2006-10-30 21:58:08 · 4 answers · asked by Raj Oberoi 1 in Science & Mathematics ➔ Mathematics
these are 2 different things
d/dx(log cot x) = 1/cotx(d/dx(cotx)) = - cosec ^2 x / cot x
for integral it is difficult
2006-10-31 00:26:10 · answer #1 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
The integral of log(cot(x) ) is not elementary. It
involves the dilogarithm function. To try to
evaluate it, note first of all that log(cot x) = -log (tan x)
Then let u = tan x, x = arctan u, dx = 1/ (u^2 + 1) du.
Now try to carry on from here!
2006-10-31 03:50:11 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
the integral will be (1-tan^2x) / log cotx because it is integral of log cotx or log (U) and the answer is U' / U !!
and the differential will be 1 / (1-tan^2x) .
by the way yo , it was the hardesst question i've ever answered in math !
good luck !
2006-10-31 00:27:08 · answer #3 · answered by arash 3 · 0⤊ 3⤋
â«(log cotx) dx
= ??????
d/dx (log cotx0
= -cosec²x/cotx
2006-10-31 00:32:02 · answer #4 · answered by penta 2 · 0⤊ 1⤋