Understanding the Behavior of Trigonometric Functions on Specific Intervals

[Miike012]

I have a question regarding what I highlighted in the paint doc.

How can they assume that the function is increasing?

As a matter of fact its not increasing on the entire interval from 0 to pi.

 

[Dick]

I have a question regarding what I highlighted in the paint doc.

How can they assume that the function is increasing?

As a matter of fact its not increasing on the entire interval from 0 to pi.

Well, clearly it isn't increasing. It isn't even continuous on [0,pi] if a=1/9. It is increasing where it is continuous as the derivative shows. There has got to be more context in the problem than you have shown. Where did you find this problem?

 

[Miike012]

Well, clearly it isn't increasing. It isn't even continuous on [0,pi] if a=1/9. It is increasing where it is continuous as the derivative shows. There has got to be more context in the problem than you have shown. Where did you find this problem?

The paint document contains the entire paragraph.

 

[haruspex]

The attempt to integrate through pi/2 this way looks doubtful. I would break the integration range at pi/4 etc.
$\int_{\frac{\pi}4}^{\frac{\pi}2}\frac{\sec^2(\theta).d\theta}{a^2+ \tan ^2(\theta)} =$ $\int_0^{\frac{\pi}4}\frac{cosec^2(\theta).d\theta}{a^2+\cot^2(\theta)} = \int_0^{\frac{\pi}4}\frac{\sec^2(\theta).d\theta}{a^2\tan^2(\theta)+1}$

 

[Dick]

The attempt to integrate through pi/2 this way looks doubtful. I would break the integration range at pi/4 etc.
$\int_{\frac{\pi}4}^{\frac{\pi}2}\frac{\sec^2(\theta).d\theta}{a^2+ \tan ^2(\theta)} =$ $\int_0^{\frac{\pi}4}\frac{cosec^2(\theta).d\theta}{a^2+\cot^2(\theta)} = \int_0^{\frac{\pi}4}\frac{\sec^2(\theta).d\theta}{a^2\tan^2(\theta)+1}$

I'm really not sure what the text Miike012 posted is getting at. But I would hope it's trying to say that if you are using the antiderivative arctan(tan(x)/a), you need to break the integration range at each point where that function is discontinuous. And arctan(tan(x)/a) is not the same thing as arctan(tan(x/a)) which is a problem with Miike012's original post.

 

[haruspex]

I'm really not sure what the text Miike012 posted is getting at. But I would hope it's trying to say that if you are using the antiderivative arctan(tan(x)/a), you need to break the integration range at each point where that function is discontinuous.

Agreed, but it didn't look to me as though the line taken in the attachment was helpful. Hence my proposed alternative.

And arctan(tan(x)/a) is not the same thing as arctan(tan(x/a)) which is a problem with Miike012's original post.

I don't see anywhere that arctan(tan(x/a)) is mentioned.

 

[Dick]

Agreed, but it didn't look to me as though the line taken in the attachment was helpful. Hence my proposed alternative.

I don't see anywhere that arctan(tan(x/a)) is mentioned.

I was trying to guess the sense of the question from the graphs posted. Since arctan(tan(x/9)) was graphed, I thought that might have something to do with what the OP thought the antiderivative was. It should have been arctan(tan(x)/9).

 

[Miike012]

I don't care about anything else other than the statement made in the attachment which was..."Therefore arctan(tan(x)/a) must increase continuously with x." This statement can be found at the bottom of the attachment of post number 3.

Which was my original question in post number 1.

 

[Dick]

I don't care about anything else other than the statement made in the attachment which was..."Therefore arctan(tan(x)/a) must increase continuously with x." This statement can be found at the bottom of the attachment of post number 3.

Which was my original question in post number 1.

Ok, fine. It doesn't. It increases from 0 to pi/2 if a is positive, then decreases discontinuously, then increases again from pi/2 to pi.