# Is indefinite integral allowed if integrand has no real numbers in its domain

I recently came across the integral:

$$\int \sqrt{(x-3)}\Big[\arcsin(\ln x) + \arccos(\ln x)\Big]dx.$$

It is easy to compute the integral without observing domain of integrand and answer comes out as $$\frac{\pi}{3(x-3)^{(3/2)}} + C$$ using $$\arcsin(t) + \arccos(t)=\pi/2$$.

So is it allowed to integrate like this or can I conclude that the function does not have any indefinite integral. Kindly help.

• How are you defining integration with a function that has no domain? – Peter Foreman Apr 23 '20 at 10:14
• usually for complex integrals especially in indefinate integrals we usually do not consider domain of integrand and apply some substitutions or properties which is what i did here. – Ginger bread Apr 23 '20 at 10:34
• I don't know who "we" refers to but it is not possible for any integral, definite or otherwise, to be performed without knowing that the integrand has a non-empty domain. – Peter Foreman Apr 23 '20 at 14:02

If $$f$$ is defined at no point, then $$\int f(x)\;dx$$ cannot be computed. But:
If $$f$$ has complex values, such as $$f(x) := \sqrt{(x-3)}\Big[\arcsin(\ln x) + \arccos(\ln x)\Big]$$ then $$\int f(x)\;dx$$ can often be computed. Of course you have to verify that $$\arcsin(\ln x) + \arccos(\ln x) = \pi/2$$ even for cases where the $$\arcsin$$ and $$\arccos$$ are non-real (by looking up the definitions in that case).