# Csc x dx

How do you find &int; cscx dx?

mathman
Look it up in an integral table? {ln(tan(x/2))}

Hurkyl
Staff Emeritus
Gold Member
By seeing the trick used to find &int; sec x dx and applying it to this problem.

(Your textbook almost certainly shows how to compute the integral of sec x, and if it doesn't, look up the answer in an integral table and work backwards to discover the trick)

reinhard_t
&int; csc x dx = &int; csc x * (csc x + cot x)/(csc x + cot x) dx
=&int; ((csc x)^2 + csc x cot x)/(csc x + cot x) dx
=-ln | csc x + cot x | + constant

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HallsofIvy
Homework Helper
Now, that's just scary! Here's how I would do it: csc(x)= 1/sin(x) so this is the same as &int; dx/sin(x)- an odd power of x. Back in calculus, I learned that when you have an odd power of either sin(x) or cos(x), you can use one of them with dx and use sin2+ cos2= 1 for the rest: here sin(x) is in the denominator so I would write it as
&int;sin(x)dx/sin2= &int; sin(x)dx/(1- cos2(x)).
Now let u= cos(x) so du= sin(x)dx and the integeral becomes
&int; du/(1-u3). The denominator factors as (1-u)(1+ u) so- partial fractions. 1/(1-u2)= (1/2)/(1-u)+ (1/2)/(1+u).
The integral is now (1/2)&int;du/(1-u)+ (1/2)&int;du/(1+u)= -(1/2)ln|1-u|+ (1/2)ln|1+u|+C= (1/2)ln(|1+u|/|1-u|)+ C= (1/2)ln((1+sin(x))/(x-sin(x)))+ C= ln(&radic;((1+sin(x))/(1-sin(x)))+ C
Hmmm, maybe Reihhard-t's solution isn't so scary after all!

reinhard_t
I do not see the flow of your intergration.

but the most important ,when you do math,you must try the most simple,short, and clear....and get the best result.

what I did actually just multiply with 1 which is &int;(csc x + cot x)/(csc x + cot x) ....and then just carry two more steps,and done...

OK how about treating &int; csc x dx as &int; 1/sinx dx?

&int; 1/sinx dx = &int; (1/sinx)*cosx/cosx dx

= &int; (1/u)(1/cosx) du = &int;1/sqrt(u^2 - u^4) du.

I don't know how to continue from there, or is it even continuable at all?

2nd approach:

&int;(1/sinx) dx = &int;sinx/sin^2x dx = &int;sinx/(1-cos^2x) dx
assuming u = cos x, du = -sinx

= &int; 1/(u^2-1) du

Then I stop there... don't know how to continue. Or is it continuable??

Hurkyl
Staff Emeritus
Gold Member
You can use partial fractions to continue that, once you learn them.

HallsofIvy lost a negative u= cos(x) so du= -sin(x)dx

Originally posted by Hurkyl
You can use partial fractions to continue that, once you learn them.

Can you teach me? Or someone please teach me?

reinhard_t
actually all equation can be integrated,but for you case optimus,your further integration will be very complicated,and frankly speaking ,I don't know how to integrate more,to get exactly like my previous answer.
so I recommend you to implement my previos method.that one is actually a math trick.it is fast and incisive

OK heres what i got after trying partial fractions.

Integral of

1/sinx = sinx/(1-cos^2x) = 1/(u^2-1) du = 2/(u-1) - 2/(u+1)

So the integral is 2*(ln(u-1) - ln(u+1)) + C

Is that correct?

Hurkyl
Staff Emeritus
Gold Member
I think those twos should be one-halves.

Also, don't forget your absolute value signs!

And, finally, you need to back substitute to change the 'u's back into 'x's.

reinhard_t
yeah.you've come with great idea optimus,just slight careless like what is mention by hurkvl.It is both(1/2) instead of 2,and ending it up with variable x together with modulus sign.

it becomes (1/2)ln |(cos x - 1)/(cos x + 1)|