Justifying this substitution:

has a solution for only when and .

The domain of the integrand is and .

does not exist only when . But since this range is undefined in the integrand to begin with, the substitution is valid.

So,

Substituting, we get

Not sure if that second last step was valid though.

(edit: oops forgot to sub back in)

And so

and by considering the appropriate right angled triangle, . Since that's always positive, the sign function on the left can be eliminated.

So we have

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