# How to integrate this : $\int \cot(5x) \tan(2x) \mathrm{d}x$

Methods to integrate this integral:

$$\int \cot(5x) \tan(2x) \mathrm{d}x$$

I have tried several methods, step by step, and they have led to invalid results. Helpful hints or processes are welcome.

I did this further :

$$\cot(5x) = \frac{1}{\tan(2x+ 3x)}$$ $$= \frac{1}{\frac{\tan 2x +\tan 3x}{1-\tan 2x\tan 3x}}$$

Simplifying it further by breaking $\tan(3x) = \tan(2x +x)$ again we get $$= \frac{1-\tan^2 x -2\tan x \tan 2x}{2\tan 2x -\tan^2 2x . \tan x +\tan x}$$

now the integral is $$= \frac{1-\tan^2 x -2\tan x \tan 2x}{2\tan 2x -\tan^2 2x . \tan x +\tan x} \tan 2x$$

Can we do something with this further thanks..

• Not a beautiful answer. Jul 22, 2015 at 13:24
• You can do the change of variable $u=\tan(x)$, exploiting the fact that $\tan(a+b)=(\tan(a)+\tan(b))/(1-\tan(a)\tan(b))$. This should give you a (messy) algebraic fraction in $u$. Jul 22, 2015 at 13:33
• @TerranDrop What about $\cot(5x)$? Do you suggest writing it as $\displaystyle \frac{1}{\tan(x+x+x+x+x)}$? Jul 22, 2015 at 13:36
• If we write $\cot(5x)$ as $\displaystyle\frac{\cos 5x}{\sin 5x}$ and use the Chebychev polynomials it'll be faster. Jul 22, 2015 at 13:42
• the integral looks ugly Jul 22, 2015 at 14:57

Letting $\tan x=z$, the integral can be rewritten as $$I=\int \frac{2z(1-10z^2+5z^4)}{(1-z^4)(5z-10z^3+z^5)}dz\\=2\int\frac{1-10z^2+5z^4}{(1-z^4)(5-10z^2+z^4)}dz\\=2\int \frac{1}{1-z^4}dz-8\int \frac{1}{5-10z^2+z^4}dz\\=2J-8K$$ where $J$ can be easily evaluated. As for $K$, it can be written as $$\int \frac{1}{(z^2-5-2\sqrt{5})(z^2-5+2\sqrt{5})}dz$$ which can then be evaluated without much effort. I admit that the final expressions will be ugly, but the evaluation process is straightforward.
• Maybe you could add what identity of $\cot 5x$ did you use. Jul 22, 2015 at 16:46
• @metacompactness, I just used the usual formula for $\tan 5x$ namely $\tan 5x=\frac{5\tan x-10\tan^3x+\tan^5x}{1-10\tan^25x+5\tan^4x}$ Jul 22, 2015 at 19:42