## Friday, February 3, 2012

### Natural log and trig relation: -ln(tan(phi/2)) = ln(csc(phi) + cot(phi)) and the integrals of csc^3(phi) and csc(phi)

$- \ln \left( \tan \dfrac{\phi}{2} \right) = - \ln \left( \dfrac{\sin \phi}{1 + \cos \phi} \right) = \ln \left( \dfrac{1 + \cos \phi}{\sin \phi} \right) = \ln \left( \dfrac{1}{\sin \phi} + \dfrac{\cos \phi}{\sin \phi} \right)$
$= \ln \left( \csc \phi + \cot \phi \right)$

However, I recently was using this in an analysis, was checking some work, and noticed that:

$\ln \left( \tan \dfrac{\phi}{2} \right) = \ln \left( \dfrac{1 - \cos \phi}{\sin \phi} \right) = \ln \left( \csc \phi - \cot \phi \right)$

I looked up the formulas again and became quite confused.

The tangent half angle relations are:

$\tan \dfrac{\phi}{2} = \dfrac{1 - \cos \phi}{\sin \phi}$

and

$\tan \dfrac{\phi}{2} = \dfrac{\sin \phi}{1 + \cos \phi}$

So, I guess the relation is:

$\ln \left( \csc \phi - \cot \phi \right) = -\ln \left( \csc \phi + \cot \phi \right)$

Yup, it is confirmed:

Links to the tangent half angle:

http://en.wikipedia.org/wiki/Tangent_half-angle_formula

http://planetmath.org/encyclopedia/DerivationOfHalfAngleFormulaeForTangent.html

However, my problem originates to the integral of $$\csc^3 \phi$$.  This integral can be solved by (at least how I did it)

$\int \csc^3 \phi \mathrm{d} \phi$

First, utilize integration by parts

$\mathrm{d}v = \csc^2 \phi \mathrm{d} \phi$

$u = \csc \phi$

$v = -\cot \phi$

$\mathrm{d}u = - \csc \phi \cot \phi \mathrm{d} \phi$

The formula for integration by parts is

$uv - \int v \mathrm{d}u$

Substituting in the correct values gives

$-\csc \phi \cot \phi - \int \csc \phi \cot^2 \phi \mathrm{d} \phi$

Using the trigonometric identity $$\cot^2 \phi = \csc^2 \phi - 1$$ transforms the relation to

$-\csc \phi \cot \phi - \int \csc \phi \left( \csc^2 \phi - 1 \right) \mathrm{d} \phi$

Multiplying through we obtain

$-\csc \phi \cot \phi - \int \csc^3 \phi - \csc \phi \mathrm{d} \phi$

Separating the RHS integrals produces

$-\csc \phi \cot \phi - \int \csc^3 \phi \mathrm{d} \phi + \int \csc \phi \mathrm{d} \phi$

Recall that this whole relation is equal to $$\int \csc^3 \phi \mathrm{d} \phi$$ which yields

$\int \csc^3 \phi \mathrm{d} \phi = -\csc \phi \cot \phi - \int \csc^3 \phi \mathrm{d} \phi + \int \csc \phi \mathrm{d} \phi$

At first this seems counter-intuitive and does not seem to help.  However, if you collect the integrals of $$\csc^3 \phi$$ the equation becomes clearer as where to go next.  So collecting the $$\csc^3 \phi$$ integrals to the LHS, we obtain

$2 \int \csc^3 \phi \mathrm{d} \phi = -\csc \phi \cot \phi + \int \csc \phi \mathrm{d} \phi$

Then we can divide by two and evaluate the left hand side which gives us

$\int \csc^3 \phi \mathrm{d} \phi = \dfrac{1}{2} \left[ -\csc \phi \cot \phi + \int \csc \phi \mathrm{d} \phi \right]$

Okay, so this is where my problem began.  I've found two answers for this integral of $$\csc \phi$$.  The first is the same as the one I got by evaluating the integral of $$\csc \phi$$

$\int \csc^3 \phi \mathrm{d} \phi = \dfrac{1}{2} \left[ -\csc \phi \cot \phi - \ln | \csc \phi - \cot \phi | \right]$

where

$\int \csc \phi \mathrm{d} \phi = - \ln | \csc \phi - \cot \phi |$

according to: http://www.math.com/tables/integrals/more/csc.htm

$\int \csc \phi \mathrm{d} \phi = - \ln | \csc \phi + \cot \phi |$

?? (I'll have to check these to make sure there isn't a mistake.

To make matters worse here: http://answers.yahoo.com/question/index?qid=20080623173220AAyRTys we get an answer for $$\int \csc^3 \phi \mathrm{d} \phi$$ equal to

$\dfrac{1}{2} \left[ -\csc \phi \cot \phi + \ln | \csc \phi - \cot \phi | \right]$

??? What the heck is going on?! (Again, when I get the time I will check these answers since the formulation is given.  However, this does give the final answer I am looking for which is

$\dfrac{1}{2} \left \lbrace -\csc \phi \cot \phi + \ln \left[ \tan \left( \dfrac{\phi}{2} \right) \right] \right \rbrace$

However again, the relation I first displayed above shows that

$\dfrac{1}{2} \left[ -\csc \phi \cot \phi - \ln | \csc \phi + \cot \phi | \right]$

$= -\dfrac{1}{2} \left[ \csc \phi \cot \phi + \ln | \csc \phi + \cot \phi | \right]$

$= -\dfrac{1}{2} \left[ \csc \phi \cot \phi - \ln | \csc \phi - \cot \phi | \right]$

$= \dfrac{1}{2} \left[ -\csc \phi \cot \phi + \ln | \csc \phi - \cot \phi | \right]$

So this tells me that the second link should be correct!  The reason for this post originates to me trying to change the angle $$\phi$$ from spherical polar coordinates to cylindrical polar coordinates $$\left( r, z \right)$$ through the relation $$\phi = \tan \left( r/z \right)$$and $$R^2 = r^2 + z^2$$.  Thus, I am posting this to keep myself from become confused!