Expert: Paul Klarreich - 10/15/2007

Question
QUESTION: Thank you in advance Mr Klarreich,
   I am going over a table of integrals, and trying to see how some of the formulas were derived so that I might remember them. Two which I am currently stumped on are the integral for 1/(sin (x)) which equals 1/2 ln|{(cos x)-1}/{(cos x) 1}| and the integral for 1/(sin^m x) which comes to -1/(m-1){cos x/(sin^(m-1) x)}  (m 2)/(m-1)*integral of 1/(cos^(m-2) x).
I am sorry if these are hard to read, but without the knowledge of how to include superscripts, I have to use ^ instead. The book I am working out of says integration by parts should help with this, but I am not seeing it. Thanks for your help,
Doug

ANSWER: Questioner:   Doug
Category:  Calculus
Private:  Yes
 
Subject:  Table of Integral Proofs
Question:  Thank you in advance Mr Klarreich,
  I am going over a table of integrals, and trying to see how some of the formulas were derived so that I might remember them. Two which I am currently stumped on are the integral for 1/(sin (x)) which equals 1/2 ln|{(cos x)-1}/{(cos x) 1}| and the integral for 1/(sin^m x) which comes to -1/(m-1){cos x/(sin^(m-1) x)}  (m 2)/(m-1)*integral of 1/(cos^(m-2) x).
I am sorry if these are hard to read, but without the knowledge of how to include superscripts, I have to use ^ instead. The book I am working out of says integration by parts should help with this, but I am not seeing it. Thanks for your help,
Doug

----------------------------------------
Hi, Doug,

First a request:

I prefer not to answer questions marked 'Private.'  That does not make sense for math questions.  I can't think of any good reason to make a math problem private, and I can think of some good reasons to make it public:

1. Other people may have the same trouble and it is much easier and quicker for them to find the solution among the archived answers than to have to send it in and wait.

2. I might get it wrong.  Others might notice and help me correct it.  Then you get a better answer.  

You may believe you are the only one having trouble with a problem, but that is not so.  If it's hard for you, it's hard for other people too.
........................................
 1
----- = csc x
sin x

{
| csc x dx =
}

Use this 'trick', normally applied to  sec x, but it works here, too.  Multiply top and bottom by  csc x + cot x:

{  csc x(csc x + cot x) dx
|  -----------------------
}    (csc x + cot x)

{  csc^2(x) + csc x cot x
|  ----------------------- dx
}      csc x + cot x

Now the numerator is exactly -1 times the derivative of the denominator.  Let  

u = csc x + cot x
du = -(csc^2(x) + csc x cot x) dx

I think you can handle the rest.
......................................
For your second example, the standard table of integrals :

http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions#Integr

gives this reduction formula:

{    dx
| -------- =
} sin^m(x)

 1        cos x         m - 2  {      1
------ ------------- +  ------- | ------------- dx
1 - m   sin^(m-1)(x)     m - 1  }  sin^(m-2)(x)

Also, the INTEGRATOR (not the name of an action movie, please) found at:

http://integrals.wolfram.com/index.jsp

tells us that

{    dx          - cos x
| -------- = - cot x = -------
} sin^2(x)          sin x

which is easily checked.

I think those two clues might give us the hint about using I.B.P.  Here's the basic clue about IBP:

Find one part (factor) that is easy to integrate.  Make it your dv.  The other part, which is easy to differentiate -- practically everything is -- becomes your u.


Since we now know how to integrate  1/sin^2(x), we can use that as the dv part.

{    dx
| -------- = the ORIGINAL INTEGRAL, to be called OI.
} sin^m(x)

{         dx
| ---------------------- =
} sin^2(x) sin^(m-2)(x)


Now let  u =  1/sin^(m-2)(x) = sin^(2-m)(x)

du = (2-m) sin^(1-m)(x) (cos x) dx
du = (2-m) cos x sin^(1-m)(x) dx

and  dv = dx/sin^2(x),  v = - cos x/sin x

So IBP gives us a 'first term' of uv:

- cos x     1
------- ------------ =  UV
 sin x sin^(m-2)(x)

- cos x  
--------------   =  UV
sin^(m-1)(x)

And the IBP gives us a 'second term' of  - | vdu    <<'Integral of vdu'
{  - cos x
|- ------- (2-m) cos x sin^(1-m)(x) dx
}    sin x

{  cos x (2-m) cos x
| --------------------   dx
}  sin x sin^(m-1)(x)

{  (2-m) cos^2(x)
| ----------------  dx
}    sin^(m)(x)

We're getting there.  [Yes, I remember the dentist saying that as he yanked and pulled on my tooth.]

Now we use an old IBP trick.  We arrange for the Original Integral to reappear on the right side, but hit with a coefficient.  We shall be able to solve algebraically for the O.I. in terms of the rest of the stuff.

{  (2-m)(1 - sin^2(x))
| --------------------  dx
}    sin^(m)(x)

Separate into two terms.  I have to type them separately -- sorry about that.


{      (2-m)
| ---------------  dx     << first term
}    sin^(m)(x)

    {    1
(2-m)| ----------  dx  = (2-m) ORIGINAL INTEGRAL
    } sin^(m)(x)

{  (2-m)(- sin^2(x))
| -------------------  dx  << second term
}    sin^(m)(x)

{  (m - 2)sin^2(x)
| -------------------  dx  << still second term
}    sin^(m)(x)

{     (m - 2)
| -------------------  dx  << still second term
}    sin^(m-2)(x)

     {         1
(m-2) | -------------------  dx  
     }    sin^(m-2)(x)

= (m-2) REDUCED INTEGRAL

So what do we have?  I shall leave it to you to put it all on one (large) sheet of paper and do the algebra:


OI = UV + (m-2)RI + (m-2)OI

and all you have to do is solve for OI:

OI - (m-2)OI = UV + (m-2)RI

(1 - m)OI  = UV + (m-2)RI
       UV    m - 2
OI =  ----- + ------ RI
     1 - m   1 - m

which you can easily compare to the formula.
.......................................
Whew!  

But once again, PLEASE don't mark questions private.  Most of the time I just reject them, but this seemed of interest, so I did it.  It would be good if you could resubmit it and then I can publish the answer.


---------- FOLLOW-UP ----------

QUESTION: Good morning Mr Klarreich,
  I understood perfectly all of the work you showed here. The second answer was very thorough; thank you. At the end of the first answer, I tried working it out, and feel like I am really close. I used substitution, and ended up getting -ln|csc (x)   cot (x)|. I noticed what is inside equals 1 cos(x)/sin (x) which equals 1-cos^2x/{sin (x)(1-cos(x))}.
From here, I just seem to be going in circles. I don't see how we get the formula 1/2ln|{(cos x)-1}/{(cos x)  1}|. Could you please show how this is finished off? Thanks,
Doug

Answer
Questioner:   Doug
Category:  Calculus
Private:  No
 
I am going over a table of integrals, and trying to see how some of the formulas were derived so that I might remember them. Two which I am currently stumped on are the integral for 1/(sin (x)) which equals 1/2 ln|{(cos x)-1}/{(cos x) 1}| and the integral for 1/(sin^m x) which comes to -1/(m-1){cos x/(sin^(m-1) x)}  (m 2)/(m-1)*integral of 1/(cos^(m-2) x).
I am sorry if these are hard to read, but without the knowledge of how to include superscripts, I have to use ^ instead. The book I am working out of says integration by parts should help with this, but I am not seeing it. Thanks for your help,
Doug

----------------------------------------
Hi, Doug,
1
----- = csc x
sin x

{
| csc x dx =
}

Use this 'trick', normally applied to  sec x, but it works here, too.  Multiply top and bottom by  csc x + cot x:

{  csc x(csc x + cot x) dx
|  -----------------------
}    (csc x + cot x)

{  csc^2(x) + csc x cot x
|  ----------------------- dx
}      csc x + cot x

Now the numerator is exactly -1 times the derivative of the denominator.  Let  

u = csc x + cot x
du = -(csc^2(x) + csc x cot x) dx

I think you can handle the rest.

---------- FOLLOW-UP ----------

QUESTION: Good morning Mr Klarreich,
 I understood perfectly all of the work you showed here. The second answer was very thorough; thank you. At the end of the first answer, I tried working it out, and feel like I am really close. I used substitution, and ended up getting -ln|csc (x)   cot (x)|. I noticed what is inside equals 1 cos(x)/sin (x) which equals 1-cos^2x/{sin (x)(1-cos(x))}.
From here, I just seem to be going in circles. I don't see how we get the formula 1/2ln|{(cos x)-1}/{(cos x)  1}|. Could you please show how this is finished off? Thanks,
Doug
 
----------------------------------------
Hi, Doug,

Bad news about integrals -- many of them exist in more than one form, especially the ones with trig and log functions.  So you want the answer to match:

=======================================================
WARNING: USE COURIER FONT TO VIEW THIS
===================================================

1     cos x - 1
--- ln ---------
2     cos x + 1

That does not seem right.  cos x - 1 < 0 for most values of x, so the argument to ln would be negative, which is not allowed.  Perhaps it was supposed to be:

1     1 - cos x
--- ln ---------
2     1 + cos x

THIS WE CAN DO.

Here it is again:
1
----- = csc x
sin x

{
| csc x dx =
}

Use this 'trick', normally applied to  sec x, but it works here, too.  Multiply top and bottom by  csc x + cot x:

{  csc x(csc x + cot x) dx
|  -----------------------
}    (csc x + cot x)

{  csc^2(x) + csc x cot x
|  ----------------------- dx
}      csc x + cot x

Now the numerator is exactly -1 times the derivative of the denominator.  Let  

u = csc x + cot x
du = -(csc^2(x) + csc x cot x) dx

Now you have:

{  -du
| ------ = - ln u =
}   u

- ln (csc x + cot x) =
        1
ln [--------------]
   csc x + cot x

         1
ln [---------------------]
   1/sin x + cos x/sin x

      sin x
ln [-------------]
    1 + cos x

      sin x(1 - cos x)
ln [------------------------]
    (1 + cos x)(1 - cos x)

    1 - cos x
ln [-----------]
      sin x

Now, since we have proved that:

 sin x      1 - cos x
---------- = ----------
1 + cos x      sin x

go back to our integral:

      sin x
ln [-------------]
    1 + cos x

         (sin x)^2
1/2 ln [-------------]
       (1 + cos x)^2

         sin x        sin x
1/2 ln [----------- -----------]
        1 + cos x   1 + cos x

         sin x      1 - cos x
1/2 ln [----------- -----------]
        1 + cos x     sin x

        1 - cos x
1/2 ln [-----------]
        1 + cos x