Question

u question for the integral of 1/sqrt(9-16x^2) dx

Answer 1

I got u = 8x
du = 8dx
1/8du = dx

\[1/8 \int\limits \frac{du}{\sqrt{a^2 -u^2}}\]????

Answer 2

1/4*

Answer 3

hmm...

Answer 4

I actually just looked it up on wolfman, seems like I was doing it correctly. Why do I need to complete the square? http://www.wolframalpha.com/input/?i=integral+of+1%2Fsqrt%289-16x%5E2%29+dx

Answer 5

you don't, that's why I deleted my remark

Answer 6

oh :P I was mainly concerned with the du, but I looked it up in the book and it seemd to be the same. I was confused at first about getting the du by itself, but I realized I can send it over to the left of the integral :).

Answer 7

but u should be 4x

Answer 8

yeah I messed up :p

Answer 9

besides that it's good though :)