Having trouble with an integral

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In summary, the conversation discusses the process of solving the integral of √(x^2 - t^2)dt/(xt) x > t > 0. The speaker applies trig substitution and rearranges the formula to obtain the integral ∫cos(θ)^2 dθ/sinθ, but is unsure of how to proceed. They then mention the possibility of using a different variable, z, and discuss the options for the multiple choice question, which include (x^2 - t^2)^(3/2) / (3t) + C, t ( sqrt(x^2 - t^2) + t tan(t/sqrt(x^2 - t^2))^-1 + C, - ((
  • #1
Temp0
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Homework Statement



∫√(x^2 - t^2)dt/(xt) x > t > 0

Homework Equations


The Attempt at a Solution



So I noticed that the integrand had the form a^2 - b^2x^2, and I can apply trig substitution, so I did this:

t = xsin(θ), dt = xcos(θ), and therefore, x^2 - t^2 = x^2 - x^2sin(θ)^2.
The last formula can be rearranged into x^2 cos(θ)^2 (From the identity 1 - sin(θ)^2)
After simplification, I obtain the integral
∫cos(θ)^2 dθ/sinθ

From here, I don't know where to go. After rearranging multiple times, there is no integral and I keep getting (cosθ) - ∫cscθdθ.

Thank you for any clarity you can provide for me.
 
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  • #2
∫cscx dx = -ln( csc x + cot x)
 
Last edited:
  • #3
Temp0 said:

Homework Statement



∫√(x^2 - t^2)dt/(xt) x > t > 0

Homework Equations





The Attempt at a Solution



So I noticed that the integrand had the form a^2 - b^2x^2, and I can apply trig substitution, so I did this:

t = xsin(θ), dt = xcos(θ), and therefore, x^2 - t^2 = x^2 - x^2sin(θ)^2.
The last formula can be rearranged into x^2 cos(θ)^2 (From the identity 1 - sin(θ)^2)
After simplification, I obtain the integral
∫cos(θ)^2 dθ/sinθ

From here, I don't know where to go. After rearranging multiple times, there is no integral and I keep getting (cosθ) - ∫cscθdθ.

Thank you for any clarity you can provide for me.

I think you should get (cosθ) + ∫cscθdθ. And ∫cscθdθ can be done. You can look it up if nothing else.
 
  • #4
That's actually where I'm stuck, because after the cosecant integral, this is for a multiple choice question and there is no choice for the - ln(csc x + cot x), so I was just wondering if there was any other way to do this.
 
  • #5
Temp0 said:
That's actually where I'm stuck, because after the cosecant integral, this is for a multiple choice question and there is no choice for the - ln(csc x + cot x), so I was just wondering if there was any other way to do this.

What are your choices? You aren't done when you express it in terms of ##\theta##. You probably want to express it in terms of t.
 
Last edited:
  • #6
Yup, but none of the choices have the term "ln" in it, and there's also a "none of the above" option, but I'm always nervous about picking that type of option.

I have:
(x^2 - t^2)^(3/2) / (3t) + C
t ( sqrt(x^2 - t^2) + t tan(t/sqrt(x^2 - t^2))^-1 + C
- ((t-x)(t+x))^(3/2) / (3t) + C
and (x^2 - t^2) + t tan(t/ sqrt(x^2-t^2))^-1 / t + C
 
  • #7
Temp0 said:

Homework Statement



∫√(x^2 - t^2)dt/(xt) x > t > 0
I interpreted that as ##\int_0^x\frac{\sqrt{x^2-t^2}}{xt}dt##, but seeing the choice of answers it must mean ##\int^t\frac{\sqrt{x^2-z^2}}{xz}dz##
after the cosecant integral, this is for a multiple choice question and there is no choice for the - ln(csc x + cot x), so I was just wondering if there was any other way to do this.
But that's a different x. You mean - ln(csc θ + cot θ) where t = x sin θ.
 
  • #8
Yeah I just have a habit of using x as my variable, but where did the z come from? I also don't know why they provided the x > t > 0, I assumed that they provided that information so that I would know the denominator wouldn't be 0, but does it have another meaning?
 
  • #9
haruspex said:
I interpreted that as ##\int_0^x\frac{\sqrt{x^2-t^2}}{xt}dt##, but seeing the choice of answers it must mean ##\int^t\frac{\sqrt{x^2-z^2}}{xz}dz##

But that's a different x. You mean - ln(csc θ + cot θ) where t = x sin θ.

I don't think x > t > 0 is supposed to be a range of integration. The answers are indefinite integrals. It's just to insure the square root is defined and there no division by zero.

Temp0 said:
Yup, but none of the choices have the term "ln" in it, and there's also a "none of the above" option, but I'm always nervous about picking that type of option.

I have:
(x^2 - t^2)^(3/2) / (3t) + C
t ( sqrt(x^2 - t^2) + t tan(t/sqrt(x^2 - t^2))^-1 + C
- ((t-x)(t+x))^(3/2) / (3t) + C
and (x^2 - t^2) + t tan(t/ sqrt(x^2-t^2))^-1 / t + C

You could try differentiating those to see if they give you anything like you want. But I think you are pretty safe with the "none of the above".
 
  • #10
Dick said:
I don't think x > t > 0 is supposed to be a range of integration. The answers are indefinite integrals. It's just to insure the square root is defined and there no division by zero.
Isn't that what I said? It looked like a range of integration until I saw the choice of answers.
Temp0 said:
where did the z come from?
##\int^t f(t).dt## is a "pun". The t in the upper bound is different from the other two t's, which are 'dummy variables'. In the present problem it seemed to me that could be a source of confusion, so I tried to make it clearer by using a different dummy variable.
 
  • #11
haruspex said:
Isn't that what I said? It looked like a range of integration until I saw the choice of answers.

##\int^t f(t).dt## is a "pun". The t in the upper bound is different from the other two t's, which are 'dummy variables'. In the present problem it seemed to me that could be a source of confusion, so I tried to make it clearer by using a different dummy variable.

I guess I'm not sure what you are trying to say. Given there is a +C in the answers it's just a plain old indefinite integral without limits. ##\int \frac{\sqrt{x^2-t^2}}{xt}dt##, isn't it? I've never seen the notation with only an upper limit.
 
  • #12
Dick said:
I guess I'm not sure what you are trying to say. Given there is a +C in the answers it's just a plain old indefinite integral without limits. ##\int \frac{\sqrt{x^2-t^2}}{xt}dt##, isn't it? I've never seen the notation with only an upper limit.

All of the proposed answers have t in them. I don't know how to get that in there without putting a t in a bound. When someone writes ∫cos(x).dx = sin(x), as they do, what is really meant is ∫x cos(t).dt = sin(x). It's still an indefinite integral because there is no lower bound specified.
Re the offered answers, I suspect t ( sqrt(x^2 - t^2) + t tan(t/sqrt(x^2 - t^2))^-1 + C is to be read as ##t (\sqrt{x^2 - t^2} + t \arctan(\frac t{\sqrt{x^2 - t^2}}) + C##, and similarly the fourth choice. (The parentheses don't match up, so I'm still not sure how to interpret it.) I mention this because arctan and ln are connected: arctan(x) = ln((1-x)/(1+x))/(2i), or something like that.
 
  • #13
haruspex said:
All of the proposed answers have t in them. I don't know how to get that in there without putting a t in a bound. When someone writes ∫cos(x).dx = sin(x), as they do, what is really meant is ∫x cos(t).dt = sin(x). It's still an indefinite integral because there is no lower bound specified.
Re the offered answers, I suspect t ( sqrt(x^2 - t^2) + t tan(t/sqrt(x^2 - t^2))^-1 + C is to be read as ##t (\sqrt{x^2 - t^2} + t \arctan(\frac t{\sqrt{x^2 - t^2}}) + C##, and similarly the fourth choice. (The parentheses don't match up, so I'm still not sure how to interpret it.) I mention this because arctan and ln are connected: arctan(x) = ln((1-x)/(1+x))/(2i), or something like that.

Well, I would say ##\int x \cos(t) dt = x sin(t) +C##, and that has a ##t## in it. Sure, there's a relation between arctan and log, but I don't see any i's around.
 
  • #14
Dick said:
Well, I would say ##\int x \cos(t) dt = x sin(t) +C##, and that has a ##t## in it.
That only works if you interpret ##\int x \cos(t) dt## as meaning ##\int^t x \cos(t) dt##, and that in turn is exactly the same as ##\int^t x \cos(u) du## (since the t in the bound does not mean the same as the t in the integrand), which is shorthand for ##\int^{u=t} x \cos(u) du##.
Sure, there's a relation between arctan and log, but I don't see any i's around.
Agreed - just noting that the absence of a ln function in the offered solutions is not quite conclusive.
 
  • #15
haruspex said:
That only works if you interpret ##\int x \cos(t) dt## as meaning ##\int^t x \cos(t) dt##, and that in turn is exactly the same as ##\int^t x \cos(u) du## (since the t in the bound does not mean the same as the t in the integrand), which is shorthand for ##\int^{u=t} x \cos(u) du##.

Agreed - just noting that the absence of a ln function in the offered solutions is not quite conclusive.

Yeah, but they don't look right. I'd check by differentiating, but as you've noted some of them may not even be transcribed right. Hardly seems worth it. On the other hand notation like ##\int x \cos(t) dt = x sin(t) +C## is what I've seen in integral tables all my life with no need for special comment or special notation.
 

FAQ: Having trouble with an integral

What is an integral?

An integral is a mathematical concept that represents the area under a curve in a graph. It is used to find the total value of a quantity that is changing over a given interval.

Why do I need to use integrals?

Integrals are used to solve problems in many areas of science, including physics, engineering, and economics. They allow us to find the total value of continuously changing quantities and make predictions about their behavior.

What are the different types of integrals?

There are two main types of integrals: definite and indefinite. Definite integrals have specific limits of integration and give a single numerical value as the result. Indefinite integrals have no limits and give a function as the result, which can then be evaluated at different points.

How do I solve an integral?

Solving an integral involves finding an antiderivative, which is the reverse process of differentiation. It is often done using integration techniques, such as substitution, integration by parts, or trigonometric identities.

What are common mistakes to avoid when working with integrals?

Some common mistakes when working with integrals include forgetting to add the constant of integration, using incorrect limits of integration, and making algebraic errors. It is also important to check for any discontinuities or special cases that may require different integration techniques.

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