Which comes out to \frac{pi^{2}}{1008}... 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
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...
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...
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. :confused:
-Ell
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...