Ask Me Help Desk - Integration of square root of tanx minus square root of cotx

Quote:

Originally Posted by manasamanju

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

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.