Calc II homework - substitution of definite and indefinite integrals

In summary, the student is stumped on how to solve an equation involving inverse sin and integration by parts. They were able to solve the equation using the notation for inverse sin and pi/6. However, the answer they got was not the right answer.
  • #1
Capella Riddle
5
0
It's been a year since I took Calc I, and I'm taking Calc II online this semester. This is technically a review problem from Calc I, and I managed the other seven, but I can't figure out how to solve this problem.

1.a Homework Statement

∫(a*sin(14x))/([itex]\sqrt{1-196x^2}[/itex] dx, evaluated at x=0 (lower limit) and x=1/28 (upper limit), where "a" is a constant.

2.a Relevant equations
I've defined u as 14x, and dx=du/14; thus u(0)=0 and u(1/28)=14


3.a The attempt at a solution
Using this definition of u, I've been able to get the problem to this point:

[itex]\frac{1}{14}[/itex]∫a*sin(u)*(1-u^2)^(-1/2), evaluated at an upper limit of 1/2 and a lower limit of 0

From here, I'm stumped. Am I using the wrong definition of u, or is there something I'm missing?

Thanks for any help,
Ell
 
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  • #2
The obvious thing to try is integration by parts, differentiating the sin and integrating the surd. That gives an integral of the form ∫cos(θ)arcsin(θ)dθ, but there I get stuck. I thought the trick would be to repeat the process and end up with an equation like
(nasty definite integral) = [some function] + c*(same nasty definite integral)
but although arcsin can be integrated it just seems to get worse.
 
  • #3
I've been stuck on it all afternoon. It comes out nice and simple if I assume sin(14x) is a typo that was instead supposed to be sin[itex]^{-1}[/itex](14x). The integral solves to [itex]\frac{225a}{7}[/itex]. That isn't, however, the right answer. :confused:
-Ell
 
  • #4
What is the supposed right answer? I think you are likely correct that there is a typo somewhere. I don't think that function has an elementary antiderivative. Maple gives a numerical answer 0.009367189922a. Is that the "right" answer?
 
  • #5
The homework is posted in an online program that tell's me when the answer's wrong, but not what the right answer is. When I plug it in my calculator (not absolutely positive I'm doing so correctly), I get 1.6702*10[itex]^{-4}[/itex]. The program wouldn't take that -with an "a" attached- as the correct answer either. As we haven't yet learned integration by parts and the homework set is supposed to be over substitution, I'm not sure haruspex is headed in the right direction.
-Ell
 
  • #6
Capella Riddle said:
I've been stuck on it all afternoon. It comes out nice and simple if I assume sin(14x) is a typo that was instead supposed to be sin[itex]^{-1}[/itex](14x). The integral solves to [itex]\frac{225a}{7}[/itex]. That isn't, however, the right answer. :confused:
-Ell
The typo is an interesting suggestion, but if that's right I would think it should be 'asin' rather than 'a*sin'. So it's ∫00.5asin(u)√(1-u2)du/14. For that, I don't get 225/7, so please post your working.
 
  • #7
Oh. I've never seen asin as a notation for inverse sin before, so I assumed "a" was used as a constant, as it was further down in the problem set. Here's how I worked it:
u=asin(14x)
du=14(1-196x[itex]^{2}[/itex])[itex]^{\frac{-1}{2}}[/itex]
u(1/28)=30
u(0)=0

∫[itex]^{30}_{0}[/itex]u([itex]\frac{du}{14}[/itex]) = [itex]\frac{1}{14}[/itex]∫[itex]^{30}_{0}[/itex]u du = ([itex]\frac{1}{14}[/itex])([itex]\frac{u^{2}}{2}[/itex])|[itex]^{30}_{0}[/itex] = [itex]\frac{30^{2}}{28}[/itex] - [itex]\frac{0^{2}}{28}[/itex] = [itex]\frac{900}{28}[/itex] = [itex]\frac{225}{7}[/itex]
 
  • #8
Capella Riddle said:
Oh. I've never seen asin as a notation for inverse sin before, so I assumed "a" was used as a constant, as it was further down in the problem set. Here's how I worked it:
u=asin(14x)
du=14(1-196x[itex]^{2}[/itex])[itex]^{\frac{-1}{2}}[/itex]
u(1/28)=30
u(0)=0

∫[itex]^{30}_{0}[/itex]u([itex]\frac{du}{14}[/itex]) = [itex]\frac{1}{14}[/itex]∫[itex]^{30}_{0}[/itex]u du = ([itex]\frac{1}{14}[/itex])([itex]\frac{u^{2}}{2}[/itex])|[itex]^{30}_{0}[/itex] = [itex]\frac{30^{2}}{28}[/itex] - [itex]\frac{0^{2}}{28}[/itex] = [itex]\frac{900}{28}[/itex] = [itex]\frac{225}{7}[/itex]

u(1/28) is asin(1/2) which is pi/6. Not 30. How did you get that?? I've seen asin instead of arcsin before. The free computer algebra system "maxima" uses it. Which is what I use.
 
Last edited:
  • #9
Which comes out to [itex]\frac{pi^{2}}{1008}[/itex]... Yes! That's the right answer! And right before the midnight deadline, too :biggrin: You guys are lifesavers;I had my calculator on degrees instead of radians, and now I know to look for the "asin" notation.
Thanks again, and ya'll rock!
-Ell
 

Related to Calc II homework - substitution of definite and indefinite integrals

1. What is substitution in the context of definite and indefinite integrals?

Substitution is a technique used to solve integrals by replacing a complex expression with a simpler one. This is done by substituting a variable with another expression that makes the integral easier to solve.

2. How do I know when to use substitution in solving integrals?

You can use substitution when you have an integral with a complicated expression, especially when it involves trigonometric functions or exponentials. Look for patterns and try to identify if a substitution can simplify the integral.

3. What is the difference between definite and indefinite integrals in relation to substitution?

A definite integral has specific limits of integration and will result in a numerical value, while an indefinite integral does not have limits and will result in a function with a constant of integration. Substitution can be used to solve both types of integrals.

4. Can substitution be used to solve integrals with multiple variables?

No, substitution can only be used for integrals with a single variable. If an integral has multiple variables, other techniques such as partial fraction decomposition or integration by parts may be used.

5. Are there any tips for choosing the appropriate substitution for an integral?

It is helpful to choose a substitution that will eliminate a complex term in the integral or make the integral simpler to solve. This can be done by looking for patterns, using trigonometric identities, or trying different substitutions until one works.

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