Question

find or evaluate the integral csc2xdx

Answer 1

is it -ln absv(cscx + cotx) x^2?

Answer 2

\[\Large\bf\sf (\tan x)'\quad=\quad \sec^2x\]\[\Large\bf\sf (\cot x)'\quad=\quad -\csc^2 x\]

Answer 3

That second derivative identity should help us out, yes? :o

Answer 4

zep zep zep!

Answer 5

It is batman!! :O Protect the streets good sir!

Answer 6

oh but i thought the integral of csc was -ln absv(cscx + cotx)???

Answer 7

Yes, when we don't have a square on it.

Answer 8

there is no square its csc 2x dx

Answer 9

Is your problem csc2x or csc^2x?

Answer 10

oh...

Answer 11

my b!

Answer 12

If you need, you can do a simple u-sub to get things to work out for you.
\[\Large\bf\sf u=2x, \qquad\qquad \frac{1}{2}du=dx\]

\[\Large\bf\sf \int\limits \csc 2x\;dx\quad=\quad \frac{1}{2}\int \csc u\; du \]Now you can apply that rule directly to the csc u

Answer 13

oh so just 1/2 csc (2x)(2)

Answer 14

No you didn't integrate yet :o
Integrate the csc u

Answer 15

confused?

Answer 16

You just posted the integral of csc a minute ago .. :P

Answer 17

OH OOPS! right so take the integral and since its a negative ln can i just move that in front of the 1/2

Answer 18

i.e. -1/2 ln absv (cscx+cotx)

Answer 19

\[\Large\bf\sf =\quad -\frac{1}{2}\ln\left|\csc u+\cot u\right|+c\]Mmmm yah looks good so far :)

Answer 20

We're integrating in u, yes?
So our stuff inside the trig functions should be u's.

Answer 21

True! and then you can substitute the u( 2x) back in?

Answer 22

ya there we go!

Answer 23

and i forget the c is substituted with du so instead of plus c it would be plus 2?

Answer 24

c is substituted with du...? :o

Answer 25

or do you just get rid of the c altogether???

Answer 26

constant?

Answer 27

indefinite integral, so you add +C if that's what you're asking o_O

Answer 28

c stays.


We made a substitution,
Integrated,
(an arbitrary constant of integration shows up)
undo our substitution,

yah the c has nothing to do with changing from x to u.
It's still arbitrary regardless of our variable we're integrating in :o

Answer 29

ok ok! and does the problem have to be simplified now or is it done???

Answer 30

bc it looks rlly messy...

Answer 31

Leave it

Answer 32

Yah it sure looks messy :)
But it can't be written much simpler than that.

That's just how some of these work out.

Answer 33

^^^^

Answer 34

LOVELY! THANKS AGAIN:-)