Question

∫xcsc^2(2x-1) dx please integrate . thank you very much :)

Answer 1

\[\int f'(u)f(u)du\]

Answer 2

prolly shoulda used more of a chainrule looking notation shouldnta i

Answer 3

and its a first thought anyways, usually off by a mile

Answer 4

Integration by parts

Answer 5

helps if you know that the anti derivative of cosecant squared is -cotangent

Answer 6

its all about integration of trigonometric functions

Answer 7

first thought was good lol

Answer 8

∫xcsc^2(2x-1) dx
=(-1/2 )∫x d[cot(2x-1)]
=(-1/2 ) [x cot(2x-1) - ∫cot(2x-1)dx
= (-1/2 ) [x cot(2x-1)] - (-1/2 ) ∫cot(2x-1)dx
= (-1/2 ) [x cot(2x-1)] +(1/2 ) ∫cos(2x-1) / sin(2x+1)dx
= (-1/2 ) [x cot(2x-1)] + (1/2)(1/2)∫ 1/sin(2x+1) d (sin(2x+1)
= (-1/2 ) [x cot(2x-1)] + (1/4) ln|sin(2x+1)| +C

Answer 9

thats doesnt look quite right

Answer 10

x csc^2(2x-1) comes from something similar to -cot(2x-1)

Answer 11

im reading the x wrong ....

Answer 12

♫♫ ...waking up is hard to do

Answer 13

Was i incorrect?

Answer 14

http://www.wolframalpha.com/input/?i=integrate+xcsc%5E2%282x-2%29

callisto is right in that this thing is ugly :/

Answer 15

you sure it aint 2x^2 - 1 ??

Answer 16

what do you mean?

Answer 17

I mean, is the problem typed correctly? because what we have as a result seems a bit messy to me ....

Answer 18

not very messy..

Answer 19

im curious about you changing csc^2 to a d[....]; ive never considered that before

Answer 20

x d[cot(2x-1)]

u = x ; du = dx

v = ...... ; dv = d/dx cot(2x-1)

im not quite sure how that would make a transition

Answer 21

v = -1/2 cot(2x-1) ; dv = csc^2(2x-1) dx

so it look like a fine manuever

Answer 22

hmm.. just put it inside [i meant the d(x)] by integrating it.. Students here just do it like that..

d/dx [cot(2x-1)] = 2csc^2 (2x-1)
therefore ∫csc^2 (2x-1) dx = (1/2)cot(2x-1) +C
and ∫xcsc^2 (2x-1) dx = (1/2)∫x d[cot(2x-1)]
eh.. i don't really know how to explain..