Dear Uncle Colin,

I tried to work out $\int \frac{1}{\sqrt{3 - 4x - 4x^2}}\dx$ going via the complex numbers and end up with a different answer compared to the ‘proper’ way – I get an $\arcosh$ instead of an $\arcsin$. Can you sort out this witchcraft?

Am Genuinely Not Expecting Strange Integrals

Hi, AGNESI, and thanks for your message!

Let’s run through this both ways and see where the problem lies.

The “proper” way

That’s an awkward quadratic inside the square root, but it turns out that $3 - 4x - 4x^2 \equiv 2^2 - (2x+1)^2$.

Then applying the rule $\int \frac{1}{\sqrt{a^2 - u^2}} du = \arcsin\br{\frac{u}{a}} + C$, you get $\frac{1}{2} \arcsin \br{x + \frac{1}{2}} + C$.

So far so tasty.

The complex way

Factoring -1 out of the bottom turns the integral into $\frac{1}{i} \int \frac{1}{\sqrt{4x^2 + 4x-3}} \dx$ – and the quadratic in the square root is now $(2x+1)^2 - 2^2$.

Using the rule $\int \frac{1}{\sqrt{u^2 - a^2}} = \arcosh\br{\frac{u}{a}}$, the integral gives $\frac{1}{2i} \arcosh \br{x + \frac{1}{2}} + c$.

Those don’t look the same! What gives?

Are they really different?

I’m going to try to work out under what circumstances these two answers would be the same.

Let the first result be $I_1$ and the second $I_2$.

Rearranging the first gives $\sin\br{2I_1 - C} = x + \frac{1}{2}$.

Rearranging the second gives $\cosh\br{2i I_2 - ic} = x + \frac{1}{2}$.

However, $\cosh(ti) \equiv \cos(t)$, so this second left-hand side is $\cos\br{2I_2 - c}$.

However, $\cos\br{t-\piby2} \equiv \sin(t)$, so if I let $c = C + \piby 2$, the second left-hand side is $\sin\br{2I_2 - C}$ – which is the same as the first answer!

So, $I_1$ and $I_2$ differ by a constant – so they’re both valid solutions to the integral!

(I’d go with the arcsine one, myself. Don’t want to be unleashing forces beyond our control, now, do we?)

Hope that helps!

- Uncle Colin