3
$\begingroup$

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..

$\endgroup$
8
  • 2
    $\begingroup$ Not a beautiful answer. $\endgroup$ – user5402 Jul 22 '15 at 13:24
  • $\begingroup$ 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 $u$. $\endgroup$ – TerranDrop Jul 22 '15 at 13:33
  • $\begingroup$ @TerranDrop What about $\cot(5x)$? Do you suggest writing it as $\displaystyle \frac{1}{\tan(x+x+x+x+x)}$? $\endgroup$ – user5402 Jul 22 '15 at 13:36
  • 1
    $\begingroup$ If we write $\cot(5x)$ as $\displaystyle\frac{\cos 5x}{\sin 5x}$ and use the Chebychev polynomials it'll be faster. $\endgroup$ – user5402 Jul 22 '15 at 13:42
  • 1
    $\begingroup$ the integral looks ugly $\endgroup$ – Dr. Sonnhard Graubner Jul 22 '15 at 14:57
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Maybe you could add what identity of $\cot 5x$ did you use. $\endgroup$ – user5402 Jul 22 '15 at 16:46
  • $\begingroup$ @metacompactness, I just used the usual formula for $\tan 5x$ namely $\tan 5x=\frac{5\tan x-10\tan^3x+\tan^5x}{1-10\tan^25x+5\tan^4x}$ $\endgroup$ – Samrat Mukhopadhyay Jul 22 '15 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.