Indefinite Integral: Justification for Dropping Absolute Value Bars

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In a book while doing an indefinite integral the author first wrote (sec2 x)1/2 = |sec x| , fine , then the author says the following :
"since we are doing an indefinite integral we can drop the absolute value bars" ,
now what is the justification for this ?
 
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No such justification exists at all.
 
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No such justification exists at all.
does this mean that the assumption is wrong?
 
Yes.

Suppose you have an indefinite integral of this sort.

First, you should see that in some regions of the number line, where sec(x)>0, you may drop the absolute value sign, and use sec(x) instead.

In other regions of the number line, you need to use |sec(x)|=-sec(x)

Now, it may well happen that that minus sign will not make the local anti-derivative different there than for the other region, but that is actually something you need to VERIFY, rather than to make it into a presupposition.

Understood?
 
Yes.

Suppose you have an indefinite integral of this sort.

First, you should see that in some regions of the number line, where sec(x)>0, you may drop the absolute value sign, and use sec(x) instead.

In other regions of the number line, you need to use |sec(x)|=-sec(x)

Now, it may well happen that that minus sign will not make the local anti-derivative different there than for the other region, but that is actually something you need to VERIFY, rather than to make it into a presupposition.

i got you that it is not correct to drop the absolute value bars arbitrarily , but rather take 2 different cases for |f(x)|>=0 and <0 , then if we get it same we can say we " drop the absolute value bars " ,
but in this case (\int|sec(x)|) i don't think that we can drop the absolute value bars in any case , or is there any such possible condition ??
 
One would have to see the context. Perhaps the task was to do the integral, so (as stated above) there are two cases, but to evaluate one of the cases is enough because of their similarity to each other.
 
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