# Thread: all in a sign

1. ## all in a sign

Hi folks,

Could someone please tell me if $\dfrac{-1}{\sqrt{a^2 - x^2}} = \dfrac{1}{\sqrt{x^2 - a^2}}$

I think the above is true. The reason I ask is that

$\int \dfrac {- dx}{\sqrt{a^2 - x^2}} = \cos^{-1}\dfrac{x}{a}$

But

$\int \dfrac {dx}{\sqrt{x^2 - a^2}} = \cosh^{-1}\dfrac{x}{a}$

these results are clearly different. This suggests the initial statement is wrong, but I can't see why.

2. ## Re: all in a sign

Surely you see that, except in the special case that $x^2= a^2$, one of $a^2- x^2$ and $x^2- a^2$ is positive and the other negative so one of the square roots is real and the other imaginary. The two cannot possibly be equal!

3. ## Re: all in a sign

No, it's false. $\sqrt{f(X)}$ is always non-negative.

It is common that we can find different ways to solve an integral, but they always differ from each other by only an additive constant. That additive constant can be difficult to see, especially when trigonometry is involved.

That said, your examples are more different than that.

4. ## Re: all in a sign

Originally Posted by s_ingram
Hi folks,

Could someone please tell me if $\dfrac{-1}{\sqrt{a^2 - x^2}} = \dfrac{1}{\sqrt{x^2 - a^2}}$

I think the above is true. The reason I ask is that

$\int \dfrac {- dx}{\sqrt{a^2 - x^2}} = \cos^{-1}\dfrac{x}{a}$

But

$\int \dfrac {dx}{\sqrt{x^2 - a^2}} = \cosh^{-1}\dfrac{x}{a}$

these results are clearly different. This suggests the initial statement is wrong, but I can't see why.
It is true that $\displaystyle \frac{1}{a - x} = \frac{1}{(-1)(x - a)} = -\frac{1}{x - a}$

but in this case we have
$\displaystyle \frac{1}{\sqrt{a^2 - x^2}} = \frac{1}{\sqrt{(-1)(x^2 - a^2)}} = \frac{1}{\sqrt{-1}} \frac{1}{\sqrt{x^2 - a^2}}$

and clearly $\displaystyle - 1 \neq \frac{1}{\sqrt{-1}}$

-Dan

5. ## Re: all in a sign

Thanks Dan. You saw why I made the mistake and what the misunderstanding was. many thanks.