I've calculated the following integral by substitution $z=x^2$

$$\int x\cdot\cos(x^2)\,\mathrm d x=\frac12\int\cos(z)\,\mathrm dz=\frac12\sin(z)+c=\frac12\cos(x^2)+c$$

with $c\in\mathbb R$.

My first question is about the constant $c$. Is it the same constant when you resubstitute?

And second, I've used the trick $z=x^2\leadsto \mathrm dz=2x\,\mathrm dx\leadsto \mathrm dx=\frac{\mathrm dz}{2x}$. Can you place $\Leftrightarrow$ between these 'transformations'? Or just $\Rightarrow$ or none of them?

  • $\begingroup$ 1) $C$ is completely arbitrary. Exact value of which can be found if any additional information is available about the antiderivative. 2) You substitution works in one way, since when doing backwards operations there might be some additional constant, which is $0$ in forward transformations. $\endgroup$ – Kaster May 2 '13 at 14:07
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    $\begingroup$ Can't put double arrow. If $z=x^2+1$ then $dz=2x\,dx$. $\endgroup$ – André Nicolas May 2 '13 at 14:09

The value $c$ is an arbitrary constant of integration. If you had a definite integral, the precise constant value would be the same under your substitution. However, there are substitutions and techniques of integration that vary in the end evaluations, but vary only in terms of the constant of integration.

Logically speaking, if $z = x^2,\;$ then $\;dz = 2x\,dx$.

But it is not necessarily the case that if $\;dz = 2x\,dx,\;$ then $\;z = x^2.\;$ This direction of the implication would only be appropriate if we qualify it: $\;$if $\;dz = 2x\, dx,\;$ then $\;z = x^2 + C\;$ for some constant $C$.

So the implication in this step is one-directional: $$z= x^2 \implies dz = 2x \,dx.$$

Your second step, however, uses merely an algebraic equivalence: $ab = c \iff b = \dfrac{c}{a}, \;\;a \neq 0$.
So in this step it's appropriate to express this as a bi-directional implication: $$dz = 2x\,dx \iff dx = \frac {1}{2x}\,dz, \quad x \neq 0$$

  • $\begingroup$ Deserves a thumbs - up! +1 $\endgroup$ – Amzoti May 3 '13 at 0:22

Yes, it is the same constant once you substitute $x^{2}$ back into the equation, but without initial values (initial conditions) it doesn't really matter because it is still an unknown constant value.

As pointed out in comments above, the derivative operation is not true both ways. The algebraic steps can be $\iff$ however. So you could write: $z=x^2\implies \mathrm dz=2x\,\mathrm dx\iff \mathrm dx=\frac{\mathrm dz}{2x}$


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