I was looking at the math qts here in answer... and algebra questions always get a lot of answers from the answerers. but very few people answer stuff abt calculus, most of the calc qts have ZERO answers...
can any1 solve this?
anti derivative of (cos 2z - tan 2z) / cos^2 2z dz
2007-11-13 14:23:21 · 8 answers · asked by jara 3 in Science & Mathematics ➔ Mathematics
Dr. D's answer is pretty convinsing but he's wrong...
dy/dz (sec z) is not tan^2 (2z) its secztanz...
(sorry dr.d... im not insulting... but i honestly think it's wrong)
i need more answer...
2007-11-13 14:39:22 · update #1
Dr. D's answer is pretty convinsing but he's wrong...
dy/dz (sec 2z) is not tan^2 (2z) its sec2ztan2z...
(sorry dr.d... im not insulting... but i honestly think it's wrong)
i need more answer...
2007-11-13 14:40:17 · update #2
dr.d was right..
sorry dr.
2007-11-13 15:10:43 · update #3
Write it as ∫ [sec(2z) - tan(2z) sec^2 (2z)] dz
For the second integral, let u = tan(2z)
du = 2 sec^2 (2z) dz
So that becomes ∫ u/2 du = u^2 / 4
= tan^2 (2z) / 4
The first one is a standard integral which evaluates to
1/2 * ln| (cosz + sinz) / (cosz - sinz)|
Answer:
1/2 * ln| (cosz + sinz) / (cosz - sinz)| - tan^2 (2z) / 4 + C
*EDIT*
My answer checks out with any integration software. The only alleged discrepancy is that some software might give sec^2 (2z) instead of tan^2 (2z) . But this is no discrepancy at all since these two terms are separated by a numerical constant.
2007-11-13 14:29:04 · answer #1 · answered by Dr D 7 · 0⤊ 0⤋
I am only looking at questions which have 1 or fewer answers in the first twenty minutes. And if the problem is not well written or well posed I will ignore it...proper English and math language is important.
2007-11-13 14:51:24 · answer #2 · answered by ted s 7 · 0⤊ 0⤋
In my case, I am careful of anything that looks like homework. I am further careful that I can answer the question unless it looks like a stupid joke. That is why I don't answer such questions. It's been too long since I've studied calculus.
2007-11-13 14:35:14 · answer #3 · answered by Jack 7 · 0⤊ 0⤋
A1. ln(y - a million) - ln 2 = x + ln x ln[(y - a million) / 2] = x + ln x ln[(y - a million) / 2] - ln x = x (y - a million)/2 = e^x y = 2e^x + a million A2. a. ln(3x) = 6 b. e^(ln x^2) = 9 c. x ln e = 2.34 d. 4x ln e = a million/2 A3. a. y =(x + 3) /(a million/x) + a million/3 = (x + 3)x + a million/3 = x^2 + x + a million/3
2016-09-29 05:03:35 · answer #4 · answered by ? 4 · 0⤊ 0⤋
I think DR D has your answer. Tomany people didn't pay attention in class and don't want to remember. Sorry.
2007-11-13 14:33:20 · answer #5 · answered by Lisa L 1 · 0⤊ 0⤋
I'm pretty sure it's becasue far fewer people take calculus than take algebra...
2007-11-13 14:37:09 · answer #6 · answered by thermodynaman2004 1 · 1⤊ 0⤋
Why is it that you people who only have 5% best answers are criticizing others for what they do and do not choose to answer?
2007-11-13 14:28:22 · answer #7 · answered by Jeƒƒ Lebowski 6 · 0⤊ 2⤋
cuz calculus iz hard
2007-11-13 14:27:21 · answer #8 · answered by HP 1 · 0⤊ 0⤋