# Talk:Integration by Substitution

Is it acceptable to leave the equality as an equality, ignoring that the change of variables might shift what the constant of integration is? Would it be better to put:

$\ds \int y\dfrac {\d y} {\d x} \rd x = \int y \rd y + C$

?--GFauxPas (talk) 18:20, 9 January 2013 (UTC)

Since the constant is arbitrary, it is meaningless to add it; after all, its value will depend on what you choose for both primitives. Any primitive for the left integrand is a primitive for the right integrand - this justifies equality IMO. A valid point to raise, though. --Lord_Farin (talk) 18:25, 9 January 2013 (UTC)
That's what I would have thougt, but I just encountered online a false proof that $1 = 0$, since clearly
$\ds \int \frac 1 x \rd x = 1 - \int \frac {-1} x \rd x$
using integration by parts, and that makes me a bit nervous. --GFauxPas (talk) 18:26, 9 January 2013 (UTC)
No problem. Recall that $\ds \int 0 \rd x = C$. --Lord_Farin (talk) 18:28, 9 January 2013 (UTC)
What are you trying to emphasize with that? --GFauxPas (talk) 18:39, 9 January 2013 (UTC)
Not sure, I'm afraid. --Lord_Farin (talk) 18:49, 9 January 2013 (UTC)

In answer to the original question, it's something that gets glossed over in the remedial classes, the teacher saying something like "It doesn't matter about the constant" but not explaining why. Recommend we add a page, appropriately titled and linked to this one. Same for integration by parts. Probably straightforward to do but I'm too lazy to think about it right now. --prime mover (talk) 20:04, 9 January 2013 (UTC)