How do I solve the trigonometric integral of f(x)=cotx+tanx?

[Mentallic]

Part 2, leading on from https://www.physicsforums.com/showthread.php?t=315803"

Homework Statement

Given $f(x)=cotx+tanx$ find
$$\int{[f(x)]^2}dx$$

The Attempt at a Solution

I've attempted many different varieties of approaches to the problem. Trying to use the substitution method for sinx, cosx, tanx... and a few others... re-arranging the function, trying to get it into a more convenient form... trying to use some of the ideas given in the first thread... No luck.

Basically, it has all been a bunch of guessing and hoping something useful will appear. Plus another bunch of frustration on my part, but I won't get into the details of that xD

 

[rock.freak667]

Expand out [f(x)]^2

use $cot^2x+1=cosec^2x$ and $tan^2x+1=sec^2x$

 

[Mentallic]

Thanks rockfreak

It turns out to be tanx-cotx. I also obtained $\int{sec^2x+cosec^2x}dx$ through another longer method but it strike me at that moment that I can take the integral of each of these. (I think I'll keep my standard integral formulas close-by next time).

Thanks again.