Substitution with sec yields different result?

[Allen_T]

This is an example from another website(http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx):

when evaluating this integral:

we made this substitution:

and got this:

and then the websites say that "since we are doing an indefinite integral we will assume that

will be positive and so we can drop the absolute value bars"

My question is, why is such assumption legal? Thanks!!!

 

[HallsofIvy]

No, it is NOT. What "websites" say that? (You could, of course, add "\(\displaystyle \pm\)" later.)

 

[Deleted member 4993]

This is an example from another website(http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx):

when evaluating this integral:

we made this substitution:

and got this:

and then the websites say that "since we are doing an indefinite integral we will assume that

will be positive and so we can drop the absolute value bars"

My question is, why is such assumption legal? Thanks!!!

As far as I know

\(\displaystyle \sqrt{x^2} \ = \ x\) and NOT = -x

 

[lookagain]

As far as I know

\(\displaystyle \sqrt{x^2} \ = \ x\) and NOT = -x

\(\displaystyle \sqrt{x^2} \ = \ |x| \ \ \ for \ \ all \ \ real \ \ x\)