=uv-\int v\,\mathrm d\mkern1mu u=\frac{x}{\sqrt{1-x^2}}-\int\frac{x^2}{(1-x^2)^{3/2}}\,\mathrm d\mkern1mu x\\
&=\frac{x}{\sqrt{1-x^2}}+\int\frac{1-x^2}{(1-x^2)^{3/2}}\,\mathrm d\mkern1mu x
-\int\frac{\mathrm d\mkern1mu x}{(1-x^2)^{3/2}}\\
&=\frac{x}{\sqrt{1-x^2}}+I-\int\frac{\mathrm d\mkern1mu x}{(1-x^2)^{3/2}}
\end{align*}
whence the result.
