Finding ∫ cscx dx: Tips and Tricks

In summary: CIn summary, to find the integral of csc x dx, one can use the trick of converting it to the integral of 1/sinx or sinx/(1-cos^2x) and then using partial fractions to solve. The final answer is (1/2)ln |(cos x - 1)/(cos x + 1)| + C.
  • #1
PrudensOptimus
641
0
How do you find ∫ cscx dx?
 
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  • #2
Look it up in an integral table? {ln(tan(x/2))}
 
  • #3
By seeing the trick used to find ∫ 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)
 
  • #4
∫ csc x dx = ∫ csc x * (csc x + cot x)/(csc x + cot x) dx
=∫ ((csc x)^2 + csc x cot x)/(csc x + cot x) dx
=-ln | csc x + cot x | + constant


hope it answers your question...
 
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  • #5
Now, that's just scary! :smile:

Here's how I would do it: csc(x)= 1/sin(x) so this is the same as ∫ 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
∫sin(x)dx/sin2= ∫ sin(x)dx/(1- cos2(x)).
Now let u= cos(x) so du= sin(x)dx and the integeral becomes
∫ 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)∫du/(1-u)+ (1/2)∫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(√((1+sin(x))/(1-sin(x)))+ C
Hmmm, maybe Reihhard-t's solution isn't so scary after all!
 
  • #6
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 ∫(csc x + cot x)/(csc x + cot x) ...and then just carry two more steps,and done...

that's math all about
 
  • #7
OK how about treating ∫ csc x dx as ∫ 1/sinx dx?

Here's what I had to far, I had 2 approaches

∫ 1/sinx dx = ∫ (1/sinx)*cosx/cosx dx

= ∫ (1/u)(1/cosx) du = ∫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:

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

= ∫ 1/(u^2-1) du

Then I stop there... don't know how to continue. Or is it continuable??
 
  • #8
You can use partial fractions to continue that, once you learn them.
 
  • #9
HallsofIvy lost a negative u= cos(x) so du= -sin(x)dx
 
  • #10
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?
 
  • #11
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
 
  • #12
OK here's 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?
 
  • #13
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.
 
  • #14
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)|
 

1. What is the definition of the CSCx function?

The CSCx function, also known as the cosecant function, is defined as the reciprocal of the sine function. It is represented as csc(x) and is equal to 1/sin(x).

2. How do you solve integrals involving CSCx dx?

To solve integrals involving CSCx dx, we use the integration by parts method. This involves choosing one part of the integral to differentiate and the other part to integrate. We can then use the formula: ∫u·dv = u·v - ∫v·du to solve the integral.

3. Can you provide an example of solving an integral with CSCx dx?

Yes, an example of solving an integral with CSCx dx is ∫csc(x) dx. Using the integration by parts method, we can let u = csc(x) and dv = dx. This gives us du = -csc(x)·cot(x) dx and v = x. Plugging these into the formula, we get ∫csc(x) dx = csc(x)·x + ∫cot(x) dx. We can then use the formula for the integral of cot(x) to get the final solution of csc(x)·x + ln|sin(x)| + C.

4. What is the importance of solving integrals with CSCx dx in science?

Solving integrals with CSCx dx is important in many areas of science, particularly in physics and engineering. It allows us to find the area under a curve and calculate important values such as work, velocity, and acceleration. These calculations are essential in understanding and predicting the behavior of systems in the natural world.

5. Are there any special cases when solving integrals with CSCx dx?

Yes, there are special cases when solving integrals with CSCx dx. One special case is when the integral is in the form of ∫csc(x)·cot(x) dx. In this case, we can use the substitution u = csc(x) to simplify the integral to ∫u du, which is easier to solve. Another special case is when the integral involves a power of CSCx, such as ∫csc^2(x) dx. In this case, we can use the trigonometric identity 1 + cot^2(x) = csc^2(x) to rewrite the integral and make it easier to solve.

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