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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.

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    $\begingroup$ How are you defining integration with a function that has no domain? $\endgroup$ – Peter Foreman Apr 23 '20 at 10:14
  • $\begingroup$ 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. $\endgroup$ – Ginger bread Apr 23 '20 at 10:34
  • $\begingroup$ 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. $\endgroup$ – Peter Foreman Apr 23 '20 at 14:02
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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).

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