Evaluating $\int \sqrt{x^2-3}\:dx$ I need to solve:
$$\int \sqrt{x^2-3} \, dx.$$
So I use the substitution:
$$x=\frac{\sqrt 3}{\cos(t)}$$
$$dx= \frac{\sqrt 3 \sin(t) \, dt}{\cos^2(t)} $$
and I get 
$$3\int \frac{\sqrt{\frac1 {\cos^2(t)}-1}\cdot\sin(t) \, dt}{\cos^2(t)}. $$
So I get 
$$3\int \frac{\sin(t)\tan(x)\,dx}{\cos^2(x)},$$
that is $(\sin(t)\sin(t)/\cos(t))/\cos^2(t)$ and finally I have 
$$3\int \frac{\sin^2(t)\,dt}{\cos^3(t)}$$
Substitute $\sin(t)=s$, so $\cos(t) \, dt=ds$. The integral becomes
$$3 \int \frac{s^2\,ds}{(1-s^2)^2};$$
$$dt= \frac{ds}{\cos t}$$
So problem is I don't know how do I get from this $\sqrt{(1/\cos^2(t))-1} = \tan(t);$
It was stupid question idk how I didn't saw that nvm, after that use partial integration $u=s$, $du=ds$, $v=1/(1-s^2)$ and i get $s/2(1-s^2)-1/2$ integral of $ds/(1-s^2)$
and the solution is $3/2(s/1-s^2-1/2\ln(1+s/1-s)$
 A: Note that $$\cos^2 \theta + \sin^2 \theta = 1$$ so dividing both sides by $\cos^2 \theta$ gives $$1 + \tan^2 \theta = \frac{1}{\cos^2 \theta} \iff \frac{1}{\cos^2 \theta} - 1 = \tan^2 \theta$$
And so, taking the square root of both sides gives you what you want. 
A: Hint. As you have suggested, the change of variable
$$
x=\frac{\sqrt{3}}{\cos t},\quad dx=\frac{\sqrt{3}\sin t}{\cos^2 t}\:dt,
$$
gives
$$
\begin{align}
\int \sqrt{x^2-3}\:dx&=\sqrt{3}\int\sqrt{\frac3{\cos^2 t}-3}\:\cdot \frac{\sin t}{\cos^2 t}\:dt
\\\\&=3\int\sqrt{\frac{1-\cos^2 t}{\cos^2 t}}\:\cdot \frac{\sin t}{\cos^2 t}\:dt
\\\\&=3\int \frac{\sin^2 t}{\cos^3 t}\:dt
\\\\&=3\int \frac{\sin^2 t\: \cos t}{(1-\sin^2 t)^2}\:dt
\\\\&=3\int \frac{s^2}{(1-s^2)^2}\:ds
\end{align}
$$ can you take it from here?
A: $3\int \frac {\sin^2(t)}{\cos^3(t)} dt$
Rather than a substitution I suggest you do this.
$3\int \tan^2 t \sec t dt$
integration by parts:
$u = \tan t, dv = \sec t \tan t dt\\
du = sec^2 t dt, v = sec t$
$3\sec t \tan t - 3\int \sec^3t dt\\
3\sec t \tan t - 3\int (\tan^2 t - 1)\sec t dt\\
3\sec t \tan t + 3 \int \sec t dt - 3\int \tan^2 t \sec t dt\\
3\sec t \tan t + 3 \ln |\sec t + \tan t| - 3\int \tan^2 t \sec t dt$
And we have come full circle:
$3\int \tan^2 t \sec t dt = 3\sec t \tan t + 3 \ln |\sec t + \tan t| - 3\int \tan^2 t \sec t dt\\
3\int \tan^2 t \sec t dt = I\\
I = 3\sec t \tan t + 3 \ln |\sec t + \tan t| - I\\
2 I = 3\sec t \tan t + 3 \ln |\sec t + \tan t|\\
I = \frac 32 \sec t \tan t + \frac 32 \ln |\sec t + \tan t|+C$
Alternatively, If you want to stick with your initial approach.
$\int \frac {s^2}{(1-s^2)^2} ds$
Find a partial fraction decomposition:
$\frac {s^2}{(1-s^2)^2} = \frac A{1-s} +\frac B{1+s} + \frac C{(1-s)^2} + \frac D{(1-s)^2}$
And that will integrate to
$-A\ln (1-s) + B\ln(1+s) + C(1-s)^{-1} - D(1+s)^{-1} + c$ (sorry do double use the same letter)
Hopefully, the two answers are equivalent.
A: First rescale
$$\int \sqrt{x^2-3} \, dx=3\int \sqrt{t^2-1} \, dt.$$
Then by parts,
$$\int \sqrt{t^2-1} \, dt=t\sqrt{t^2-1}-\int\frac{t^2}{\sqrt{t^2-1}}dt=t\sqrt{t^2-1}-t-\int\frac{dt}{\sqrt{t^2-1}}.$$
In the last integral, you should recognize the derivative of $\text{arcosh }t$.
