Solving $\int \frac{1}{\sqrt{x^2 - c}} dx$ I want to solve $$\int \frac{1}{\sqrt{x^2 - c}} dx\quad\quad\text{c is a constant}$$
How do I do this?
It looks like it is close to being an $\operatorname{arcsin}$?

I would have thought I could just do:
$$\int \left(\sqrt{x^2 - c}\right)^{-\frac12}\, dx=\frac{2\sqrt{c+x^2}}{2x}\text{????}$$
But apparently not. 
 A: To answer this question. You need to use a technique called Trigonometric Substitution (of course there are other ways, but this way is the easiest in my opinion).
First draw a right triangle. Call the height $\sqrt{x^2-c}$, the base $\sqrt{c}$, and the hypotenuse $x$. Call the angle between the base and the hypotenuse $\theta$. We know that $$\sec(\theta) = \frac{x}{\sqrt{c}}$$
$$\sqrt{c}\sec(\theta)  = x$$
$$dx = \sqrt{c}\sec(\theta)\tan(\theta)d\theta$$
So we substitute this into the integral, and the integral simplifies to $$\begin{align} 
\int \frac{dx}{\sqrt{x^2-c}} &= \int \frac{\sqrt{c}\sec(\theta)\tan(\theta)} {\sqrt{(\sqrt{c}\sec(\theta))^2-c}}d\theta \\
&= \int \frac{\sqrt{c}\sec(\theta)\tan(\theta)}{\sqrt{c}\tan(\theta)}d\theta \\
&= \int \sec(\theta)d\theta \\
&= \ln|\sec(\theta)+\tan(\theta)| + C \\
&= \ln\left|\frac{x}{\sqrt{c}} + \frac{\sqrt{x^2-c}}{\sqrt{c}} \right| + C
\end{align}$$
A: HINT:
$$\frac{1}{\sqrt{x^2-c}} = \frac{1}{\sqrt{c}\sqrt{\frac{x^2}{c} - 1}} = \frac{1}{\sqrt{c}}\cdot \frac{1}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2 - 1}}$$
A: The trick is to use trig substitution:
HINT:
Let $x = \sqrt{c}  \sec u \implies x^2 = c \sec^2u \implies x^2 - c = c ( \sec^2 u - 1 ) =  c \tan^2 u$
and
$$ dx  = \sqrt{c} \sec u \tan u $$
Hence, your integral looks like
$$ \int \frac{\sqrt{c} \sec u \tan u}{\sqrt{c } \tan u } = ... $$
A: If you try this derivative (which is not helpful in itself),
$$(\sqrt{x^2-1})'=\frac x{\sqrt{x^2-1}},$$
you can notice that by adding an $x$ term,
$$(x+\sqrt{x^2-1})'=1+\frac x{\sqrt{x^2-1}}=\frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}},$$
and
$$\frac{(x+\sqrt{x^2-1})'}{x+\sqrt{x^2-1}}=\frac1{\sqrt{x^2-1}}.$$
From this you can conclude (and adapt for positive or negative $c$).
A: Let us set $c=1$ for a moment.
From hyperbolic trigonometry, we know that $\cosh^2t-1=\sinh^2t$ (equivalent of $1-cos^2=\sin^2t$) and the substitution $x=\cosh t$ seems natural. Then
$$\int\frac{dx}{\sqrt{x^2-1}}=\int\frac{\sinh t\ dt}{\sinh t}=t\text{ (!)}$$
Similarly, set $c=-1$ and use $x=\sinh t$:
$$\int\frac{dx}{\sqrt{x^2+1}}=\int\frac{\cosh t\ dt}{\cosh t}=t\text{ (!)}$$
Scaling $x$ by $1/\sqrt c$ gives you the final solution, one of $\text{arcosh}(x/\sqrt c)$ or  $\text{arsinh}(x/\sqrt{-c})$.
A: $$ \int \frac{1}{\sqrt{x^2-c}}dx = \int \frac{1}{\sqrt{c\left(\frac{x^2}{c}-1\right)}}dx=\frac{1}{\sqrt{c}} \int \frac{1}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2-1}}dx $$
Let $u=\frac{x}{\sqrt{c}}$, then $du=\frac{1}{\sqrt{c}} dx$. So now we have
$$ \int \frac{1}{\sqrt{u^2-1}}du =\mathrm{arcosh}(u)+C = \mathrm{arcosh}\left( \frac{x}{\sqrt{c}}\right)+C $$
