Absolute integrals

[fred2028]

How would you integrate the indefinite integral
\(\displaystyle \[\int {\frac{{\cos x}}{{\left. {\left| {\cos x} \right.} \right|}}} dx\]\)?
I know how to integrate this definitely since you have to break it up into intervals to make sure cos is always positive, but if it's an indefinite integral, my textbook just acts as if the absolute value bars aren't there?

 

[pka]

Go to this website http://www.wolframalpha.com/]
In the input window, type in this exact expression: integrate cos[x]/abs[cos[x]]dx.
(you can copy & paste)
Click the equals bar at the end if the input window.
You will see that there is no standard result.
But the graph may be helpful.

 

[fred2028]

Go to this website http://www.wolframalpha.com/]
In the input window, type in this exact expression: integrate cos[x]/abs[cos[x]]dx.
(you can copy & paste)
Click the equals bar at the end if the input window.
You will see that there is no standard result.
But the graph may be helpful.

So I guess there is no "official" integral for this? The graph seems weird; the it looks similar to sinx but with straight lines ... I can't seem to interpret the graph beyond that lol

 

[pka]

So I guess there is no "official" integral for this? The graph seems weird; the it looks similar to sinx but with straight lines ... I can't seem to interpret the graph beyond that lol

It should not see off.
For \(\displaystyle \frac{-\pi}{2}<x<\frac{\pi}{2}\) we are integrating 1
and between \(\displaystyle \frac{\pi}{2}<x<\frac{3\pi}{2}\) we are integrating -1.

 

[fred2028]

So I guess there is no "official" integral for this? The graph seems weird; the it looks similar to sinx but with straight lines ... I can't seem to interpret the graph beyond that lol

It should not see off.
For \(\displaystyle \frac{-\pi}{2}<x<\frac{\pi}{2}\) we are integrating 1
and between \(\displaystyle \frac{\pi}{2}<x<\frac{3\pi}{2}\) we are integrating -1.

So is the indefinite integral = 0 then, if you keep integrating 1 and -1?

 

[BigGlenntheHeavy]

|cos(x)| = cos(x) if cos(x) ? 0
|cos(x)| = -cos(x) if cos(x) < 0

Hence, if cos(x) ? 0, then \(\displaystyle \int \frac{cos(x)}{|cos(x)|}dx \ = \ \int dx \ = \ x \ + \ C.\)

However if cos(x) < 0, then \(\displaystyle \int \frac{cos(x)}{|cos(x)|} \ = \ -\int dx \ = \ -x \ + \ C.\)


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