Understanding Trigonometric Substitution

[Astro]

When using trigonometric substitution in calculus you're supposed to always keep in mind the domain of the angle. In the case of √(x²-a²) (where "a" is a number >0) you use x=a⋅arcsec Θ for the substitution.

For trigonometric substitution, textbooks state that the domain of Θ must be {0≤Θ<π/2} ∪ {π/2<Θ≤π} when using the x=a⋅arcsec Θ substitution. Some textbooks alternatively give the allowable domain as {0≤Θ<π/2} ∪ {π≤Θ<3π/2} instead. In either case, there are issues which don't seem to make sense to me.

Please see the attached PDF file for my questions.

Thank you.

 

[Mark44]

When using trigonometric substitution in calculus you're supposed to always keep in mind the domain of the angle. In the case of √(x²-a²) (where "a" is a number >0) you use x=a⋅arcsec Θ for the substitution.

For trigonometric substitution, textbooks state that the domain of Θ must be {0≤Θ<π/2} ∪ {π/2<Θ≤π} when using the x=a⋅arcsec Θ substitution. Some textbooks alternatively give the allowable domain as {0≤Θ<π/2} ∪ {π≤Θ<3π/2} instead. In either case, there are issues which don't seem to make sense to me.

Please see the attached PDF file for my questions.

Thank you.

Question1: The presence of $\sqrt{x^2 - a^2}$ in the integrand leads to the substitution $x = a\sec(\theta)$, as you know. For the square root expression to be real, it must be true that $x^2 \ge a^2$, or equivalently, $x \ge a$ or $x \le -a$.
In the unit circle, with x representing the hypotenuse of a reference triangle, and a representing the horizontal base, a can be positive or negative, but we always have $|a| \le x$. So in the context of the unit circle, I don't see that $x \le -a$ can occur.
Question : The usual domain for the one-to-one cosine function (denoted Cos(x) instead of cos(x)) is $[0, \pi]$. This is essentially the same domain as for the one-to-one secant function (denoted Sec(x)), except that $\pi/2$ is removed, so the two disjoint sets would be the range for the arcsec function. I don't know why they would pick the interval $[\pi, 3\pi/2]$ to be part of the range.

 

[Astro]

Thank you for trying to answer my two questions. Below is my reply.

Question1: The presence of $\sqrt{x^2 - a^2}$ in the integrand leads to the substitution $x = a\sec(\theta)$, as you know. For the square root expression to be real, it must be true that $x^2 \ge a^2$, or equivalently, $x \ge a$ or $x \le -a$.
In the unit circle, with x representing the hypotenuse of a reference triangle, and a representing the horizontal base, a can be positive or negative, but we always have $|a| \le x$. So in the context of the unit circle, I don't see that $x \le -a$ can occur.

You say that in the context of the unit circle x ≤ -a cannot occur and I agree. However, since that's true, then why does the textbook say that

"If x ≤ -a , then π/2 < Θ = sec^-1x/a = arccos a/x ≤ π, and tan Θ ≤ 0." ?

(See the PDF I originaly posted for a scan of that part of the textbook and a drawn depiction of the unit circle).

Question : The usual domain for the one-to-one cosine function (denoted Cos(x) instead of cos(x)) is $[0, \pi]$. This is essentially the same domain as for the one-to-one secant function (denoted Sec(x)), except that $\pi/2$ is removed, so the two disjoint sets would be the range for the arcsec function. I don't know why they would pick the interval $[\pi, 3\pi/2]$ to be part of the range.

Does anyone else know why?

 

[Mark44]

You say that in the context of the unit circle x ≤ -a cannot occur and I agree. However, since that's true, then why does the textbook say that

"If x ≤ -a , then π/2 < Θ = sec^-1x/a = arccos a/x ≤ π, and tan Θ ≤ 0." ?

(See the PDF I originaly posted for a scan of that part of the textbook and a drawn depiction of the unit circle).

In the PDF they don't refer to the unit circle, and are just looking at the algebra of the expression $\sqrt{x^2 - a^2} = a\sqrt{\sec^2(\theta) - 1}$. The expression on the right is real if $\sec(\theta) \ge 1$ or if $\sec(\theta) \le -1$. The first inequality is satisfied for $\theta \in [0, \pi/2)$, and the second is satisfied for $\theta \in (\pi/2, \pi]$, meaning that $\theta$ is in the second quadrant, which is where $\tan(\theta) \le 0$.

 

[Mark44]

Question : The usual domain for the one-to-one cosine function (denoted Cos(x) instead of cos(x)) is $[0, \pi]$. This is essentially the same domain as for the one-to-one secant function (denoted Sec(x)), except that $\pi/2$ is removed, so the two disjoint sets would be the range for the arcsec function. I don't know why they would pick the interval $[\pi, 3\pi/2]$ to be part of the range.

Does anyone else know why?

I think the simplest answer is that someone writing a textbook is free to pick whatever intervals they want, as long as they choose them in a way that makes the Cosine function one-to-one. As I said before, the usual domain for this function is $[0, \pi]$, but someone could also opt for $[0, \pi/2) \cup [\pi, 3\pi/2)$.

 

[2nafish117]

Question1: The presence of $\sqrt{x^2 - a^2}$ in the integrand leads to the substitution $x = a\sec(\theta)$, as you know. For the square root expression to be real, it must be true that $x^2 \ge a^2$, or equivalently, $x \ge a$ or $x \le -a$.
In the unit circle, with x representing the hypotenuse of a reference triangle, and a representing the horizontal base, a can be positive or negative, but we always have $|a| \le x$. So in the context of the unit circle, I don't see that $x \le -a$ can occur.
Question : The usual domain for the one-to-one cosine function (denoted Cos(x) instead of cos(x)) is $[0, \pi]$. This is essentially the same domain as for the one-to-one secant function (denoted Sec(x)), except that $\pi/2$ is removed, so the two disjoint sets would be the range for the arcsec function. I don't know why they would pick the interval $[\pi, 3\pi/2]$ to be part of the range.

When using trigonometric substitution in calculus you're supposed to always keep in mind the domain of the angle. In the case of √(x²-a²) (where "a" is a number >0) you use x=a⋅arcsec Θ for the substitution.

For trigonometric substitution, textbooks state that the domain of Θ must be {0≤Θ<π/2} ∪ {π/2<Θ≤π} when using the x=a⋅arcsec Θ substitution. Some textbooks alternatively give the allowable domain as {0≤Θ<π/2} ∪ {π≤Θ<3π/2} instead. In either case, there are issues which don't seem to make sense to me.

Please see the attached PDF file for my questions.

Thank you.

the unit circle diagram has the radius as x .why is that?? shouldn't it have radius one
and shouldn't the adjacent be(in your pdf diagram) a/x.
then secθ=1/(a/x)=x/a??
and concerning your second question ,choosing any interval is ok as longas the function remains bijective for inverse to exist
for ex cotangent can be defined as follows
(π,2π)
even though it is usually taken as(0,pi)

 

[Mark44]

the unit circle diagram has the radius as x .why is that?? shouldn't it have radius one
and shouldn't the adjacent be(in your pdf diagram) a/x.
then secθ=1/(a/x)=x/a??

My comment was in reference to the diagram in the PDF. The circle is identified as a unit circle, so the radius should be 1 (not x) as you point out.

and concerning your second question ,choosing any interval is ok as longas the function remains bijective for inverse to exist
for ex cotangent can be defined as follows
(π,2π)
even though it is usually taken as(0,pi)

 

[Astro]

Regarding my first question:

the unit circle diagram has the radius as x .why is that?? shouldn't it have radius one
and shouldn't the adjacent be(in your pdf diagram) a/x.
then secθ=1/(a/x)=x/a??
and concerning your second question ,choosing any interval is ok as longas the function remains bijective for inverse to exist
for ex cotangent can be defined as follows
(π,2π)
even though it is usually taken as(0,pi)

You're right; I shouldn't have said unit circle.

The reason the radius in the circle is x and not 1 is because:
--- By definition: cos θ=ajd/hyp and sec θ = 1/cos θ. Since cos θ=ajd/hyp, therefore sec θ=hyp/ajd.
--- Also, by definition, θ=arcsec hyp/adj
--- In the question x=a secθ and therefore rearranging we get θ=x/a . Let's assume the triangle is in the 2nd quadrant. Therefore, the adjacent side (in this case "a") is less than 1.

That being said, Mark44 said:

In the PDF they don't refer to the unit circle, and are just looking at the algebra of the expression $\sqrt{x^2 - a^2} = a\sqrt{\sec^2(\theta) - 1}$. The expression on the right is real if $\sec(\theta) \ge 1$ or if $\sec(\theta) \le -1$. The first inequality is satisfied for $\theta \in [0, \pi/2)$, and the second is satisfied for $\theta \in (\pi/2, \pi]$, meaning that $\theta$ is in the second quadrant, which is where $\tan(\theta) \le 0$.

This makes sense; I understand that. However, I'm still having trouble understanding this geometrically with the context of a trigonometric circle (regardless of whether it's a unit circle or not). After all, as soon as we use trig substitution, don't we have to consider the variables geometrically as well, even though initially there isn't any need? (See attached JPEG image.)

---------------------
Regarding my second question:

That makes sense about the function being one-to-one.

 

[Mark44]

In your latest drawing, in case 2, there are several things that I think are incorrect.
1. Your angle $\theta$ is not in the right place. You show it as an acute angle in your reference triangle. It should instead be the angle from the positive x-axis to the ray that extends out to the circle. Keep in mind that $\theta$ is in the second quadrant, so $\pi/2 < \theta < \pi$.
2. In case 2, you say that $x \le -a$. This will never happen once you make x the length of the hypotenuse of the right triangle, which is by convention nonnegative.
3. In that same picture, you show the horizontal leg being -a. It should be a, where a is itself negative. Then $\theta = arcsec(x/a)$. Since x > 0 and a < 0, x/a < 0 and arcsec(x/a) results in an angle in the 2nd quadrant.

 

[Astro]

In your latest drawing, in case 2, there are several things that I think are incorrect.
1. Your angle $\theta$ is not in the right place. You show it as an acute angle in your reference triangle. It should instead be the angle from the positive x-axis to the ray that extends out to the circle. Keep in mind that $\theta$ is in the second quadrant, so $\pi/2 < \theta < \pi$.

Ah, yes, I see that now.

2. In case 2, you say that $x \le -a$. This will never happen once you make x the length of the hypotenuse of the right triangle, which is by convention nonnegative.

Actually, this is what the textbook says (see Page 1 of the attachment in my first post for the scan of that part of the textbook). This is why I'm confused and the gist of my Question #1. The textbook says "If x ≤ -a , then π/2 < Θ = sec-1x/a = arccos a/x ≤ π, and tan Θ ≤ 0." I agree that as the hypotenuse, x cannot be less than 0 but then why is the textbook allowed to say this? Although x≥a and x≤-a makes sense in the context of √(a²-x²), as soon as we use trig substitution, don't we have to consider the variables geometrically as well, even though initially there isn't any need?

3. In that same picture, you show the horizontal leg being -a. It should be a, where a is itself negative. Then $\theta = arcsec(x/a)$. Since x > 0 and a < 0, x/a < 0 and arcsec(x/a) results in an angle in the 2nd quadrant.

You say that "a is itself negative". I'm guessing this is by convention in trigonometry? I think I understand but I find this a bit confusing since the x-axis in the 2nd quadrant is -ve.

 

[Mark44]

Actually, this is what the textbook says (see Page 1 of the attachment in my first post for the scan of that part of the textbook). This is why I'm confused and the gist of my Question #1. The textbook says "If x ≤ -a , then π/2 < Θ = sec-1x/a = arccos a/x ≤ π, and tan Θ ≤ 0." I agree that as the hypotenuse, x cannot be less than 0 but then why is the textbook allowed to say this?

As I said earlier, I believe they're looking at this relationship purely from an algebra perspective. Once you bring in the geometry, you're placing additional constraints on any variables.

Although x≥a and x≤-a makes sense in the context of √(a²-x²), as soon as we use trig substitution, don't we have to consider the variables geometrically as well, even though initially there isn't any need?

I don't think so.

You say that "a is itself negative". I'm guessing this is by convention in trigonometry?

No. You don't make a variable negative by merely tacking a minus sign on its front. For example, if a < 0, then -a > 0.

I think I understand but I find this a bit confusing since the x-axis in the 2nd quadrant is -ve.

 

[Astro]

Ok, so like you said, when the textbook talks about x≤-a, it's only looking at this relationship purely from an algebra perspective. Fair enough; that makes sense.

Perhaps I should rephrase my question:

Although the textbook is only looking at the x≤-a relationship purely from an algebraic perspective, in the same sentence, the textbook also says that when x≤-a then π/2<θ≤π. If the textbook is only looking at the values of "x" and "a" algebraically, then how does the textbook arrive to the conclusion that when x≤-a, that θ is in Quadrant #2? In other words, why would it mix up two concepts in the same sentence?

 

[Mark44]

Going back to your original substitution, you have $\sec(\theta) = \frac x a$. Assuming that x > 0, if a > 0, $\theta$ is in the first or fourth quadrant. If a < 0 (and x is still pos.), $\theta$ will be in the second or third quadrant. For the secant function to be one-to-one, we need to restrict the angle to, say, just the first and second quadrants.