Calc II homework - substitution of definite and indefinite integrals

[Capella Riddle]

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))/(\sqrt{1-196x^2} 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:

\frac{1}{14}∫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

 

[haruspex]

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.

 

[Capella Riddle]

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^{-1}(14x). The integral solves to \frac{225a}{7}. That isn't, however, the right answer.
-Ell

 

[LCKurtz]

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?

 

[Capella Riddle]

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^{-4}. 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

 

[haruspex]

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^{-1}(14x). The integral solves to \frac{225a}{7}. That isn't, however, the right answer.
-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 ∫₀^0.5asin(u)√(1-u²)du/14. For that, I don't get 225/7, so please post your working.

 

[Capella Riddle]

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^{2})^{\frac{-1}{2}}
u(1/28)=30
u(0)=0

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

 

[Dick]

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^{2})^{\frac{-1}{2}}
u(1/28)=30
u(0)=0

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

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.

 

[Capella Riddle]

Which comes out to \frac{pi^{2}}{1008}... Yes! That's the right answer! And right before the midnight deadline, too 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