integration of square root of tan x minus square root of cot x

integration of square root of tan x minus square root of cot x

\int \sqrt{tan(x)}dx-\int\sqrt{cot(x)}dx

This one is a booger. To do this the easy way, see this site:

Integral - Wolfram|Alpha (http://www.wolframalpha.com/input/?i=Integral)

Type in sqrt[Tan[x]]-sqrt[Cot[x]], the click on the little equals sign. It will give you the indefinite integral. It is long. This site will also show the steps, but with this one it is rather arduous.

But, here's a start the top of my head:

For \int\sqrt{tan(x)}dx

Let u=tan(x), \;\ x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du

This leads to:

\int\frac{\sqrt{u}}{u^{2}+1}du

Now, let w=\sqrt{u}, \;\ w^{2}=u, \;\ du=2wdw

Then, we get:

2\int\frac{w^{2}}{w^{4}+1}dw

For \int\sqrt{cot(x)}dx=\int\frac{1}{\sqrt{tan(x)}}dx

Let u=tan(x), \;\ x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du

Then, we get:

\int\frac{1}{\sqrt{u}(u^{2}+1)}du

Another sub similar to the first one gives:

2\int\frac{1}{w^{4}+1}dw

Now, \int\frac{1}{w^{4}+1}dw can be tackled thusly:

\frac{1}{w^{4}+1}=\frac{1}{2}\left[\frac{w^{2}+1-w^{2}+1}{w^{4}+1}\right]=\frac{1}{2}\left[\frac{w^{2}+1}{w^{4}+1}-\frac{w^{2}-1}{w^{4}+1}\right]

Divide top and bottom by w^{2}:

\frac{1}{2}\left\int\frac{(1+\frac{1}{w^{2}})dw}{w ^{2}+\frac{1}{w^{2}}}-\int\frac{(1-\frac{1}{w^{2}})dw}{w^{2}+\frac{1}{w^{2}}}\right]

But, this looks odd. Let's make the sub:

u=w-\frac{1}{w}, \;\ du=\left(1+\frac{1}{w^{2}}\right)dw

Let v=w+\frac{1}{w}, \;\ dv=\left(1-\frac{1}{w^{2}}\right)dw

u^{2}=w^{2}-2+\frac{1}{w^{2}}, \;\ u^{2}+2=w^{2}+\frac{1}{w^{2}}

v^{2}=w^{2}+2+\frac{1}{w^{2}}, \;\ v^{2}-2=w^{2}+\frac{1}{w^{2}}

This gives:

\frac{1}{2}\left[\int\frac{1}{u^{2}+2}du-\int\frac{1}{v^{2}-1}dv\right]

With some ingenuity, the first integral, \int\frac{w^{2}}{w^{4}+1}dw, can be done this way.

That is all I have time for now. As I said, this is doable, but tedious. That is what technology is for.