Which solution to the integral $\int \frac {dx}{x \sqrt {x^2-1}}$ is correct and why?

Question

I have this integral: $$\int \frac {dx}{x \sqrt {x^2-1}}.$$ I solved it like this: $$ \sqrt {x^2-1} = t \to x = \sqrt {t^2+1} \to $$ $$dx = \frac {t}{\sqrt {t^2+1}}\,dt .$$ Solving for t: $$ \int \frac {dt}{t^2+1} = \arctan(t) + c .$$ In the solution I have, however, it is solved in another way: $$ x = \sec(t), \ \sqrt{x^2-1}=\tan(t), $$ $$dx=\sec(t)\tan(t)\,dt \to $$ $$\int \frac {\sec(t)\tan(t)} {\sec(t)\tan(t)}\,dt= \int dt = t +c= \operatorname{arcsec}(x) + c.$$ Which solution is correct and why?

Answer 1

If you differentiate $\operatorname{arcsec} x$, you get $1/\sqrt{1-x^2}$. So it's not correct.

Trying the first, we get $(\arctan\sqrt{x^2-1})'=1/x^2(1/2)1/(\sqrt{x^2-1})(2x)=1/(x\sqrt{x^2-1})$.

So the first is correct.

I think, in your calculation for the second one, you would get $\sqrt{x^2-1}=|\tan x|$.

Answer 2

Both your answers are correct but you didn't finish the first one, you still have $t$ in it instead of $x$, that would give $\arctan(\sqrt{x^2-1})$.

You can verify that this is equal to $\operatorname{arcsec}(x)$ on $\mathbb R^+$.

On $\mathbb R^-$ you have the sign to change too $\pi-\arctan(...)$

Answer 3

In both of your solutions, you make the assumption that $x>1$. In the first one, this happens when you write $x=\sqrt{t^2+1}$ instead of $x=-\sqrt{t^2+1}$. In the second one, you're implicitly taking $t\in(0,\pi/2)$ which leads you to $\sqrt{x^2-1}=\tan(t)$ and not $\sqrt{x^2-1}=-\tan(t)$. Both of your answers are correct in the interval $x>1$ because $\arctan\left(\sqrt{x^2-1}\right)=\operatorname{arcsec}(x)$ for $x>1$.

So, to consider the case $x<-1$, in the first solution, you instead take $x=-\sqrt{t^2+1}$, and for the second, we need $t\in(\pi/2,\pi)$ which makes $\tan(t)<0$ and $\sqrt{x^2-1}=-\tan(t)$. Because $\operatorname{arcsec}(-x)=\pi-\operatorname{arcsec}(x)$, there is a compact way to write the final asnwer without the need for branches: $$\int\frac{dx}{x\sqrt{x^2-1}}=\operatorname{arcsec}\left(|x|\right)+C $$