$\int \tan x dx$

By Liwei Deng (P2942) on Monday, February 5, 2001 - 10:54 pm :

I tried to integrate tan x using integration by parts, but got some strange results.

$\int \tan x dx = \int \sin x \sec x dx$

Let $\sin x$ be $du / dx$ , and $\sec x$ be $v$

so $u = - \cos x$ , and $dv / dx = \tan x \times \sec x$
using integration by parts,

$\int \sin x \sec x = - \cos x \times \sec x - \int - \cos x \times \tan x \times \sec x dx$
it simplifies to

$\int \tan x = - 1 + \int \tan x$
so 0 = -1 !!!

What is going on here? Can anyone spot any mistakes in the calculations?

By James Lingard (Jchl2) on Monday, February 5, 2001 - 11:30 pm :

Hi,

When you integrate by parts, as with any other (indefinite, i.e. without limits specified in the integral) integration, you have to introduce an arbitrary constant, so all that will happen is that the constant will cancel with the -1 - you've managed to prove that $\int \tan x dx = \int \tan x dx$ !

If you want to integrate $\tan x$ , the best way is to notice that $\tan x = \sin x / \cos x$ , where $\sin x = - d / dx (\cos x)$ , so you can substitute $t = \cos x$ , and you get

$\int \tan x dx = - \ln (\cos x) + c$

James.

By Liwei Deng (P2942) on Wednesday, February 7, 2001 - 08:11 am :

thanks, James,
but surely there should be a constant of integration on each side, and they cancel out, so the result is still -1=0?

By James Lingard (Jchl2) on Wednesday, February 7, 2001 - 09:33 am :

No. There should be a constant of integration on each side, but they don't have to be the same, so they don't cancel out. So it is conventional to amalgamate them and just put in one constant.

For example, when solving the differential equation $dy / dx = x$ , you get

$\int dy / dx dx = \int x dx$

$y + c_{1} = (1 / 2) x^{2} + c_{2}$

$y = (1 / 2) x^{2} + (c_{2} - c_{1})$

$y = (1 / 2) x^{2} + c$

where we write $c$ for the $c_{2} - c_{1}$ , the combined constant of integration.

Here's another simple ''example'', which is essentially the same as your first problem:

$x + 1 = \int d / dx (x + 1) dx$

$x + 1 = \int 1 dx$

$x + 1 = x$

$1 = 0$

This shows why you really do need the constant of integration. And here's another ''example'' where we integrate on both sides:

$\int d / dx (x) dx + 1 = \int d / dx (x + 1) dx$

$\int 1 dx + 1 = \int 1 dx$

$x + 1 = x$

$1 = 0$

So here we need a constant of integration, which is not the same on both sides.

James.

By Liwei Deng (P2942) on Wednesday, February 7, 2001 - 10:30 am :

Thanks, James.
I think I understand now.