Can't Get Integral To Come Out Right

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In summary: Then the integral becomes∫ h^3 dt / (h^2 \cosh^2 t )^(3/2) = ∫ h dt / (h \cosh t )^3 = h/2 ∫ dt / \cosh^3 tCan you finish it from here?In summary, the original integral can be solved using integration by parts, with a substitution of u = r^2 + h^2 and dv = r(r^2 + h^2)^(-3/2) dr. However, another approach is to split the integrand into two fractions and use a substitution of r = h sinh(t). This leads to the integral ∫ h dt / (h cos
  • #1
Meadman23
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Homework Statement



What is ∫r^3/(r^2 + h^2) ^ (3/2)

Homework Equations



uv - ∫v du

The Attempt at a Solution



Using Maxima gives me (r^2 + 2h^2) / sqrt(r^2 + h^2), but when I solve it using integration by parts, I get (3r^2 + 4h^2) / sqrt (r^2 + h^2).

I do:

dv = r(r^2 + h^2) ^ (-3/2)
v = -1/sqrt(r^2 + h^2)

u = r^2
du = 2r dr

Then I plug it into the int. by parts formula.

uv - ∫v du = -r^2/ sqrt (r^2 + h^2) - ∫ - 2r dr / (r^2 + h^2)

And then I integrate once more and I'm left with:

-r^2 / sqrt (r^2 + h^2) + 4*sqrt (r^2 + h^2)

I know that this should work out almost exactly like example 7 from this page, but it just won't. What's going on??
 
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  • #2
Try a straight u substitution instead of parts. Write your numerator as ##r^2\cdot r\, dr## and let ##u=r^2+h^2##.
 
  • #3
Doesn't seem to work.

Letting u = r ^2 + h^2 takes care of r dr but not r^2.

I even tried setting r^2 = u - h^2 , but that doesn't get me anywhere either.
 
  • #4
Meadman23 said:
...

I even tried setting r2 = u - h2 , but that doesn't get me anywhere either.
That should (and does) work !

Break the integrand into the sum/difference of two fractions.
 
  • #5
Meadman23 said:
Doesn't seem to work.

Letting u = r ^2 + h^2 takes care of r dr but not r^2.

Sure it does. Put ##u - h^2## in for that ##r^2##.
 
  • #6
Okay, thanks guys. I split the fraction up into two.

∫u/(2*u^(3/2)) - ∫h^2/(2*u^(3/2)

I wasn't able to get it to come out right until I considered the h^2 as a constant and factored it out along with the 1/2.

1/2∫u/(u^(3/2)) - h^2/2 ∫1/(2*u^(3/2)

Why can I consider h^2 as a constant here? Also, why did integration by parts provide the wrong answer for me??
 
  • #7
Meadman23 said:
Okay, thanks guys. I split the fraction up into two.

∫u/(2*u^(3/2)) - ∫h^2/(2*u^(3/2)

I wasn't able to get it to come out right until I considered the h^2 as a constant and factored it out along with the 1/2.

1/2∫u/(u^(3/2)) - h^2/2 ∫1/(2*u^(3/2)

Why can I consider h^2 as a constant here? Also, why did integration by parts provide the wrong answer for me??

Well, h is not a function of r, is it ?
 
  • #8
I mean, h^2 = u - r^2 which means to me that the value of h^2 is dependent on the value of r. How could it not be?
 
  • #9
Meadman23 said:
I mean, h^2 = u - r^2 which means to me that the value of h^2 is dependent on the value of r. How could it not be?

No you're confusing yourself. u and r^2 are both functions of r, but their difference is not.

Consider this trivial example. Let [itex]y= 1 - x^2[/itex]. Now y is a function of x, obviously. But x^2 is also a function of x.

So if we rearrange this to get [itex]1 = y - x^2[/itex], does this mean that "1" is a function of x? No, certainly not.
 
Last edited:
  • #10
I might be reaching here but let's see.

I have:

u = h^2 + r^2

and

r^2

u - r^2 = (h^2 + r^2) - r^2 = h^2

Thus, r^2 vanishes and h^2 is no longer of a function of r. Thus, I can factor it out of my integral.

Am I seeing this right?
 
  • #11
Hmm, let [itex] r = h \sinh t [/itex] as a change of variable.
 

1. Why is it important to get the integral to come out right?

The integral is an essential mathematical concept used in many fields of science, including physics, engineering, and economics. It allows us to calculate the area under a curve, which is crucial for solving real-world problems and making accurate predictions.

2. What are some common mistakes that can lead to incorrect integrals?

Some common mistakes include forgetting to apply the correct integration rules, using the wrong limits of integration, and making algebraic errors while simplifying the integrand. It's also crucial to carefully check the signs and constants in the integrand to avoid errors.

3. How can I improve my ability to solve integrals?

Practice is key when it comes to solving integrals. The more you work on different types of integrals, the more familiar you will become with the various techniques and strategies. It's also helpful to review and understand the fundamental concepts of integration, such as the power rule and substitution.

4. Are there any useful tools or resources for solving integrals?

Yes, there are various helpful tools and resources available for solving integrals. Many online calculators can quickly solve integrals and show the step-by-step process. Additionally, there are many textbooks, online tutorials, and practice problems that can assist in improving your understanding and skills in solving integrals.

5. Can integrals be solved analytically or only numerically?

Integrals can be solved both analytically and numerically. Analytical solutions involve using integration rules and techniques to find an exact solution, while numerical solutions involve using numerical methods to approximate the solution. In most cases, it's preferable to find an analytical solution, but sometimes numerical methods are necessary, especially for more complex integrals.

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